Sửa đề: 8x-1

=>2(8x^2-x)(8x^2-x+2)-126=0

=>2[(8x^2-x)^2+2(8x^2-x)]-126=0

=>(8x^2-x)^2+2(8x^2-x)-63=0

=>(8x^2-x+9)(8x^2-x-7)=0

=>8x^2-x-7=0

=>x=1 hoặc x=-7/8

Đặt \(\sqrt{x^2+9}=t>0\) ta được:

\(t^2+8x=\left(x+8\right)t\Leftrightarrow t^2-\left(x+8\right)t+8x=0\)

\(\Leftrightarrow t^2-tx-8t+8x=0\)

\(\Leftrightarrow t\left(t-x\right)-8\left(t-x\right)=0\)

\(\Leftrightarrow\left(t-x\right)\left(t-8\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2+9}=x\left(x\ge0\right)\\\sqrt{x^2+9}=8\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2+9=x^2\left(vn\right)\\x^2=55\end{matrix}\right.\)

\(\Rightarrow x=\pm\sqrt{55}\)

\(\left(x^2+8x+8\right)^2=\left(4x+6\right)\left(2x^2+12x+10\right)\)

\(\left(x^2+8x+8\right)^2-\left[\left(4x+6\right)\left(2x^2+12x+10\right)\right]=0\)

\(\left(x^2+4x+2\right)^2=0\)

\(x^2+4x=-2\)

\(x\left(x+4\right)=-2\)

\(x=\pm\sqrt{2}-2\)

a:Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)

=>3x-9-10x+2=-4

=>-7x-7=-4

=>-7x=3

=>x=-3/7

b: =>\(\dfrac{5-x}{4x\left(x-2\right)}+\dfrac{7}{8x}=\dfrac{x-1}{2x\left(x-2\right)}+\dfrac{1}{8\left(x-2\right)}\)

=>\(2\left(5-x\right)+7\left(x-2\right)=4\left(x-1\right)+x\)

=>10-2x+7x-14=4x-4+x

=>5x-4=5x-4

=>0x=0(luôn đúng)

Vậy: S=R\{0;2}

Bài 1: 

b: \(\Leftrightarrow x-2=0\)

hay x=2

\(\Rightarrow2x\left(x-4\right)-x\left(x-2\right)=8x+8\)

\(\Leftrightarrow2x^2-8x-x^2+2x=8x+8\)

\(\Leftrightarrow x^2-14x-8=0\)

\(\Delta'=\left(-7\right)^2-1.\left(-8\right)=57\)

\(\sqrt{\Delta}=\sqrt{57}\)\(\Rightarrow\)Phương trình có 2 nghiệm phân biệt

\(x_1=\frac{7+\sqrt{57}}{1}=7+\sqrt{57}\)                                          \(x_2=\frac{7-\sqrt{57}}{1}=7-\sqrt{57}\)

\(4\left(\frac{x^2}{2}+5x+4\right)^2\)=\(4\left(2x+1\right)\left(x^2+8x+7\right)\)

\(\Leftrightarrow\left(x^2+10x+8\right)^2=4\left(2x+1\right)\left(x^2+8x+7\right)\)

dat \(2x+1=a,x^2+8x+7=b\) \(\Rightarrow a+b=x^2+10x+8\)

pt tro thanh

\(\left(a+b\right)^2=4ab\Rightarrow a^2+2ab+b^2-4ab=0\)

\(\Leftrightarrow\left(a-b\right)^2=0\Leftrightarrow a=b\Leftrightarrow2x+1=x^2+8x+1\)

                                                  \(\Leftrightarrow x^2+6x=0\Leftrightarrow x\left(x+6\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=-6\end{cases}}\)

cảm ơn bn nhìu