Quay lại đáp án

	5	bớt 36 ký tự trong đáp án		Sửa 20-03-16 09:39 PM Minh................ 1
Dk $\left\{ \begin{array}{l} 1-x\geq 0\\ 2x-1\geq 0\end{array} \right.$ $\Rightarrow$ X$\in$ $\left[ {} \right.$$\frac{1}{2}$,1]PT $\Leftrightarrow$ 2$\sqrt{1-x}$+3$\sqrt{2x-1}$=7x+6$\sqrt{(1-x)(2x-1}$ -4DẶt t=2$\sqrt{1-x}$+3$\sqrt{2x-1}$ $\Leftrightarrow$ $t^{2}$=4(1-x)+9(2x-1)+12$\sqrt{(1-x)(2x-1)}$$\Leftrightarrow$ $t^{2}$=14x-5+12$\sqrt{(1-x)(2x-1)}$ $\Leftrightarrow$ $t^{2}$+5=2(7x+6 $\sqrt{(1-x)(2x-1)}$)$\Rightarrow$ $\frac{t^{2+}+5}{2}$=7x+6$\sqrt{(1-x)(2x-1)}$PT ban đầu trở thành t=$\frac{t^{2}+5}{2}$ - 4 $\Leftrightarrow$ $t^{2}$ -2t -3 =0 $\Rightarrow$ t=3 và t=-1với t=3 $\Rightarrow$ 2$\sqrt{1-x}$ +3$\sqrt{2x-1}$=3 $\Rightarrow$ x=...tương tự vơi t=-1đúng thì tích v vote up nka mn.....!!!!!!!!!!!!!!!! Dk $\left\{ \begin{array}{l} 1-x\geq 0\\ 2x-1\geq 0\end{array} \right.$ $\Rightarrow$ X$\in$ $\left[ {} \right.$$\frac{1}{2}$,1]PT $\Leftrightarrow$ 2$\sqrt{1-x}$+3$\sqrt{2x-1}$=7x+6$\sqrt{(1-x)(2x-1}$ -4DẶt t=2$\sqrt{1-x}$+3$\sqrt{2x-1}$ $\Leftrightarrow$ $t^{2}$=4(1-x)+9(2x-1)+12$\sqrt{(1-x)(2x-1)}$$\Leftrightarrow$ $t^{2}$=14x-5+12$\sqrt{(1-x)(2x-1)}$ $\Leftrightarrow$ $t^{2}$+5=2(7x+6 $\sqrt{(1-x)(2x-1)}$)$\Rightarrow$ $\frac{t^{2+}+5}{2}$=7x+6$\sqrt{(1-x)(2x-1)}$PT ban đầu trở thành t=$\frac{t^{2}+5}{2}$ - 4 $\Leftrightarrow$ $t^{2}$ -2t -3 =0 $\Rightarrow$ t=3 và t=-1với t=3 $\Rightarrow$ 2$\sqrt{1-x}$ +3$\sqrt{2x-1}$=3 $\Rightarrow$ x= 1(tm)tương tự vơi t=-1 $\Rightarrow$ x=$\frac{3}{7}$(l)đúng thì tích v vote up nka mn.....!!!!!!!!!!!!!!!! Dk $\left\{ \begin{array}{l} 1-x\geq 0\\ 2x-1\geq 0\end{array} \right.$ $\Rightarrow$ X$\in$ $\left[ {} \right.$$\frac{1}{2}$,1]PT $\Leftrightarrow$ 2$\sqrt{1-x}$+3$\sqrt{2x-1}$=7x+6$\sqrt{(1-x)(2x-1}$ -4DẶt t=2$\sqrt{1-x}$+3$\sqrt{2x-1}$ $\Leftrightarrow$ $t^{2}$=4(1-x)+9(2x-1)+12$\sqrt{(1-x)(2x-1)}$$\Leftrightarrow$ $t^{2}$=14x-5+12$\sqrt{(1-x)(2x-1)}$ $\Leftrightarrow$ $t^{2}$+5=2(7x+6 $\sqrt{(1-x)(2x-1)}$)$\Rightarrow$ $\frac{t^{2+}+5}{2}$=7x+6$\sqrt{(1-x)(2x-1)}$PT ban đầu trở thành t=$\frac{t^{2}+5}{2}$ - 4 $\Leftrightarrow$ $t^{2}$ -2t -3 =0 $\Rightarrow$ t=3 và t=-1với t=3 $\Rightarrow$ 2$\sqrt{1-x}$ +3$\sqrt{2x-1}$=3 $\Rightarrow$ x=...tương tự vơi t=-1đúng thì tích v vote up nka mn.....!!!!!!!!!!!!!!!!

	4	bớt 1 ký tự trong đáp án		Sửa 20-03-16 09:32 PM Minh................ 1
Dk $\left\{ \begin{array}{l} 1-x\geq 0\\ 2x-1\geq 0\end{array} \right.$ $\Rightarrow$ X$\in$ $\left[ {} \right.$$\frac{1}{2}$,1]PT $\Leftrightarrow$ 2$\sqrt{1-x}$+3$\sqrt{2x-1}$=7x+6$\sqrt{(1-x)(2x-1}$ -4DẶt t=2$\sqrt{1-x}$+3$\sqrt{2x-1}$ $\Leftrightarrow$ $t^{2}$=4(1-x)+9(2x-1)+12$\sqrt{(1-x)(2x-1)}$$\Leftrightarrow$ $t^{2}$=14x-5+12$\sqrt{(1-x)(2x-1)}$ $\Leftrightarrow$ $t^{2}$+5=2(7x+6 $\sqrt{(1-x)(2x-1)}$)$\Rightarrow$ $\frac{t^{2+}+5}{2}$=7x+6$\sqrt{(1-x)(2x-1)}$PT ban đầu trở thành t=$\frac{t^{2}+5}{2}$ - 4 $\Leftrightarrow$ $t^{2}$ -2t -3 =0 $\Rightarrow$ t=3 và t=-1với t=3 $\Rightarrow$ 2$\sqrt{1-x}$ +3$\sqrt{2x-1}$=3 $\Rightarrow$ x= 1(tm)tương tự vơi t=-1 $\Rightarrow$ x=$\frac{3}{7}$(l)đúng thì tích v vote up nka mn.....!!!!!!!!!!!!!!!! Dk $\left\{ \begin{array}{l} 1-x\geq 0\\ 2x-1\geq 0\end{array} \right.$ $\Rightarrow$ X$\in$ $\left[ {} \right.$$\frac{1}{2}$,1]PT $\Leftrightarrow$ 2$\sqrt{1-x}$+3$\sqrt{2x-1}$=7x+6$\sqrt{(1-x)(2x-1}$ -4DẶt t=2$\sqrt{1-x}$+3$\sqrt{2x-1}$ $\Leftrightarrow$ $t^{2}$=4(1-x)+9(2x-1)+12$\sqrt{(1-x)(2x-1)}$$\Leftrightarrow$ $t^{2}$=14x-5+12$\sqrt{(1-x)(2x-1)}$ $\Leftrightarrow$ $t^{2}$+5=2(7x+6 $\sqrt{(1-x)(2x-1)}$)$\Rightarrow$ $\frac{t^{2+}+5}{2}$=7x+6$\sqrt{(1-x)(2x-1)}$PT ban đầu trở thành t=$\frac{t^{2}+5}{2}$ - 4 $\Leftrightarrow$ $t^{2}$ -2t -13 =0 $\Rightarrow$ t=3 và t=-1với t=3 $\Rightarrow$ 2$\sqrt{1-x}$ +3$\sqrt{2x-1}$=3 $\Rightarrow$ x= 1(tm)tương tự vơi t=-1 $\Rightarrow$ x=$\frac{3}{7}$(l)đúng thì tích v vote up nka mn.....!!!!!!!!!!!!!!!! Dk $\left\{ \begin{array}{l} 1-x\geq 0\\ 2x-1\geq 0\end{array} \right.$ $\Rightarrow$ X$\in$ $\left[ {} \right.$$\frac{1}{2}$,1]PT $\Leftrightarrow$ 2$\sqrt{1-x}$+3$\sqrt{2x-1}$=7x+6$\sqrt{(1-x)(2x-1}$ -4DẶt t=2$\sqrt{1-x}$+3$\sqrt{2x-1}$ $\Leftrightarrow$ $t^{2}$=4(1-x)+9(2x-1)+12$\sqrt{(1-x)(2x-1)}$$\Leftrightarrow$ $t^{2}$=14x-5+12$\sqrt{(1-x)(2x-1)}$ $\Leftrightarrow$ $t^{2}$+5=2(7x+6 $\sqrt{(1-x)(2x-1)}$)$\Rightarrow$ $\frac{t^{2+}+5}{2}$=7x+6$\sqrt{(1-x)(2x-1)}$PT ban đầu trở thành t=$\frac{t^{2}+5}{2}$ - 4 $\Leftrightarrow$ $t^{2}$ -2t -3 =0 $\Rightarrow$ t=3 và t=-1với t=3 $\Rightarrow$ 2$\sqrt{1-x}$ +3$\sqrt{2x-1}$=3 $\Rightarrow$ x= 1(tm)tương tự vơi t=-1 $\Rightarrow$ x=$\frac{3}{7}$(l)đúng thì tích v vote up nka mn.....!!!!!!!!!!!!!!!!

	3	bớt 98 ký tự trong đáp án		Sửa 20-03-16 09:32 PM Minh................ 1
Dk $\left\{ \begin{array}{l} 1-x\geq 0\\ 2x-1\geq 0\end{array} \right.$ $\Rightarrow$ X$\in$ $\left[ {} \right.$$\frac{1}{2}$,1]PT $\Leftrightarrow$ 2$\sqrt{1-x}$+3$\sqrt{2x-1}$=7x+6$\sqrt{(1-x)(2x-1}$ -4DẶt t=2$\sqrt{1-x}$+3$\sqrt{2x-1}$ $\Leftrightarrow$ $t^{2}$=4(1-x)+9(2x-1)+12$\sqrt{(1-x)(2x-1)}$$\Leftrightarrow$ $t^{2}$=14x-5+12$\sqrt{(1-x)(2x-1)}$ $\Leftrightarrow$ $t^{2}$+5=2(7x+6 $\sqrt{(1-x)(2x-1)}$)$\Rightarrow$ $\frac{t^{2+}+5}{2}$=7x+6$\sqrt{(1-x)(2x-1)}$PT ban đầu trở thành t=$\frac{t^{2}+5}{2}$ - 4 $\Leftrightarrow$ $t^{2}$ -2t -13 =0 $\Rightarrow$ t=3 và t=-1với t=3 $\Rightarrow$ 2$\sqrt{1-x}$ +3$\sqrt{2x-1}$=3 $\Rightarrow$ x= 1(tm)tương tự vơi t=-1 $\Rightarrow$ x=$\frac{3}{7}$(l)đúng thì tích v vote up nka mn.....!!!!!!!!!!!!!!!! Dk $\left\{ \begin{array}{l} 1-x\geq 0\\ 2x-1\geq 0\end{array} \right.$ $\Rightarrow$ X$\in$ $\left[ {} \right.$$\frac{1}{2}$,1]PT $\Leftrightarrow$ 2$\sqrt{1-x}$+3$\sqrt{2x-1}$=7x+6$\sqrt{(1-x)(2x-1}$ -4DẶt t=2$\sqrt{1-x}$+3$\sqrt{2x-1}$ $\Leftrightarrow$ $t^{2}$=4(1-x)+9(2x-1)+12$\sqrt{(1-x)(2x-1)}$$\Leftrightarrow$ $t^{2}$=14x-5+12$\sqrt{(1-x)(2x-1)}$ $\Leftrightarrow$ $t^{2}$+5=2(7x+6 $\sqrt{(1-x)(2x-1)}$)$\Rightarrow$ $\frac{t^{2+}+5}{2}$=7x+6$\sqrt{(1-x)(2x-1)}$PT ban đầu trở thành t=$\frac{t^{2}+5}{2}$ - 4 $\Leftrightarrow$ $t^{2}$ -2t -13 =0 $\Rightarrow$ t=1+$\sqrt{14}$ và t=1-$\sqrt{14}$với t=1+$\sqrt{14}$ $\Rightarrow$ 2$\sqrt{1-x}$ +3$\sqrt{2x-1}$=1+$\sqrt{14}$ $\Rightarrow$ x= $\frac{10+\sqrt{14}}{7}$ (ko tm)tương tự vơi t=1-$\sqrt{14}$ $\Rightarrow$ x=$\frac{10-\sqrt{14}}{7}$ (tm)đúng thì tích v vote up nka mn.....!!!!!!!!!!!!!!!! Dk $\left\{ \begin{array}{l} 1-x\geq 0\\ 2x-1\geq 0\end{array} \right.$ $\Rightarrow$ X$\in$ $\left[ {} \right.$$\frac{1}{2}$,1]PT $\Leftrightarrow$ 2$\sqrt{1-x}$+3$\sqrt{2x-1}$=7x+6$\sqrt{(1-x)(2x-1}$ -4DẶt t=2$\sqrt{1-x}$+3$\sqrt{2x-1}$ $\Leftrightarrow$ $t^{2}$=4(1-x)+9(2x-1)+12$\sqrt{(1-x)(2x-1)}$$\Leftrightarrow$ $t^{2}$=14x-5+12$\sqrt{(1-x)(2x-1)}$ $\Leftrightarrow$ $t^{2}$+5=2(7x+6 $\sqrt{(1-x)(2x-1)}$)$\Rightarrow$ $\frac{t^{2+}+5}{2}$=7x+6$\sqrt{(1-x)(2x-1)}$PT ban đầu trở thành t=$\frac{t^{2}+5}{2}$ - 4 $\Leftrightarrow$ $t^{2}$ -2t -13 =0 $\Rightarrow$ t=3 và t=-1với t=3 $\Rightarrow$ 2$\sqrt{1-x}$ +3$\sqrt{2x-1}$=3 $\Rightarrow$ x= 1(tm)tương tự vơi t=-1 $\Rightarrow$ x=$\frac{3}{7}$(l)đúng thì tích v vote up nka mn.....!!!!!!!!!!!!!!!!

	2	thêm 141 ký tự trong đáp án		Sửa 20-03-16 09:13 PM Minh................ 1
Dk $\left\{ \begin{array}{l} 1-x\geq 0\\ 2x-1\geq 0\end{array} \right.$ $\Rightarrow$ X$\in$ $\left[ {} \right.$$\frac{1}{2}$,1]PT $\Leftrightarrow$ 2$\sqrt{1-x}$+3$\sqrt{2x-1}$=7x+6$\sqrt{(1-x)(2x-1}$ -4DẶt t=2$\sqrt{1-x}$+3$\sqrt{2x-1}$ $\Leftrightarrow$ $t^{2}$=4(1-x)+9(2x-1)+12$\sqrt{(1-x)(2x-1)}$$\Leftrightarrow$ $t^{2}$=14x-5+12$\sqrt{(1-x)(2x-1)}$ $\Leftrightarrow$ $t^{2}$+5=2(7x+6 $\sqrt{(1-x)(2x-1)}$)$\Rightarrow$ $\frac{t^{2+}+5}{2}$=7x+6$\sqrt{(1-x)(2x-1)}$PT ban đầu trở thành t=$\frac{t^{2}+5}{2}$ - 4 $\Leftrightarrow$ $t^{2}$ -2t -13 =0 $\Rightarrow$ t=1+$\sqrt{14}$ và t=1-$\sqrt{14}$với t=1+$\sqrt{14}$ $\Rightarrow$ 2$\sqrt{1-x}$ +3$\sqrt{2x-1}$=1+$\sqrt{14}$ $\Rightarrow$ x= $\frac{10+\sqrt{14}}{7}$ (ko tm)tương tự vơi t=1-$\sqrt{14}$ $\Rightarrow$ x=$\frac{10-\sqrt{14}}{7}$ (tm)đúng thì tích v vote up nka mn.....!!!!!!!!!!!!!!!! PT $\Leftrightarrow$ 2$\sqrt{1-x}$+3$\sqrt{2x-1}$=7x+6$\sqrt{(1-x)(2x-1}$ -4DẶt t=2$\sqrt{1-x}$+3$\sqrt{2x-1}$ $\Leftrightarrow$ $t^{2}$=4(1-x)+9(2x-1)+12$\sqrt{(1-x)(2x-1)}$$\Leftrightarrow$ $t^{2}$=14x-5+12$\sqrt{(1-x)(2x-1)}$ $\Leftrightarrow$ $t^{2}$+5=2(7x+6 $\sqrt{(1-x)(2x-1)}$)$\Rightarrow$ $\frac{t^{2+}+5}{2}$=7x+6$\sqrt{(1-x)(2x-1)}$PT ban đầu trở thành t=$\frac{t^{2}+5}{2}$ - 4 $\Leftrightarrow$ $t^{2}$ -2t -13 =0 $\Rightarrow$ t=1+$\sqrt{14}$ và t=1-$\sqrt{14}$với t=1+$\sqrt{14}$ $\Rightarrow$ 2$\sqrt{1-x}$ +3$\sqrt{2x-1}$=1+$\sqrt{14}$ $\Rightarrow$ x= $\frac{10+\sqrt{14}}{7}$tương tự vơi t=1-$\sqrt{14}$ $\Rightarrow$ x=$\frac{10-\sqrt{14}}{7}$đúng thì tích v vote up nka mn.....!!!!!!!!!!!!!!!! Dk $\left\{ \begin{array}{l} 1-x\geq 0\\ 2x-1\geq 0\end{array} \right.$ $\Rightarrow$ X$\in$ $\left[ {} \right.$$\frac{1}{2}$,1]PT $\Leftrightarrow$ 2$\sqrt{1-x}$+3$\sqrt{2x-1}$=7x+6$\sqrt{(1-x)(2x-1}$ -4DẶt t=2$\sqrt{1-x}$+3$\sqrt{2x-1}$ $\Leftrightarrow$ $t^{2}$=4(1-x)+9(2x-1)+12$\sqrt{(1-x)(2x-1)}$$\Leftrightarrow$ $t^{2}$=14x-5+12$\sqrt{(1-x)(2x-1)}$ $\Leftrightarrow$ $t^{2}$+5=2(7x+6 $\sqrt{(1-x)(2x-1)}$)$\Rightarrow$ $\frac{t^{2+}+5}{2}$=7x+6$\sqrt{(1-x)(2x-1)}$PT ban đầu trở thành t=$\frac{t^{2}+5}{2}$ - 4 $\Leftrightarrow$ $t^{2}$ -2t -13 =0 $\Rightarrow$ t=1+$\sqrt{14}$ và t=1-$\sqrt{14}$với t=1+$\sqrt{14}$ $\Rightarrow$ 2$\sqrt{1-x}$ +3$\sqrt{2x-1}$=1+$\sqrt{14}$ $\Rightarrow$ x= $\frac{10+\sqrt{14}}{7}$ (ko tm)tương tự vơi t=1-$\sqrt{14}$ $\Rightarrow$ x=$\frac{10-\sqrt{14}}{7}$ (tm)đúng thì tích v vote up nka mn.....!!!!!!!!!!!!!!!!

	1	tạo mới		Trả lời 20-03-16 09:07 PM Minh................ 1
PT $\Leftrightarrow$ 2$\sqrt{1-x}$+3$\sqrt{2x-1}$=7x+6$\sqrt{(1-x)(2x-1}$ -4DẶt t=2$\sqrt{1-x}$+3$\sqrt{2x-1}$ $\Leftrightarrow$ $t^{2}$=4(1-x)+9(2x-1)+12$\sqrt{(1-x)(2x-1)}$$\Leftrightarrow$ $t^{2}$=14x-5+12$\sqrt{(1-x)(2x-1)}$ $\Leftrightarrow$ $t^{2}$+5=2(7x+6 $\sqrt{(1-x)(2x-1)}$)$\Rightarrow$ $\frac{t^{2+}+5}{2}$=7x+6$\sqrt{(1-x)(2x-1)}$PT ban đầu trở thành t=$\frac{t^{2}+5}{2}$ - 4 $\Leftrightarrow$ $t^{2}$ -2t -13 =0 $\Rightarrow$ t=1+$\sqrt{14}$ và t=1-$\sqrt{14}$với t=1+$\sqrt{14}$ $\Rightarrow$ 2$\sqrt{1-x}$ +3$\sqrt{2x-1}$=1+$\sqrt{14}$ $\Rightarrow$ x= $\frac{10+\sqrt{14}}{7}$tương tự vơi t=1-$\sqrt{14}$ $\Rightarrow$ x=$\frac{10-\sqrt{14}}{7}$đúng thì tích v vote up nka mn.....!!!!!!!!!!!!!!!!

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hoahoa.nhynhay: . 11/5/2018 1:39:46 PM
hoahoa.nhynhay: . 11/5/2018 1:39:46 PM
hoahoa.nhynhay: . 11/5/2018 1:39:46 PM
hoahoa.nhynhay: . 11/5/2018 1:39:46 PM
hoahoa.nhynhay: . 11/5/2018 1:39:47 PM
hoahoa.nhynhay: . 11/5/2018 1:39:47 PM
hoahoa.nhynhay: . 11/5/2018 1:39:47 PM
hoahoa.nhynhay: . 11/5/2018 1:39:47 PM
hoahoa.nhynhay: . 11/5/2018 1:39:48 PM
hoahoa.nhynhay: . 11/5/2018 1:39:48 PM
hoahoa.nhynhay: . 11/5/2018 1:39:48 PM
hoahoa.nhynhay: . 11/5/2018 1:39:48 PM
hoahoa.nhynhay: . 11/5/2018 1:39:48 PM
hoahoa.nhynhay: . 11/5/2018 1:39:49 PM
hoahoa.nhynhay: . 11/5/2018 1:39:49 PM
hoahoa.nhynhay: . 11/5/2018 1:39:49 PM
hoahoa.nhynhay: . 11/5/2018 1:39:49 PM
hoahoa.nhynhay: . 11/5/2018 1:39:50 PM
hoahoa.nhynhay: . 11/5/2018 1:39:50 PM
hoahoa.nhynhay: ..................... 11/5/2018 1:39:52 PM
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