`x^4+3x^2-2=0`

Đặt `x^2=t(t>=0)`

`pt<=>t^2+3t-2=0`

`<=>t^2+3t+9/4=17/4`

`<=>(t+3/2)^2=17/4`

`<=>t+3/2=sqrt{17}/2(do \ t>=0=>t+3/2>=3/2)`

`<=>t=(sqrt{17}-3)/2`

`<=>x^2=(sqrt{17}-3)/2`

`<=>x=+-sqrt{(sqrt{17}-3)/2}`

\(a,M=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{2}{x}-\dfrac{2-x}{x\sqrt{x}+x}\right)\left(x>0;x\ne1\right)\\ M=\dfrac{x+\sqrt{x}+x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{2\sqrt{x}+2-2+x}{x\left(\sqrt{x}+1\right)}\\ M=\dfrac{2x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{x\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\\ M=\dfrac{2x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(b,M=-\dfrac{1}{2}\Leftrightarrow\dfrac{2x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=-\dfrac{1}{2}\\ \Leftrightarrow-4x=x+\sqrt{x}-2\\ \Leftrightarrow5x+\sqrt{x}-2=0\)

Đặt \(\sqrt{x}=t\)

\(\Leftrightarrow5t^2+t-2=0\\ \Delta=1^2-4\cdot5\left(-2\right)=41\\ \Leftrightarrow\left[{}\begin{matrix}t=\dfrac{-1-\sqrt{41}}{10}\\t=\dfrac{-1+\sqrt{41}}{10}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\left(1+\sqrt{41}\right)^2}{100}=\dfrac{-42-2\sqrt{41}}{100}\\x=\dfrac{\left(\sqrt{41}-1\right)^2}{100}=\dfrac{42-2\sqrt{41}}{100}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-21-\sqrt{41}}{50}\left(L\right)\\x=\dfrac{21-\sqrt{41}}{50}\left(N\right)\end{matrix}\right.\\ \Leftrightarrow x=\dfrac{21-\sqrt{41}}{50}\)

a: Ta có: \(M=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{2}{x}+\dfrac{x-2}{x\sqrt{x}+x}\right)\)

\(=\dfrac{x+\sqrt{x}+x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{2\sqrt{x}+2+x-2}{x\left(\sqrt{x}+1\right)}\)

\(=\dfrac{2x}{\sqrt{x}-1}\cdot\dfrac{x}{\sqrt{x}\left(\sqrt{x}+2\right)}\)

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

\(\left|2x-\frac{1}{2}\right|+1=3x\)

\(\Leftrightarrow\left|2x-\frac{1}{2}\right|=3x-1\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{3}{10}\end{cases}}\)

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

3x²+3x-2x²-4x=1+x²

3x²+3x-2x²-4x-x²=1

x=-1

vậy............

1. Rút gọn biểu thức :

\(M=4.\left(2-3x\right)-\left|2x-3\right|\) (*)

- Xét 2 TH :

+ Trường hợp 1 : \(\left|2x-3\right|=\left(2x-3\right)\) thì (*) trở thành :

\(M=4.\left(2-3x\right)-\left(2x-3\right)\)

\(\Rightarrow M=8-12x-2x+3\)

\(\Rightarrow M=-14x+11\)

+ Trường hợp 2 : \(\left|2x-3\right|=\left(3-2x\right)\) thì (*) trở thành :

\(M=4.\left(2-3x\right)-\left(3-2x\right)\)

\(\Rightarrow M=8-12x-3+2x\)

\(\Rightarrow M=-10x+5\)

Cái bài giải phương trình ở lớp 8 mới học nhé bạn.

a: \(=\dfrac{x-2x-1}{x+1}=\dfrac{-\left(x+1\right)}{x+1}=-1\)

b: \(=\dfrac{2+2x}{x\left(x+1\right)}=\dfrac{2\left(x+1\right)}{x\left(x+1\right)}=\dfrac{2}{x}\)

c: \(=\dfrac{3x-1}{2\left(3x+1\right)}+\dfrac{3x+1}{2\left(3x-1\right)}-\dfrac{6x}{\left(3x-1\right)\left(3x+1\right)}\)

\(=\dfrac{9x^2-6x+1+9x^2+6x+1-12x}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{18x^2-12x+2}{2\left(3x-1\right)\left(3x+1\right)}\)

\(=\dfrac{2\left(3x-1\right)^2}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{3x-1}{3x+1}\)

a) (x-2)³+6(x+1)²-x³+12=0

\(\Rightarrow\)x³-6x²+12x-8+6(x²+2x+1)-x³+12=0

\(\Rightarrow\)x³-6x²+12x-8+6x²+12x+6-x³+12=0

\(\Rightarrow\)24x+10=0

\(\Rightarrow\)24x=-10

\(\Rightarrow\)x=\(\dfrac{-10}{24}=\dfrac{-5}{12}\)

b)(x-5)(x+5)-(x+3)²+3(x-2)²=(x+1)²-(x-4)(x+4)+3x²

\(\Rightarrow\)x²-25-(x²+6x+9)+3(x²-4x+4)=x²+2x+1-(x²-16)+3x²

\(\Rightarrow\)x²​-25-x²-6x-9+3x²-12x+12=x²+2x+1-x²+16+3x²

\(\Rightarrow\)3x²-18x-22=3x²+2x+17

\(\Rightarrow\)3x²-18x-22-3x²-2x-17=0

\(\Rightarrow\)-20x-39=0

\(\Rightarrow\)-20x=39

\(\Rightarrow\)x=\(-\dfrac{39}{20}\)

c) (2x+3)² +(x-1)(x+1)=5(x+2)²-(x-5)(x+1)+(x+4)

⇒4x²+12x+9+x²-1=5(x²+4x+4)-(x²+x-5x-5)+x+4

⇒5x²+12x+8=5x²+20x+20-x²-x+5x+5+x+4

⇒5x²+12x+8-5x²-20x-20+x²+x-5x-5-x-4=0

⇒x²-13x-21=0

a: M(1)=-3

=>1-2m+m=3

=>1-m=3

hay m=-2

b: M(x)=x²+4x-2

Đặt M(x)=0

=>x²+4x+4-6=0

=>(x+2)²=6

hay \(x\in\left\{\sqrt{6}-2;-\sqrt{6}-2\right\}\)