b) Ta có: \(x^3+x^2-x+2=0\)

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

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

\(\Leftrightarrow x+2=0\)

hay x=-2

a/

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

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

\(\Leftrightarrow6-6x=0\)

=> x=1

b, \(4x^2+12x+9-3\left(x^2-16\right)=x^2-4x+4\)

\(\Leftrightarrow12x+9+48=-4x+4\Leftrightarrow16x=-53\Leftrightarrow x=-\dfrac{53}{16}\)

c, \(\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5x+2\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+3-5x-2\right)=0\Leftrightarrow\left(x-3\right)\left(-4x+1\right)=0\Leftrightarrow x=3;x=\dfrac{1}{4}\)

d, \(x^2\left(x+4\right)-9\left(x+4\right)=0\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x+4\right)=0\Leftrightarrow x=-3;3;-4\)

a: \(x^2-2-x+\sqrt{2}=0\)

=>\(\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)-\left(x-\sqrt{2}\right)=0\)

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

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

b: \(\left(1-\sqrt{2}\right)x^2-2\left(1+\sqrt{2}\right)x+1+3\sqrt{2}=0\)

\(\Delta=\left(-2-2\sqrt{2}\right)^2-4\left(1-\sqrt{2}\right)\left(1+3\sqrt{2}\right)\)

\(=12+8\sqrt{2}+4\left(\sqrt{2}-1\right)\left(3\sqrt{2}+1\right)\)

\(=12+8\sqrt{2}+4\left(6+\sqrt{2}-3\sqrt{2}-1\right)\)

\(=12+8\sqrt{2}+24-8\sqrt{2}-4=32>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{2\left(1+\sqrt{2}\right)-4\sqrt{2}}{2\left(1-\sqrt{2}\right)}=1\\x_2=\dfrac{2\left(1+\sqrt{2}\right)+4\sqrt{2}}{2\left(1-\sqrt{2}\right)}=-7-4\sqrt{2}\end{matrix}\right.\)

a) ĐKXĐ: \(x^2-1\ge0\)

Đặt \(\sqrt{x^2-1}=t\left(t\ge0\right)\)

\(\Rightarrow t=t^2\Rightarrow t\left(t-1\right)=0\Rightarrow\left[{}\begin{matrix}t=0\\t=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\sqrt{x^2-1}=0\\\sqrt{x^2-1}=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\pm1\\x=\pm\sqrt{2}\end{matrix}\right.\)

b) ĐKXĐ: \(x\ge2\)

Ta có: \(\sqrt{x-2}+\sqrt{x-3}\ge0\) mà \(\sqrt{x-2}+\sqrt{x-3}=-5< 0\Rightarrow\) không có x thỏa

c) \(\sqrt{x^2+4x+4}+\left|x-4\right|=0\)

\(\Rightarrow\left|x+2\right|+\left|x-4\right|=0\) mà \(\left|x+2\right|+\left|x-4\right|\ge0\Rightarrow\left\{{}\begin{matrix}x+2=0\\x-4=0\end{matrix}\right.\)

\(\Rightarrow\) không có x thỏa

a, \(\dfrac{x+1}{x+3}>1\Leftrightarrow\dfrac{x+1}{x+3}-1>0\Leftrightarrow\dfrac{x+1-x-3}{x+3}>0\)

\(\Rightarrow x+3< 0\)do  -2 < 0 

\(\Rightarrow x< -3\)Vậy tập nghiệm BFT là S = { x | x < -3 } 

b, \(\dfrac{2x-1}{x-3}\le2\Leftrightarrow\dfrac{2x-1}{x-3}-2\le0\Leftrightarrow\dfrac{2x-1-2x+6}{x-3}\le0\)

\(\Rightarrow x-3\le0\)do 5 > 0 

\(\Rightarrow x\le3\)Vậy tập nghiệm BFT là S = { x | x \(\le\)3 } 

c, \(\dfrac{x^2+2x+2}{x^2+3}\ge1\Leftrightarrow\dfrac{x^2+2x+2}{x^2+3}-1\ge0\)

\(\Leftrightarrow\dfrac{x^2+2x+2-x^2-3}{x^2+3}\ge0\Rightarrow2x-1\ge0\)do x^2 + 3 > 0 

\(\Rightarrow x\ge\dfrac{1}{2}\)Vậy tập nghiệm BFT là S = { x | x \(\ge\)1/2 } 

mình ko chắc nên mình đăng sau :> 

d, \(\dfrac{2x+1}{x^2+2}\ge1\Leftrightarrow\dfrac{2x+1}{x^2+2}-1\ge0\Leftrightarrow\dfrac{2x+1-x^2-2}{x^2+2}\ge0\)

\(\Rightarrow-x^2+2x-1\ge0\Rightarrow-\left(x-1\right)^2\ge0\)vô lí 

a)

\(2x-1+5\left(3-x\right)>0\\ 2x-2+15-5x>0\\ -3x+13>0\\ x< \dfrac{13}{3}.\)

\(a)2x-3=4x+6\\ \Rightarrow2x=-9\\ \Rightarrow x=-\dfrac{9}{2}\\ c)x\left(x-1\right)+x\left(x+3\right)=0\\ \Rightarrow x^2-x+x^2+3x=0\\ \Rightarrow2x^2+2x=0\\ \Rightarrow2x\left(x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x=0\\x+1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

`a)2x-3=4x+6`

`<=>2x-4x=6+3`

`<=>-2x=9`

`<=>x=-9/2`

Vậy `S={-9/2}`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`c)x(x-1)+x(x+3)=0`

`<=>x(x-1+x+3)=0`

`<=>x(2x+2)=0`

`@TH1:x=0`

`@TH2:2x+2=0<=>2x=-2<=>x=-1`

Vậy `S={-1;0}`

aa)2x – 3 = 4x + 6 

\(=>2x-4x=6+3\)

\(=>-2x=9\)

\(=>x=-\dfrac{9}{2}\)

c) x(x – 1) + x(x + 3) = 0      

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

\(x\left(2x+2\right)=0\)

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

a) (3x + 2)² - (3x - 2)² = 5x + 38

<=> 6x.4 = 5x + 38 <=> 19x = 38 <=> x = 2

b) 3(x - 2)² + 9(x - 1) = 3(x² + x - 3)

<=> 3x² - 12x + 12 + 9x - 9 = 3x² + 3x - 9

<=> -6x = -12 <=> x = 2

c) (x + 3)² - (x - 3)² = 6x + 8

<=> 2x.6 = 6x + 8 <=> 6x = 8 <=> x = 4/3

d) (x - 1)³ - x(x + 1)² = 5x(2 - x) - 11(x + 2)

<=> x³ - 3x² + 3x - 1 - x³ - 2x² - x = 10x - 5x² - 11x - 22

<=> 3x = -21 <=> x = -7

e) (x + 1)(x² - x + 1) - 2x = x(x - 1)(x + 1)

<=> x³ - 1 - 2x = x³ - x

<=> x = -1