1) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Ta có: \(\dfrac{1-6x}{x-2}+\dfrac{9x+4}{x+2}=\dfrac{x\left(3x-2\right)+1}{x^2-4}\)

\(\Leftrightarrow\dfrac{\left(1-6x\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(9x+4\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{3x^2-2x+1}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(\left(1-6x\right)\left(x+2\right)+\left(9x+4\right)\left(x-2\right)=3x^2-2x+1\)

\(\Leftrightarrow x+2-6x^2-12x+9x^2-18x+4x-8-3x^2+2x-1=0\)

\(\Leftrightarrow-23x-7=0\)

\(\Leftrightarrow-23x=7\)

\(\Leftrightarrow x=-\dfrac{7}{23}\)(nhận)

Vậy: \(S=\left\{-\dfrac{7}{23}\right\}\)

2) ĐKXĐ: \(x\notin\left\{\dfrac{2}{3};-\dfrac{2}{3}\right\}\)

Ta có: \(\dfrac{3x+2}{3x-2}-\dfrac{6}{2-3x}=\dfrac{9x^2}{9x^2-4}\)

\(\Leftrightarrow\dfrac{3x+2}{3x-2}+\dfrac{6}{3x-2}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

\(\Leftrightarrow\dfrac{3x+8}{3x-2}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

\(\Leftrightarrow\dfrac{\left(3x+8\right)\left(3x+2\right)}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

Suy ra: \(9x^2+6x+24x+16=9x^2\)

\(\Leftrightarrow30x+16=0\)

\(\Leftrightarrow30x=-16\)

hay \(x=-\dfrac{8}{15}\)(nhận)

Vậy: \(S=\left\{-\dfrac{8}{15}\right\}\)

giúp mình với ạ câu nào cũng được

a) ĐKXĐ: \(x\ne\pm4\)

\(5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}-\frac{3x-1}{4-x}\)

<=> \(5+\frac{96}{\left(x-4\right)\left(x+4\right)}=\frac{2x-1}{x+4}-\frac{3x-1}{4-x}\)

<=> 5(x - 4)(x + 4) + 96(x - 4) = (2x - 1)(x - 4)(4 - x) - (3x - 1)(x + 4)(4 - x)

<=> 20x² - 16x + 64 = 18x² + 8x

<=> 20x² - 16x + 64 - 18x² - 8x = 0

<=> 2x² - 24x + 64 = 0

<=> 2(x² - 12x + 32) = 0

<=> 2(x - 8)(x - 4) = 0

<=> (x - 8)(x - 4) = 0

<=> x - 8 = 0 hoặc x - 4 = 0

<=> x = 8 (tm) hoặc x - 4 = 0 (ktm)

=> x = 8

b) ĐKXĐ: \(x\ne\pm\frac{2}{3}\)

\(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4}\)

<=> \(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-2^2}\)

<=> \(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

<=> (2 + 3x)² - 6(3x - 2) = 9x²

<=> 16 - 6x + 9x² = 9x²

<=> 16 - 6x + 9x² - 9x² = 0

<=> 16 - 6x = 0

<=> -6x = 0 - 16

<=> -6x = -16

<=> x = -16/-6 = 8/3

=> x = 8/3

Hãy nha

a)    \(2\left(9x^2+6x+1\right)=\left(3x+1\right)\left(x-2\right)\)

\(\Leftrightarrow\)\(2\left(3x+1\right)^2-\left(3x+1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\)\(\left(3x+1\right)\left[2\left(3x+1\right)-\left(x-2\right)\right]=0\)

\(\Leftrightarrow\)\(\left(3x+1\right)\left(6x+2-x+2\right)=0\)

\(\Leftrightarrow\)\(\left(3x+1\right)\left(5x+4\right)=0\)

đến đây tự lm nha

b)   \(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)    (1)

ĐKXĐ:    \(x\ne\pm\frac{1}{3}\)

\(\left(1\right)\)\(\Leftrightarrow\)\(\frac{12}{\left(1-3x\right)\left(1+3x\right)}=\frac{\left(1-3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}-\frac{\left(1+3x\right)^2}{\left(1-3x\right)\left(1+3x\right)}\)

\(\Rightarrow\)\(\left(1-3x\right)^2-\left(1+3x\right)^2=12\)

\(\Leftrightarrow\)\(\left(1-3x-1-3x\right)\left(1-3x+1+3x\right)=12\)

\(\Leftrightarrow\)\(-12x=12\)

\(\Leftrightarrow\)\(x=-1\)   (t/m ĐKXĐ)

Vậy....

a) \(2\left(9x^2+6x+1\right)=\left(3x+1\right)\left(x-2\right)\)

\(\Leftrightarrow2\left(3x+1\right)^2-\left(3x+1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left[2\left(3x+1\right)-\left(x-2\right)\right]=0\)

\(\Leftrightarrow\left(3x+1\right)\left(6x+2-x+2\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(5x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x+1=0\\5x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x=-\frac{4}{5}\end{cases}}}\)

b) ĐKXĐ: \(x\ne\pm\frac{1}{3}\)

\(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)

\(\Leftrightarrow\frac{12}{\left(1-3x\right)\left(1+3x\right)}=\frac{\left(1-3x\right)^2}{\left(1-3x\right)\left(1+3x\right)}-\frac{\left(1+3x\right)^2}{\left(1-3x\right)\left(1+3x\right)}\)

\(\Leftrightarrow\left(1-3x\right)^2-\left(1+3x\right)^2=12\)

\(\Leftrightarrow\left(1-3x-1-3x\right)\left(1-3x+1+3x\right)=12\)

\(\Leftrightarrow-12x=12\)

\(\Leftrightarrow x=-1\) (thỏa mãn)

Vậy x = -1

a: \(27x^2\left(x+3\right)-12\left(x^2+3x\right)=0\)

=>\(27x^2\left(x+3\right)-12x\left(x+3\right)=0\)

=>\(\left(x+3\right)\cdot\left(27x^2-12x\right)=0\)

=>3x(x+3)(9x-4)=0

=>x(x+3)(9x-4)=0

=>\(\left[\begin{array}{l}x=0\\ x+3=0\\ 9x-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=-3\\ x=\frac49\end{array}\right.\)

b: \(\left(x-2\right)\left(3x+5\right)=\left(2x-4\right)\left(x+1\right)\)

=>\(\left(x-2\right)\left(3x+5\right)=2\left(x-2\right)\left(x+1\right)\)

=>(x-2)(3x+5)-(x-2)(2x+2)=0

=>(x-2)(3x+5-2x-2)=0

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

=>\(\left[\begin{array}{l}x-2=0\\ x+3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=-3\end{array}\right.\)

c: \(2\left(9x^2+6x+1\right)=\left(3x+1\right)\left(x-2\right)\)

=>\(2\left(3x+1\right)^2-\left(3x+1\right)\left(x-2\right)=0\)

=>(3x+1)(6x+2)-(3x+1)(x-2)=0

=>(3x+1)(6x+2-x+2)=0

=>(3x+1)(5x+4)=0

=>\(\left[\begin{array}{l}3x+1=0\\ 5x+4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac13\\ x=-\frac45\end{array}\right.\)

=33 nha mn

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