Bài 102406

Question

Giải các phương trình:
a/\(
1+\frac{2a}{x-a}=\frac{b^{2}-a^{2}}{a^{2}+x^{2}-2ax}
\)
b/\(
\frac{x^{2}}{ab-2b^{2}}=\frac{a-b}{ac^{2}-2bc^{2}}+\frac{x}{bc}
\)
c/ \(
\frac{x^{2}+1}{n^{2}-2n}-\frac{1}{2-nx}=\frac{x}{n}
\)
d/\(
\frac{x+x^{2}}{1-x^{2}}:\frac{1-a^{2}}{\left ( 1+ax \right )^{2}\left ( a+x \right )^{2}}=\frac{ab}{\left ( b-a \right )^{2}}
\)

score 0 · Answer 1

a/\(
1+\frac{2a}{x-a}=\frac{b^{2}-a^{2}}{a^{2}+x^{2}-2ax}
\)
Điều kiện \( x\neq a\)
\(
1+\frac{2a}{x-a}=\frac{b^{2}-a^{2}}{a^{2}+x^{2}-2ax}
\)
\(
\Leftrightarrow 1+\frac{2a}{x-a}=\frac{b^{2}-a^{2}}{\left ( x -a\right )^{2}}
\)
\(
\Leftrightarrow \left ( x -a\right )^{2}-2a\left ( x -a\right )=b^{2}-a^{2}
\)
\(
\Leftrightarrow a^{2}+x^{2}-2ax-2ax+2a^{2}=b^{2}-a^{2}
\)
\(
\Leftrightarrow x^{2}-4ax+4a^{2}-b^{2}=0
\)
\(
\Delta'= 4a^{2} -4a^{2}+b^{2}=b^{2}\Leftrightarrow \begin{cases}x_{1}=2a-b \\ x_{2}=2a+b \end{cases} \) nhận vì thỏa mãn điều kiện.

b/\(
\frac{x^{2}}{ab-2b^{2}}=\frac{a-b}{ac^{2}-2bc^{2}}+\frac{x}{bc} (*)
\)
\(
ab-2b^{2}= b\left ( a-2b \right )
\)
\(
ac^{2}-2bc^{2}=c^{2}\left ( a-2b \right )
\)
\(
\Rightarrow \) mẫu chung: \( bc^{2}\left ( a-2b \right ) \)
Điều kiện cho các chữ a,b, c: \( b \neq 0, c\neq 0, a\neq 2b\)
\(
(*)\Leftrightarrow c^{2}x^{2}-c\left ( a-2b \right )-b\left ( a-b \right )=0
\)
\(
\Delta=c^{2}\left ( a-2b \right )^{2} + 4bc^{2}\left ( a-b \right )= c^{2}[ a^{2}-4ab+4b^{2}+4ab-4b^{2}] =c^{2}.a^{2}
\)
\(
\left[ \begin{array}{l}x_{1} = \frac{c\left ( a-2b \right )-ac}{2c^{2}}=\frac{-b}{c}\\x_{2} =\frac{c\left ( a-2b \right )+ac}{2c^{2}}=\frac{a-b}{c} \end{array} \right.
\)

c/ \(
\frac{x^{2}+1}{n^{2}-2n}-\frac{1}{2-nx}=\frac{x}{n}
\)
\(
\Leftrightarrow \frac{x^{2}+1}{n\left ( nx-2 \right )}+\frac{1}{nx-2}=\frac{x}{n}
\)
Điều kiện: \( n\neq 0, x\neq \frac{2}{n} \)
Khử mẫu số: \(
x^{2}+1+n=x\left ( nx-2 \right )\Leftrightarrow \left ( n-1 \right )x^{2} -2x-\left ( n+1\right )=0
\)
\(
\Delta'= 1+\left (n^{2}-1 \right )=n^{2}
\) Nghiệm:
\(\left[ \begin{array}{l}x_{1}=-1\\x_{2} = \frac{n+1}{n-1}\end{array} \right.
\)

d/\(
\frac{x+x^{2}}{1-x^{2}}:\frac{1-a^{2}}{\left ( 1+ax \right )^{2}\left ( a+x \right )^{2}}=\frac{ab}{\left ( b-a \right )^{2}}
\)
\(
\Leftrightarrow \frac{x\left (1+ x \right )}{\left ( 1-x \right )\left ( 1+x \right )}.\frac{\left ( 1+ax+a+x \right )\left ( 1+ax-a-x \right )}{1-a^{2}}=\frac{ab}{\left ( b-a \right )^{2}}
\)
Điều kiện:
\(
\begin{cases}a \neq b, a\neq 1 \\ x\neq \pm 1 \end{cases}
\)
\(
\Leftrightarrow \frac{x}{\left ( 1-x \right )}.\frac{\left ( 1-a^{2} \right )\left ( 1-x^{2} \right )}{1-a^{2}}=\frac{ab}{\left ( b-a \right )^{2}}
\)
\(
\Leftrightarrow x\left ( 1-x \right )=\frac{ab}{\left ( b-a \right )^{2}}\Leftrightarrow x^{2}+x-\frac{ab}{\left ( b-a \right )^{2}}=0
\)
\(
\Delta=1+4\frac{ab}{\left ( b-a \right )^{2}}=\left ( \frac{a+b}{a-b} \right )^{2}
\)
\(
\Leftrightarrow \left[ \begin{array}{l}x_{1} = \frac{-1-\left ( \frac{a+b}{a-b} \right )}{2}=\left ( \frac{a}{b-a} \right )\\x_{2}= \frac{-1+\left ( \frac{a+b}{a-b} \right )}{2}=\left ( \frac{b}{a-b} \right )\end{array} \right.
\)

Bài 102406

1 Đáp án

Lý thuyết liên quan

Bài 102406

1 Đáp án

Liên quan

Lý thuyết liên quan