⇔ 

Với t = 3 ⇒ x = - 1/2

Với t = - 3 ⇒ x = - 5/4

Vậy tập nghiệm của phương trình là S = { - 1/2; - 5/4 }

\(2x\left(8x-1\right)^2\left(4x-1\right)=9\)

\(\Leftrightarrow8x\left(8x-1\right)^2\left(8x-2\right)=72\)(nhân hai vế với 8)

Đặt \(8x-1=y\). Khi đó, pt được viết lại:

\(\left(y+1\right)y^2\left(y-1\right)=72\)

\(\Leftrightarrow y^2\left(y^2-1\right)=72\)

\(\Leftrightarrow y^4-y^2-72=0\)

\(\Leftrightarrow y^4+3y^3-3y^3-9y^2+8y^2+24y-24y-72=0\)

\(\Leftrightarrow y^3\left(y+3\right)-3y^2\left(y+3\right)+8y\left(y+3\right)-24\left(y+3\right)=0\)

\(\Leftrightarrow\left(y+3\right)\left(y^3-3y^2+8y-24\right)=0\)

\(\Leftrightarrow\left(y+3\right)\left(y^2\left(y-3\right)+8\left(y-3\right)\right)=0\)

\(\Leftrightarrow\left(y+3\right)\left(y-3\right)\left(y^2+8\right)=0\)

Mà \(y^2+8\ge8>0\)

\(\Rightarrow\orbr{\begin{cases}y+3=0\\y-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}y=-3\\y=3\end{cases}}}\)

TH1: \(y=-3\)

\(\Rightarrow8x-1=-3\)

\(\Leftrightarrow8x=-2\)

\(\Leftrightarrow x=\frac{-1}{4}\)

TH2: \(y=3\)

\(\Rightarrow8x-1=3\)

\(\Leftrightarrow8x=4\)

\(\Leftrightarrow x=\frac{1}{2}\)

Vậy tập nghiệm của pt là S={\(\frac{-1}{4};\frac{1}{2}\)}

Ta có: \(2x\left(8x-1\right)^2\left(4x-1\right)=9\)

=>\(\left(8x^2-2x\right)\left(64x^2-16x+1\right)=9\)

=>\(\left(8x^2-2x\right)\cdot\left\lbrack8\left(8x^2-2x\right)+1\right\rbrack-9=0\)

=>\(8\cdot\left(8x^2-2x\right)^2+\left(8x^2-2x\right)-9=0\)

=>\(8\cdot\left(8x^2-2x\right)^2+9\left(8x^2-2x\right)-8\left(8x^2-2x\right)-9=0\)

=>\(\left(8x^2-2x\right)\left\lbrack8\left(8x^2-2x\right)+9\right\rbrack-\left\lbrack8\left(8x^2-2x\right)+9\right\rbrack=0\)

=>\(\left(8x^2-2x-1\right)\left(64x^2-16x+9\right)=0\)

mà \(64x^2-16x+9=64x^2-2\cdot8x\cdot1+1+8=\left(8x-1\right)^2+8>0\forall x\)

nên \(8x^2-2x-1=0\)

=>\(8x^2-4x+2x-1=0\)

=>(2x-1)(4x+1)=0

=>\(\left[\begin{array}{l}2x-1=0\\ 4x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac12\\ x=-\frac14\end{array}\right.\)

a)2x(8x-1)² (4x-1)=9

\(\Leftrightarrow\text{ (64x}^2\text{-16x+1)(8x}^2\text{-2x)=9}\)

\(\Leftrightarrow\text{ 512x}^4\text{-256x}^3\text{+40x}^2\text{-2x=9}\)

\(\Leftrightarrow\text{ 512x}^4\text{-256x}^3\text{+40x}^2\text{-2x-9=0}\)

\(\Leftrightarrow\text{ 512x}^4\text{-128x}^3\text{-64x}^2\text{-128x}^3\text{+32x}^2\text{+16x+72x}^2\text{-18x-9=0}\)

\(\Leftrightarrow\text{ (512x}^4\text{-128x}^3\text{-64x}^2\text{)-(128x}^3\text{-32x}^2\text{-16x)+(72x}^2\text{-18x-9)=0}\)

\(\Leftrightarrow\text{ 64x}^2\text{(8x}^2\text{-2x-1)-16x(8x}^2\text{-2x-1)+9(8x}^2\text{-2x-1)=0}\)

\(\Leftrightarrow\text{ (64x}^2\text{-16x+9)(8x}^2\text{-2x-1)=0}\)

\(\Leftrightarrow\text{ (64x}^2\text{-16x+9)(8x}^2\text{-4x+2x-1)=0}\)

\(\Leftrightarrow\text{ (64x}^2\text{-16x+9)(2x-1)(4x+1)=0}\)

\(\Rightarrow\left\{{}\begin{matrix}2x-1=0\\4x+1=0\end{matrix}\right.\) (Vì 64x² -16x+9 =0 )

\(\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{1}{4}\end{matrix}\right.\)

Ta có : \(2x\left(8x-1\right)^2\left(4x-1\right)=9\)

=> \(\left(8x^2-2x\right)\left(64x^2-16x+1\right)=9\)

=> \(512x^4-128x^3+8x^2-128x^3+32x^2-2x-9=0\)

=> \(512x^4-256x^3+40x^2-2x-9=0\)

=> \(512x^4+128x^3-384x^3-96x^2+136x^2+34x-36x-9=0\)

=> \(128x^3\left(4x+1\right)-96x^2\left(4x+1\right)+34x\left(4x+1\right)-9\left(4x+1\right)=0\)

=> \(\left(4x+1\right)\left(128x^3-96x^2+34x-9\right)=0\)

=> \(\left(4x+1\right)\left(128x^3-64x^2-32x^2+16x+18x-9\right)=0\)

=> \(\left(4x+1\right)\left(64x^2\left(2x-1\right)-16x\left(2x-1\right)+9\left(2x-1\right)\right)=0\)

=> \(\left(4x+1\right)\left(2x-1\right)\left(64x^2-16x+9\right)=0\)

Ta thấy : \(64x^2-16x+9\)

\(=\left(64x^2-2.8.x+1\right)+8\)

\(=\left(8x-1\right)^2+8>0\)

=> \(\left(4x+1\right)\left(2x-1\right)=0\)

=> \(\left[{}\begin{matrix}4x+1=0\\2x-1=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=-\frac{1}{4}\\x=\frac{1}{2}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là \(S=\left\{-\frac{1}{4};\frac{1}{2}\right\}\)

a: ĐKXĐ: x<>4

Ta có: \(1+\frac{14}{\left(\left.x-4\right.\right)^2}=\frac{-9}{x-4}\)

=>\(\frac{\left(x-4\right)^2+14}{\left(x-4\right)^2}=\frac{-9\left(x-4\right)}{\left(x-4\right)^2}\)

=>\(\left(x-4\right)^2+14=-9\left(x-4\right)\)

=>\(\left(x-4\right)^2+9\left(x-4\right)+14=0\)

=>(x-4+2)(x-4+7)=0

=>(x-2)(x+3)=0

=>x=2(nhận) hoặc x=-3(nhận)

b: ĐKXĐ: x∉{1/2;-1/2}

Ta có: \(\frac{1+8x}{1+2x}-\frac{2x}{2x-1}+\frac{12x^2-9}{1-4x^2}=0\)

=>\(\frac{8x+1}{2x+1}-\frac{2x}{2x-1}-\frac{12x^2-9}{\left(2x-1\right)\left(2x+1\right)}=0\)

=>\(\frac{\left(8x+1\right)\left(2x-1\right)-2x\left(2x+1\right)-12x^2+9}{\left(2x-1\right)\left(2x+1\right)}=0\)

=>\(\left(8x+1\right)\left(2x-1\right)-2x\left(2x+1\right)-12x^2+9=0\)

=>\(16x^2-8x+2x-1-4x^2-2x-12x^2+9=0\)

=>-8x+8=0

=>-8x=-8

=>x=1(nhận)

c: ĐKXĐ: x∉{3;1}

\(\frac{1}{2x-6}-\frac{3x-5}{x^2-4x+3}=\frac12\)

=>\(\frac{1}{2\left(x-3\right)}-\frac{3x-5}{\left(x-1\right)\left(x-3\right)}=\frac12\)

=>\(\frac{x-1}{2\left(x-3\right)\left(x-1\right)}-\frac{2\left(3x-5\right)}{2\left(x-1\right)\left(x-3\right)}=\frac{x^2-4x+3}{2\left(x-1\right)\left(x-3\right)}\)

=>\(x^2-4x+3=x-1-2\left(3x-5\right)=x-1-6x+10=-5x+9\)

=>\(x^2+x-6=0\)

=>(x+3)(x-2)=0

=>x=-3(nhận) hoặc x=2(nhận)

x = -0.25 và hình như còn 1 giá trị nữa hay sao ý ạ