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

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

2x(8x-1)²(4x-1)= 9

<=> 2x(64x²-16x+1)(4x-1)=9

<=>(128x³  - 32x² + 2x)(4x-1)=9

<=>512x⁴ - 256x³ + 40x²  - 2x=9

<=>64x⁴ - 32x³ + 5x² - 0,25x - 1,125=0

<=>64x³(x-0,5) + 5x(x-0,5) + 2,5x  -0,25x - 1,125 = 0

<=> (x-0,5)(64x³ + 5x - 2,25) = 0

<=> (x-0,5)(64x³  + 16x² - 16x² - 4x + 9x - 2,25)=0

<=>(x-0,5)[64x² (x + 0,25 ) -16x(x + 0,25) + 9(x + 0,25) = 0

<=> (x-0,5)(x+0,25)(64x² -16x +9) = 0  (vì 64x² -16x +9 > 0)

<=>\(\orbr{\begin{cases}x-0,5=0\\x+0,25=0\end{cases}}\)

<=>\(\orbr{\begin{cases}x=0,5\\x=-0,25\end{cases}}\)

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

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

\(\Leftrightarrow8x\left(8x-1\right)^2\left(8x-2\right)=72\)(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={\(\frac{-1}{4};\frac{1}{2}\)}

X= 10000000

Ghi lời giải giùm mình được không?

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)