a) 2x⁴ - x³ -2x² -x +2=0

=> (2x⁴- 2x³) +(x³-x²) -(x² -x) -(2x-2)=0

=>(x-1)(2x³+x²-x-2)=0

=>(x-1)²( 2x²+3x+2)=0 ( vì 2x²+3x+2>0)

=> x-1=0 => x =1

chia cho x² , rồi đặt ẩn

a, Ta có : \(\frac{x+1}{2}+\frac{x-2}{4}=1-\frac{2\left(x-1\right)}{3}\)

=> \(\frac{6\left(x+1\right)}{12}+\frac{3\left(x-2\right)}{12}=\frac{12}{12}-\frac{8\left(x-1\right)}{12}\)

=> \(6\left(x+1\right)+3\left(x-2\right)=12-8\left(x-1\right)\)

=> \(6x+6+3x-6=12-8x+8\)

=> \(17x=20\)

=> \(x=\frac{20}{17}\)

b, Ta có : \(\frac{5x-1}{6}+x=\frac{6-x}{4}\)

=> \(\frac{5x-1+6x}{6}=\frac{6-x}{4}\)

=> \(4\left(11x-1\right)=6\left(6-x\right)\)

=> \(44x-4-36+6x=0\)

=> \(\)\(50x=40\)

=> \(x=\frac{4}{5}\)

c, Ta có : \(\frac{5\left(1-2x\right)}{3}+\frac{x}{2}=\frac{3\left(x-5\right)}{4}-2\)

=> \(\frac{20\left(1-2x\right)}{12}+\frac{6x}{12}=\frac{9\left(x-5\right)}{12}-\frac{24}{12}\)

=> \(20\left(1-2x\right)+6x=9\left(x-5\right)-24\)

=> \(20-40x+6x-9x+45+24=0\)

=> \(43x=89\)

=> \(x=\frac{89}{43}\)

a, \(\left(x^2-2x+1\right)-4=0\)

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

\(x^2-2x-3=0\)

\(\Delta=b^2-4ac=\left(-2\right)^2-4.1.3=4-12=-8< 0\)

Nên pt vô nghiệm 

b, \(\left| 5x-5\right|=0\)

\(\Leftrightarrow5x-5=0\Leftrightarrow5x=5\Leftrightarrow x=1\)

c, ĐKXĐ : \(\hept{\begin{cases}x+2\ne0\\x-2\ne0\\x^2-4\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne-2\\x\ne2\\x\ne\pm2\end{cases}\Rightarrow}x\ne\pm2}\)

\(\frac{x-2}{x+2}+\frac{3}{x-2}=\frac{x^2-11}{x^2-4}\)

\(\frac{\left(x-2\right)^2\left(x^2-4\right)}{\left(x+2\right)\left(x-2\right)\left(x^2-4\right)}+\frac{3\left(x+2\right)\left(x^2-4\right)}{\left(x-2\right)\left(x+2\right)\left(x^2-4\right)}=\frac{\left(x^2-11\right)\left(x+2\right)\left(x-2\right)}{\left(x^2-4\right)\left(x+2\right)\left(x-2\right)}\)

\(\left(x-2\right)^2\left(x^2-4\right)+3\left(x+2\right)\left(x^2-4\right)=\left(x^2-11\right)\left(x+2\right)\left(x-2\right)\)

\(\left(x-2\right)^2+3\left(x+2\right)=x^2-11\)

\(x^2-x+10=x^2-11\)

\(x^2-x+10-x^2+11=0\)

\(-x+21=0\Leftrightarrow x-21=0\Leftrightarrow x=21\)Theo ĐKXĐ : => tm 

a, \(\left(x^2-2x+1\right)-4=0\) \(\Leftrightarrow\left(x-1\right)^2=4=\left(\pm2\right)^2\)

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

Vậy phương trình có 2 nghiệm x=(3; -1)

b, \(\left|5x-5\right|=0\Leftrightarrow5x-5=0\)

                                 \(\Leftrightarrow5x=5\Rightarrow x=1\)

Vậy phương trình có nghiệm x=1

c, \(\frac{x-2}{x+2}+\frac{3}{x-2}=\frac{x^2-11}{x^2-4}\)\(\left(x\ge0;x\ne2\right)\) \(\Leftrightarrow\frac{\left(x-2\right)^2}{\left(x-2\right).\left(x+2\right)}+\frac{3.\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{x^2-11}{\left(x-2\right).\left(x+2\right)}\)

                                                                  \(\Leftrightarrow\left(x-2\right)^2+3.\left(x+2\right)=x^2-11\)

                                                                 \(\Leftrightarrow x^2-4x+4+3x+6=x^2-11\)

                                                                 \(\Leftrightarrow x=21\left(TM\right)\)

Vậy phương trình có nghiệm x=21

sr giải delta sai rồi.

a, \(\left(x^2-2x+1\right)-4=0\)

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

\(x^2-2x-3=0\)

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

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

Câu hỏi: Giải phương trình sau ....

Trả lời: Đây là bài lp 9

Mk lp 7 nên ko bt

ĐKXĐ: ...

Đặt \(x-\frac{1}{x}=a\Rightarrow a^3=x^3-\frac{1}{x^3}-3\left(x-\frac{1}{x}\right)\Rightarrow x^3-\frac{1}{x^3}=a^3+3a\)

Phương trình trở thành:

\(a^3+3a-2a-2=0\Leftrightarrow a^3+a-2=0\)

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

\(\Rightarrow a=1\Rightarrow x-\frac{1}{x}=1\Rightarrow x^2-x-1=0\)

ĐKXĐ: ...

- Với \(x=0\) không phải nghiệm

- Với \(x\ne0\)

\(\Leftrightarrow\frac{1}{x+\frac{3}{x}+24}-\frac{1}{x+\frac{3}{x}+25}=-1\)

Đặt \(x+\frac{3}{x}+24=t\)

\(\Leftrightarrow\frac{1}{t}-\frac{1}{t+1}=-1\)

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

\(\Leftrightarrow t^2+t+1=0\Leftrightarrow\left(t+\frac{1}{2}\right)^2+\frac{3}{4}=0\)

Pt đã cho vô nghiệm