|
|
a/ \( \left(x+4\right)\left(x+5\right)\left(x+7\right)\left(x+8\right)=4\)
\( \Leftrightarrow \left(x+4\right)\left(x+8\right)\left(x+5\right)\left(x+7\right)=4\)
\( \Leftrightarrow \left(x^{2} +12x+32\right)\left(x^{2}+12x+35\right) =4\)
Đặt \( x^{2} +12x+32=t, t \geq -4\) được phương trình:
\(t\left(t+3\right) -4=0 \Leftrightarrow t^{2} +3t-4=0 \Leftrightarrow t=1, t=-4\)
Với \( t=1: x^{2}+12x+32=1 \Leftrightarrow x^{2}+12x+31=0 \Leftrightarrow x=-6-\sqrt{5}, x=-6+\sqrt{5} \)
Với \( t=-4: x^{2} +12x+32=-4 \Leftrightarrow x^{2} +12x+36=0 \Leftrightarrow \left( x+6\right)^{2} =0 \Leftrightarrow x=-6\)
b/ \( \left(x-2\right)^{2} +\left(x-3\right)^{4}=1 \)
Đặt \( t= x+ \frac{-3-2}{2} =x-\frac{5}{2} \Rightarrow x= t+\frac{5}{2}\)
Thay ẩn t:
\( \left(t+\frac{5}{2} -2\right)^{4} + \left(t+\frac{5}{2} -3\right)^{4}=1\)
\( \Leftrightarrow \left(t+\frac{1}{2}\right)^{4} +\left(t-\frac{1}{2}\right)^{4}=1\)
\( \Leftrightarrow [\left(t+\frac{1}{2}\right)^{2} -\left(t-\frac{1}{2}\right)^{2}]^{2}+ 2\left(t+\frac{1}{2}\right)^{2} .\left(t-\frac{1}{2}\right)^{2}=1\)
\( 4t^{2} +3t^{2}-\frac{7}{8}=0 \Leftrightarrow t^{2}=\frac{1}{4} \Leftrightarrow t=\neq \frac{1}{2}
t =\frac{1}{2} \Rightarrow x=3, t=-\frac{1}{2} \Rightarrow x=2\)
c/ \( \left(x^{2}-4x+5\right)^{2}-\left(x-1\right)\left(x-3\right)=4\)
\( \Leftrightarrow \left(x^{2}-4x+5\right)^{2}- \left(x^{2}-4x+5\right)-2=0\)
Đặt \( \left(x^{2}-4x+5\right)=t, t\geq 1\)
Có phương trình trung gian: \( t^{2}-t-2=0 \Leftrightarrow t=-1 (L), t=2\)
\( \left(x^{2}-4x+5\right)=2 \Leftrightarrow x^{2}-4x+3=0 \Leftrightarrow x=1, x=3\)
d/ \( \left(x^{2}+3x-4\right)^{2}+3\left(x^{2}+3x-4\right)=x+4\)
\( \Leftrightarrow \left(x^{2}+3x-4\right)^{2}+4\left(x^{2}+3x-4\right)=x^{2}+4x\)
\( \Leftrightarrow [\left(x^{2}+3x-4\right)^{2} –x^{2}] +4[ \left(x^{2}+3x-4\right) –x] =0\)
\( \Leftrightarrow [\left(x^{2}+3x-4\right) –x] [ x^{2}+3x-4+x+4] =0\)
\( \Leftrightarrow \left(x^{2}+3x-4\right)\left( x^{2}+4x\right) =0\)
\( \Leftrightarrow x_{1}=-1-\sqrt{5}, x_{2}=-1+\sqrt{5}, x_{3}=0, x_{4}=-4\)
e/ \( 6x^{4} -35x^{3} +62x^{2}-35x+6=0\)
\( \Leftrightarrow 6x^{2} -35x+62-35.\frac{1}{x}+\frac{6}{x^{2}}=0\)
\( \Leftrightarrow 6\left( x^{2}+\frac{1}{x^{2}}\right) -35\left( x+\frac{1}{x}\right)+62=0\)
đặt \( x+\frac{1}{x}=t\) thì \( |t|\geq 2 \Rightarrow x^{2}+\frac{1}{x^{2}}=t^{2}-2\)
được phương trình trung gian \(6\left( t^{2}-2\right) -35t+62=0 \Leftrightarrow 6t^{2} -35t+50=0\)
\( t_{1} = \frac{35-5}{12}=\frac{5}{2}, t_{2}=\frac{35+5}{12}=\frac{10}{3}\)
\( x+\frac{1}{x}=\frac{5}{2} \Leftrightarrow x^{2}-\frac{5}{2}x+1=0 \Leftrightarrow x=2, x=\frac{1}{2}\)
\(x+\frac{1}{x}=\frac{10}{3} \Leftrightarrow x^{2}-\frac{10}{3}x+1=0 \Leftrightarrow x=3, x=\frac{1}{3}\)
|