\(\left(x^2+2x\right)^2-6x^2+12x+9=0\Leftrightarrow x^4+4x^3+4x^2-6x^2+12x+9=0\\ \Leftrightarrow x^4+4x^3-2x^2+12x+9=0\Leftrightarrow x^2+4x-2+\frac{12}{x}+\frac{9}{x^2}=0\\ \Leftrightarrow\left(x^2+\frac{9}{x^2}\right)+4\left(x+\frac{3}{x}\right)-2=0\)

Đặt \(k=x+\frac{3}{x}\Rightarrow x^2+\frac{9}{x^2}=k^2-6\)

Ta đc \(k^2-6+4k-2=0\Leftrightarrow k^2+4k-8=0\)

\(\left(x^2+2x\right)^2\)\(-6x^2\)\(+12x+9\)=0

⇔\(\left(x^2\right)^2\)\(+2.2x.x^2\)+\(2x^2\)-6x²+12x+9=0

⇔ x⁴+ 4x³+2x²-6x²+12x+9=0

⇔ x²+4x³-4x² +12x=-9

⇔x²+ 4x(x-x+3)=-9

⇔x²+12x=-9

⇔x(x+12)=-9

⇔ {x=-9 hoặc x+12=-9}

⇔ {x=-9 hoặc x=-21}

S={-9;-21}

\(\left(x+1\right)^2=4\left(x^2-2x+1\right)^2\\\Leftrightarrow\left(x+1\right)^2=4\left(x-1\right)^2\\\Leftrightarrow \left(x+1\right)^2-4\left(x-1\right)^2=0\\\Leftrightarrow \left(x+1\right)^2-\left(2x-2\right)^2=0\\\Leftrightarrow \left[\left(x+1\right)+\left(2x-2\right)\right]\left[\left(x+1\right)-\left(2x-2\right)\right] =0\\ \Leftrightarrow\left(x+1+2x-2\right)\left(x+1-2x+2\right)=0\\\Leftrightarrow \left(3x-1\right)\left(3-x\right)=0\\\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\3-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=3\end{matrix}\right. \)

Vậy phương trình có tập nghiệm \(S=\left\{\frac{1}{3};3\right\}\)

\(\left(2x+7\right)^2=9\left(x+2\right)^2\\ \Leftrightarrow\left(2x+7\right)^2-9\left(x+2\right)^2=0\\ \Leftrightarrow\left(2x+7\right)^2-\left(3x+6\right)^2=0\\ \Leftrightarrow\left[\left(2x+7\right)+\left(3x+6\right)\right]\left[\left(2x+7\right)-\left(3x+6\right)\right]=0\\ \Leftrightarrow\left(2x+7+3x+6\right)\left(2x+7-3x-6\right)=0\\ \Leftrightarrow\left(5x+13\right)\left(1-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}5x+13=0\\1-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-13}{5}\\x=1\end{matrix}\right.\)

Vậy phương trình có tập nghiệm \(S=\left\{\frac{-13}{5};1\right\}\)

\(4\left(2x+7\right)^2=9\left(x+3\right)^2\\\Leftrightarrow 4\left(2x+7\right)^2-9\left(x+3\right)=0\\ \Leftrightarrow\left(4x+14\right)^2-\left(3x+9\right)^2=0\\\Leftrightarrow \left[\left(4x+14\right)+\left(3x+9\right)\right]\left[\left(4x+14\right)-\left(3x+9\right)\right]=0\\\Leftrightarrow \left(4x+14+3x+9\right)\left(4x+14-3x-9\right)=0\\\Leftrightarrow \left(7x+23\right)\left(x+5\right)=0\\\Leftrightarrow\left[{}\begin{matrix}7x+23=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-23}{7}\\x=-5\end{matrix}\right. \)

Vậy phương trình có tập nghiệm \(S=\left\{\frac{-23}{7};-5\right\}\)

b) \(7x\left(x-2\right)-\left(x-2\right)=0\) 

<=>  \(\left(7x-1\right)\left(x-2\right)=0\)

=> x=1/7  hoặc x=2

c) <=>  (2x-1)³   =0 

=> x=1/2

d)<=>  \(\left(2x-3\right)\left(2x+3\right)-x\left(2x-3\right)=0\)

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

=> x=3/2  hoặc x=-3

e) <=>\(x^2\left(x+5\right)+9\left(x+5\right)=0\)

<=> \(\left(x+5\right)\left(x^2+9\right)=0\)

=> x=-5

f) \(x^3-6x^2-x+30=0\)

<=>\(x^3+2x^2-8x^2-16x+15x+30=0\)

<=>\(x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)=0\)

<=>\(\left(x+2\right)\left(x^2-5x-3x+15\right)=0\)

<=> \(\left(x+2\right)\left(x-5\right)\left(x-3\right)=0\)

=> x=-2 hoặc x=5 hoặc x=3

c/ đk: x khác 1; x khác -3

\(\dfrac{3x-1}{x-1}+\dfrac{2x+5}{x+3}+\dfrac{4}{x^2+2x-3}=1\)

\(\Rightarrow\left(3x+1\right)\left(x+3\right)+\left(2x+5\right)\left(x-1\right)+4=x^2+2x-3\)

\(\Leftrightarrow3x^2+10x+3+2x^2+3x-5+4=x^2+2x-3\)

\(\Leftrightarrow4x^2+11x+5=0\)

\(\Leftrightarrow\left(4x^2+2\cdot2x\cdot\dfrac{11}{4}+\dfrac{121}{16}\right)-\dfrac{41}{16}=0\)

\(\Leftrightarrow\left(2x+\dfrac{11}{4}\right)^2=\dfrac{41}{16}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{11}{4}=\dfrac{\sqrt{41}}{4}\\2x+\dfrac{11}{4}=-\dfrac{\sqrt{41}}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-11+\sqrt{41}}{8}\\x=\dfrac{-11-\sqrt{41}}{8}\end{matrix}\right.\)

Vậy.........

d/ \(\dfrac{12x+1}{6x-2}-\dfrac{9x-5}{3x+1}=\dfrac{108x-36x^2-9}{4\left(9x^2-1\right)}\)

đk: \(x\ne\pm\dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{12x+1}{2\left(3x-1\right)}-\dfrac{9x-5}{3x+1}=\dfrac{108x-36x^2-9}{4\left(3x-1\right)\left(3x+1\right)}\)

\(\Rightarrow2\left(12x+1\right)\left(3x+1\right)-4\left(9x-5\right)\left(3x-1\right)=108x-36x^2-9\)

\(\Leftrightarrow72x^2+24x+6x+2-108x^2+36x-60x-20-108x+36x^2+9=0\)

\(\Leftrightarrow-102x-9=0\)

\(\Leftrightarrow-102x=9\Leftrightarrow x=-\dfrac{3}{34}\)(TM)

Vậy.........

a/ \(\left(x+1\right)^2\left(x+2\right)+\left(x+1\right)^2\left(x-2\right)=-24\)

\(\Leftrightarrow\left(x+1\right)^2\left(x+2+x-2\right)=-24\)

\(\Leftrightarrow2x\left(x^2+2x+1\right)=-24\)

\(\Leftrightarrow2x^3+4x^2+2x+24=0\)

\(\Leftrightarrow2x^3-2x^2+8x+6x^2-6x+24=0\)

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

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

\(\Leftrightarrow2\left(x^2-x+4\right)\left(x+3\right)=0\)

Ta thấy: \(x^2-x+4=\left(x^2-2x\cdot\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{15}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{15}{4}>0\)

=> x+ 3 = 0 <=> x= -3

Vậy......

b/ \(2x^3+3x^2+6x+5=0\)

\(\Leftrightarrow2x^3+x^2+5x+2x^2+x+5=0\)

\(\Leftrightarrow x\left(2x^2+x+5\right)+\left(2x^2+x+5\right)=0\)

\(\Leftrightarrow\left(2x^2+x+5\right)\left(x+1\right)=0\)

Ta thấy: \(2x^2+x+5=\left(\sqrt{2}x+2\cdot\sqrt{2}x\cdot\dfrac{\sqrt{2}}{4}+\dfrac{1}{8}\right)+\dfrac{39}{8}=\left(\sqrt{2}x+\dfrac{\sqrt{2}}{4}\right)^2+\dfrac{39}{8}>0\)

=> x + 1 = 0 <=> x = -1

Vậy....

a: \(=\dfrac{\left(x^4-y^4\right)^2}{x^2+y^2}=\left(x^2-y^2\right)^2\cdot\left(x^2+y^2\right)\)

b: \(=\dfrac{\left(4x+3\right)\left(16x^2-12x+9\right)}{16x^2-12x+9}=4x+3\)

1.

\(x^4-6x^2-12x-8=0\)

\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)

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

\(\Leftrightarrow x=1\pm\sqrt{5}\)

3.

ĐK: \(x\ge-9\)

\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)

\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)

Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)

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

\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)

\(\Leftrightarrow...\)

2.

ĐK: \(x\ne\dfrac{2\pm\sqrt{2}}{2};x\ne\dfrac{-2\pm\sqrt{2}}{2}\)

\(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{1}{2x+\dfrac{1}{x}+4}+\dfrac{1}{2x+\dfrac{1}{x}-4}=\dfrac{3}{5}\)

Đặt \(2x+\dfrac{1}{x}+4=a;2x+\dfrac{1}{x}-4=b\left(a,b\ne0\right)\)

\(pt\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{3}{5}\left(1\right)\)

Lại có \(a-b=8\Rightarrow a=b+8\), khi đó:

\(\left(1\right)\Leftrightarrow\dfrac{1}{b+8}+\dfrac{1}{b}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{2b+8}{\left(b+8\right)b}=\dfrac{3}{5}\)

\(\Leftrightarrow10b+40=3\left(b+8\right)b\)

\(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=-\dfrac{20}{3}\end{matrix}\right.\)

TH1: \(b=2\Leftrightarrow...\)

TH2: \(b=-\dfrac{20}{3}\Leftrightarrow...\)

1) x^2 - 6x = 0

⇔ x ( x - 6 ) = 0

⇔ \(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

Vậy x = 0 hoặc x = 6

2) 2x^3 - 5x^2 - 12x = 0

⇔ 2x^3 - 8x^2 + 3x^2 - 12x = 0

⇔ 2x^2 ( x - 4 ) + 3x ( x - 4 ) = 0

⇔ ( 2x^2 + 3x ) ( x - 4 ) = 0

⇔ x ( 2x + 3 ) ( x - 4 ) = 0

⇔ \(\left[{}\begin{matrix}x=0\\2x+3=0\\x-4=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=0\\x=-1,5\\x=4\end{matrix}\right.\)

Vậy x = 0 , x = -1,5 hoặc x = 4

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

⇔ ( x + 2 ) ( x + 1 - x - 3 ) = 0

⇔ -2 ( x + 2 ) = 0

⇔ x = - 2

Vậy x = -2