Ta có: \(\left|x+1\right|+3\left|x-1\right|=x+2+\left|x\right|+2\left|x-2\right|\)

=>\(\left|x+1\right|-\left|x\right|+3\left|x-1\right|-2\left|x-2\right|=x+2\) (1)

TH1: x<-1

=>x+1<0; x<0; x-1<0; x-2<0

(1) sẽ trở thành:

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

=>-x-1+x-3x+3+2x-4=x+2

=>-x-2=x+2

=>-2x=4

=>x=-2(nhận)

TH2: -1<=x<0

=>x+1>=0; x<0; x-1<0; x-2<0

(1) sẽ trở thành:

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

=>x+1+x-3x+3+2x-4=x+2

=>x=x+2

=>0=2(vô lý)

TH3: 0<=x<1

=>x+1>0; x>=0; x-1<0; x-2<0

(1) sẽ trở thành:

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

=>1-3x+3+2x-4=x+2

=>-x=x+2

=>-2x=2

=>x=-1(loại)

TH4: 1<=x<2

=>x+1>0; x>0; x-1>=0; x-2<0

(1) sẽ trở thành:

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

=>1+3x-3+2x-4=x+2

=>5x-6=x+2

=>4x=8

=>x=2(loại)

TH5: x>=2

=>x+1>0; x>0; x-1>0; x-2>=0

(1) sẽ trở thành:

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

=>1+3x-3-2x+4=x+2

=>x+2=x+2(luôn đúng)

=>x>=2

Vậy: x>=2 hoặc x=-2

2(x²+x+1)²-7(x-1)²=13(x³-1)

<=> 2(x²+x+1)²-7(x-1)²-13(x³-1)=0

<=>2(x²+x+1)²-14(x³-1)+(x³-1)-7(x-1)²=0

<=> 2(x²+x+1)(x²+x+1-7x+7)+(x-1)(x²+x+1-7x+7)=0

<=> (2x²+2x+2)(x²-6x+8)+(x-1)(x²-6x+8)=0

<=> (x²-6x+8)(2x²+3x+1)=0

<=> (x²-4x-2x+8)(2x²+2x+x+1)=0

<=> [x(x-4)-2(x-4)][2x(x+1)+(x+1)]=0

<=> (x-4)(x-2)(x+1)(2x+1)=0

Đến đây dễ rồi nhé bạn

x\(\approx\)-0,84;-11,9

Ta có:

$\dfrac{x^4}{2x^2+1}+\dfrac{2x^2+1}{x^4}=2$

Điều kiện: $x\ne0$.

Đặt $t=\dfrac{x^4}{2x^2+1}>0$.

Khi đó:

$t+\dfrac{1}{t}=2$

$t^2-2t+1=0$

$(t-1)^2=0$

$t=1$

$\dfrac{x^4}{2x^2+1}=1$

$x^4=2x^2+1$

$x^4-2x^2-1=0$

Đặt $y=x^2\ge0$:

$y^2-2y-1=0$

$y=1+\sqrt2$ (do $y\ge0$)

$x^2=1+\sqrt2$

$x=\pm\sqrt{1+\sqrt2}$

Vậy $x=\pm\sqrt{1+\sqrt2}$.

Ta có:

$\left(\dfrac{x}{x-1}\right)^2+\left(\dfrac{x}{x+1}\right)^2=\dfrac{10}{9}$

Điều kiện: $x\ne\pm1$.

Quy đồng:

$\dfrac{x^2(x+1)^2+x^2(x-1)^2}{(x^2-1)^2}=\dfrac{10}{9}$

$\dfrac{x^2[(x+1)^2+(x-1)^2]}{(x^2-1)^2}=\dfrac{10}{9}$

$\dfrac{2x^2(x^2+1)}{(x^2-1)^2}=\dfrac{10}{9}$

$18x^2(x^2+1)=10(x^2-1)^2$

$9x^2(x^2+1)=5(x^2-1)^2$

$9x^4+9x^2=5x^4-10x^2+5$

$4x^4+19x^2-5=0$

Đặt $t=x^2\ge0$:

$4t^2+19t-5=0$

$(4t-1)(t+5)=0$

$t=\dfrac14$ (do $t\ge0$)

$x^2=\dfrac14$

$x=\pm\dfrac12$

Vậy $x=\pm\dfrac12$.

Ta có:

$x^3+3x^2-10x-24=0$

Nhóm các hạng tử:

$x^2(x+3)-10(x+3)=0$

$(x+3)(x^2-10)=0$

$x+3=0$ hoặc $x^2-10=0$

$x=-3$ hoặc $x=\pm\sqrt{10}$

Vậy $x\in{-3,-\sqrt{10},\sqrt{10}}$.

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

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

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

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

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

\(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}1-x\ge0\\x^2+x+2=\left(1-x\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\3x=-1\end{matrix}\right.\) \(\Rightarrow x=-\frac{1}{3}\)

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

=>\(\dfrac{1}{x-4}-\dfrac{1}{x-1}=2\)

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

=>2x^2-10x+8=3

=>2x^2-10x+5=0

=>\(x=\dfrac{5\pm\sqrt{15}}{2}\)