\(\left(3x-1\right)^2-\left(x+3\right)\left(2x-1\right)=7\left(x+1\right)\left(x-2\right)-3x\)

\(\Leftrightarrow9x^2-6x+1-\left(2x^2+6x-x-3\right)=7\left(x^2-x-2\right)-3x\)

\(\Leftrightarrow7x^2-11x+4=7x^2-10x-14\)

\(\Leftrightarrow-x=-28\Leftrightarrow x=28\)

\(5x-24+4x^2+3x=4-6x+4x^2+2\)

\(5x-24+4x^2+3x-4-6x+4x^2+2=0\)

\(2x-26+8x^2=0\)

\(2x+8x^2=26\)

\(5x-4\left(6-x\right)\left(x+3\right)=\left(4-2x\right)\left(3-2x\right)+2\)

\(\Leftrightarrow5x-4\left(6x-x^2+18-3x\right)=\left(12-6x-8x+4x^2\right)+2\)

\(\Leftrightarrow5x-24x+4x^2-72+12x=14-14x+4x^2\)

\(\Leftrightarrow4x^2-72-7x=14-14x+4x^2\)

\(\Leftrightarrow7x=86\Leftrightarrow x=\frac{86}{7}\)

a) 6x² - 5x + 3 = 2x - 3x(2 - x)

<=> 6x² - 5x + 3 = 2x - 6x + 3x²

<=> 6x² - 5x + 3 = -4x + 3x²

<=> 6x² - 5x + 3 + 4x - 3x² = 0

<=> 3x² - x + 3 = 0

=> Pt vô nghiệm

b) 25x² - 9 = (5x + 3)(2x + 1)

<=> 25x² - 9 = 10x² + 5x + 6x + 3

<=> 25x² - 9 = 10x² + 11x + 3

<=> 25x² - 9 - 10x² - 11x - 3 = 0

<=> 15x² - 12 - 11x = 0

<=> 15x² + 9x - 20x - 12 = 0

<=> 3x(5x + 3) - 4(5x + 3) = 0

<=> (5x + 3)(3x - 4) = 0

<=> 5x + 3 = 0 hoặc 3x - 4 = 0

<=> x = -3/5 hoặc x = 4/3

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cảm ơn

\(ĐKXĐ:x\ne2\)

\(\frac{9x^2}{x^3-8}+\frac{6}{x^2+2x+4}=\frac{3}{x-2}\)  

\(\Leftrightarrow\frac{9x^2}{x^3-8}+\frac{6\left(x-2\right)}{x^3-8}=\frac{3\left(x^2+2x+4\right)}{x^3-8}\)

\(\Rightarrow9x^2+6x-12=3x^2+6x+12\)

\(\Leftrightarrow9x^2-3x^2+6x-6x-12-12=0\)

\(\Leftrightarrow6x^2-24=0\)

\(\Leftrightarrow6\left(x^2-4\right)=0\)

\(\Leftrightarrow x^2-4=0\)

\(\Leftrightarrow x^2=4\)

\(\Leftrightarrow x=\pm2\) (loại x = 2)

vậy x = -2

\(\frac{9x^2}{x^3-8}+\frac{6}{x^2+2x+4}=\frac{3}{x-2}\)

=>\(\frac{9x^2}{x^3-8}+\frac{6\left(x-2\right)}{x^3-8}-\frac{3\left(x^2+2x+4\right)}{x^3+8}=0\)

=>\(9x^2+6x-12-3x^2-6x-24=0\)

=>\(6x^2-36\)\(6x^2-6\)

=>\(\left(6x-6\right)\left(6x+6\right)\)

=> \(6\left(x-1\right)6\left(x+1\right)\)

=>\(\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

#kenz