đôi lúc là +- 10 thì sao 

đây là pt đỗi xứng bậc chẵn bạn ơi
cos cachs giải đó bạn

(+) Kiểm tra x = 0 , sau đó chia cả hai vế cho x^2

(+) đặt x- 1/x = a => x^2 + 1/x^2 = a^2 + 2 

Thay vô giải pt bậc hai 

Ta có:

$10x^2-x-11=0$

$\Delta=(-1)^2-4\cdot10\cdot(-11)=441=21^2$

$x=\dfrac{1\pm21}{20}$

$x=1$ hoặc $x=-\dfrac{11}{10}$

Vậy $x\in\left{1,-\dfrac{11}{10}\right}$.

$2x^2-3x-2=0$

$\Delta=(-3)^2-4\cdot2\cdot(-2)=25=5^2$

$x=\dfrac{3\pm5}{4}$

$x=2$ hoặc $x=-\dfrac12$

Vậy $x\in\left{2,-\dfrac12\right}$.

$2x^2-8=0$

$2(x^2-4)=0$

$(x-2)(x+2)=0$

$x=2$ hoặc $x=-2$

Vậy $x=\pm2$.

$3x^2-5x=0$

$x(3x-5)=0$

$x=0$ hoặc $x=\dfrac53$

Vậy $x\in\left{0,\dfrac53\right}$.

a)x⁵+x-1=0

<=>(x⁵+x⁴+x³+x²+x)-(x⁴+x³+x²+x+1)=0

<=>(x⁴+x³+x²+x+1)(x-1)=0

Do x⁴+x³+x²+x+1>0

=>x+1=0

<=>x=1

a. \(\Leftrightarrow\left(2x-5\right)\left(2x+5\right)\left(x+1\right)\left(2x-9\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x-5=0\\2x+5=0\\x+1=0\\2x-9=0\end{matrix}\right.\) \(\Rightarrow x=\)

b. \(\Leftrightarrow x^3+x+3x^2+3=0\)

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

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

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

c. \(\Leftrightarrow2x\left(3x-1\right)^2-\left(9x^2-1\right)=0\)

\(\Leftrightarrow\left(6x^2-2x\right)\left(3x-1\right)-\left(3x-1\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(6x^2-5x-1\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x-1\right)\left(6x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x-1=0\\6x+1=0\end{matrix}\right.\)

d.

\(\Leftrightarrow x^3-3x^2+2x-3x^2+9x-6=0\)

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

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

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

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

e.

\(\Leftrightarrow x^3+2x^2+x+3x^2+6x+3=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+1=0\end{matrix}\right.\)