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

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

Đề bài là gì vậy ?

(5x-9)^3 +  (11-5x)^3    =386 

\(\Rightarrow\)5x-9+11-5x=386

\(\Rightarrow\)2  =  386

Ta có: 

\(\left(5x-9\right)^3+\left(11-5x\right)^3=386\)

\(\Leftrightarrow125x^3-675x^2+1215x-729+1331-1815x+825x^2-125x^3-386=0\)

\(\Leftrightarrow150x^2-600x+216=0\)

\(\Leftrightarrow25x^2-100x+36=0\)

\(\Leftrightarrow\left(5x-18\right)\left(5x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{18}{5}\\x=\frac{2}{5}\end{cases}}\)

=x(x+1)+2(x+1)

khi và chỉ khi x+1=0 hoặc x+2=0

suy ra x= -1 hoặc x =-2

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

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

2x²+4x-2x-4=x²+3x+2x+6

2x²+2x-4=x²+5x+6

2x²-x²+2x-5x=6+4

x²-3x=10

x(x-3)=10

\(\Rightarrow\orbr{\begin{cases}x=10\\x=13\end{cases}}\)

Vậy............................................................

Ta có: \(x^3-7x^2=3x^2-12x\)

\(\Leftrightarrow x^3-10x^2+12x=0\)

\(\Leftrightarrow x\left(x^2-10x+12\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-10x+12=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\\left(x-5\right)^2=13\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x-5=\pm\sqrt{13}\end{cases}}\)

\(\Rightarrow x\in\left\{0;5-\sqrt{13};5+\sqrt{13}\right\}\)

\(x^3-7x^2=3x^2-12x\)

\(\Leftrightarrow x^3-7x^2-3x^2+12x=0\)

\(\Leftrightarrow x^3-10x^2+12x=0\)

\(\Leftrightarrow x\left(x^2-10x+12\right)=0\Leftrightarrow x=0\)

\(x\left(x-3\right)+5x=x^2-8\)

\(\Leftrightarrow x^2-3x+5x=x^2-8\)

\(\Leftrightarrow2x=-8\)

\(\Leftrightarrow x=-4\)

\(x\left(x-3\right)+5x=x^2-8\)

\(\Leftrightarrow x^2-3x+5x=x^2-8\)

\(\Leftrightarrow x^2+2x-x^2+8=0\Leftrightarrow2x+8=0\)

\(\Leftrightarrow2\left(x+4\right)=0\Leftrightarrow x=-4\)