\(3x^2-6xy+3y^2-12z^2\)

\(=3\left(x^2-2xy+y^2-4z^2\right)\)

\(=3\left[\left(x-y\right)^2-4z^2\right]=3\left(x-y-2z\right)\left(x-y+2z\right)\)

Ta có: \(3x^2-6xy+3y^2-12z^2\)

    \(=3.\left(x^2-2xy+y^2-4z^2\right)\)

    \(=3.\left[\left(x-y\right)^2-4z^2\right]\)

    \(=3.\left(x-y-2z\right).\left(x-y+2z\right)\)

Theo bài ra ta có : 

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

Gớm Tú ơi, làm gì mà Dis nhiều thế :)) Nghiếp khiếp vậy mày:))))

a) \(A=\left(2\sqrt{3}-5\sqrt{27}+4\sqrt{12}\right):\sqrt{3}\)

\(=2\sqrt{3}:\sqrt{3}-5\sqrt{27}:\sqrt{3}+4\sqrt{12}:\sqrt{3}\)

\(=2\sqrt{3:3}-5\sqrt{27:3}+4\sqrt{12:3}\)

\(=2\sqrt{1}-5\sqrt{9}+4\sqrt{4}=2.1-5.3+4.2=2-15+8=-5\)

\(B=\frac{\left(2+\sqrt{3}\right)\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}}=\frac{\left(2+\sqrt{3}\right).\left(\sqrt{2-\sqrt{3}}\right)^2}{\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}}\)

\(=\frac{\left(2+\sqrt{3}\right).\left(2-\sqrt{3}\right)}{\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}=\frac{4-3}{\sqrt{4-3}}=\frac{1}{\sqrt{1}}=1\)

b) \(ĐKXĐ:x\ge\frac{7}{2}\)

Thay \(A=-5\), \(B=1\)vào biểu thức ta được:

\(1-3\sqrt{2x-7}=-5\)\(\Leftrightarrow3\sqrt{2x-7}=6\)

\(\Leftrightarrow\sqrt{2x-7}=2\)\(\Leftrightarrow2x-7=4\)

\(\Leftrightarrow2x=11\)\(\Leftrightarrow x=\frac{11}{2}\)( thỏa mãn ĐKXĐ )

Vậy \(x=\frac{11}{2}\)

Ta có: \(2x^2+5x+3=0\)

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

\(\Leftrightarrow2x\left(x+1\right)+3\left(x+3\right)=0\)

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

\(\Rightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=-1\end{cases}}\)

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

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

\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)=0\)

\(\Leftrightarrow\left(2x+3\right)\left(x+1\right)=0\Leftrightarrow x=-\frac{3}{2}orx=-1\)

Vậy nghiệm của phương trình là x = -3/2 ; -1 

\(\left(2x-5\right)^2=\left(x+2\right)^2\)

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

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

\(\Leftrightarrow\left(x-7\right).\left(3x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\3x-3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=7\\3x=3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=7\\x=1\end{cases}}\)

Vậy tập nghiệm của phương trình là \(S=\left\{7;1\right\}\)

\(\left(x+1\right)^2=4.\left(x^2-2x+1\right)\)

\(\Leftrightarrow\left(x+1\right)^2=4.\left(x-1\right)^2\)

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

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

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

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

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

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

a) Ta có: \(\left(2x-5\right)^2=\left(x+2\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}2x-5=x+2\\2x-5=-x-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\3x=3\end{cases}}\Rightarrow\orbr{\begin{cases}x=7\\x=1\end{cases}}\)

b) Ta có: \(\left(x+1\right)^2=4\left(x^2-2x+1\right)\)

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

\(\Leftrightarrow\orbr{\begin{cases}x+1=2x-2\\x+1=2-2x\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\3x=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{1}{3}\end{cases}}\)

a) đk: \(x\ne\left\{1;2\right\}\)

Ta có: \(1+\frac{2x-5}{x-2}-\frac{3x-5}{x-1}=0\)

\(\Leftrightarrow\frac{\left(x-1\right)\left(x-2\right)+\left(2x-5\right)\left(x-1\right)-\left(3x-5\right)\left(x-2\right)}{\left(x-1\right)\left(x-2\right)}=0\)

\(\Rightarrow x^2-3x+2+2x^2-7x+5-3x^2+11x-10=0\)

\(\Leftrightarrow x-3=0\Rightarrow x=3\)

Vậy x = 3

b) đk: \(x\ne\left\{0;2\right\}\)

Ta có: \(\frac{x+2}{x-2}-\frac{2}{x^2-2x}=\frac{1}{x}\)

\(\Leftrightarrow\frac{\left(x+2\right)x-2}{\left(x-2\right)x}=\frac{x-2}{\left(x-2\right)x}\)

\(\Rightarrow x^2+2x-2=x-2\)

\(\Leftrightarrow x^2+x=0\Leftrightarrow x\left(x+1\right)=0\) vì x khác 0 nên

=> \(x+1=0\Rightarrow x=-1\)

Vậy x = -1

c) đk: \(x\ne\pm3\)

Ta có: \(\frac{x+2}{x-3}+\frac{x-2}{x+3}-\frac{2\left(x^2+6\right)}{x^2-9}=0\)

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

\(\Rightarrow x^2+5x+6+x^2-5x+6-2x^2-12=0\)

\(\Leftrightarrow0x=0\) (luôn đúng)

Vậy mọi x thực thỏa mãn đk thì PT luôn có nghiệm

d) đk: \(x\ne-1\)

Ta có: \(\frac{-7x^2+4}{x^3+1}=\frac{5}{x^2-x+1}-\frac{1}{x+1}\)

\(\Leftrightarrow\frac{-7x^2+4}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{5\left(x+1\right)-\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(\Rightarrow-7x^2+4=5x+5-x^2+x-1\)

\(\Leftrightarrow6x^2+6x=0\)

\(\Leftrightarrow6x\left(x+1\right)=0\) vì x + 1 khác 0

=> x = 0

Vậy x = 0

1) Ta có: \(\left(x^2-5x\right)^2+10\left(x^2-5x\right)+24=0\)

\(\Leftrightarrow\left(x^2-5x+6\right)\left(x^2-5x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-5x+6=0\\x^2-5x+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\left(x-2\right)\left(x-3\right)=0\\\left(x-1\right)\left(x-4\right)=0\end{cases}}\Rightarrow x\in\left\{1;2;3;4\right\}\)

2) Ta có: \(x\left(x+1\right)\left(x^2+x+1\right)=42\)

\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x+1\right)-42=0\)

\(\Leftrightarrow\left(x^2+x\right)^2+\left(x^2+x\right)-42=0\)

\(\Leftrightarrow\left(x^2+x-6\right)\left(x^2+x+7\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+3\right)\left(x^2+x+7\right)=0\)

Vì \(x^2+x+7=\left(x^2+x+\frac{1}{4}\right)+\frac{27}{4}=\left(x+\frac{1}{2}\right)^2+\frac{27}{4}>0\left(\forall x\right)\)

\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)