\(x=3\ge2\Leftrightarrow y=3+1=4\\ x=-1< 2\Leftrightarrow y=\left(-1\right)^2-2=1-2=-1\\ x=2\ge2\Leftrightarrow y=2+1=3\)

a: \(=6x^3-10x^2+6x\)

b: \(=-2x^4-10x^3+6x^2\)

c: \(=-x^5+2x^3-\dfrac{3}{2}x^2\)

d: \(=2x^3+10x^2-8x-x^2-5x+4=2x^3+9x^2-13x+4\)

** Bạn lưu ý lần sau viết đề bằng công thức toán để được hỗ trợ tốt hơn.

Lời giải:

$\frac{a+b}{c}+\frac{a+c}{b}+\frac{b+c}{a}=-2$

$\Leftrightarrow \frac{a+b}{c}+1+\frac{a+c}{b}+1+\frac{b+c}{a}=0$

$\Leftrightarrow (a+b+c)(\frac{1}{c}+\frac{1}{b})+\frac{b+c}{a}=0$

$\Leftrightarrow \frac{(a+b+c)(b+c)}{bc}+\frac{b+c}{a}=0$

$\Leftrightarrow (b+c)(\frac{a+b+c}{bc}+\frac{1}{a})=0$

$\Leftrightarrow (b+c).\frac{a(a+b+c)+bc}{abc}=0$

$\Leftrightarrow \frac{(b+c)(a+b)(a+c)}{abc}=0$

$\Rightarrow (a+b)(b+c)(c+a)=0$

$\Rightarrow a+b=0$ hoặc $b+c=0$ hoặc $c+a=0$

Không mất tổng quát giả sử $a+b=0\Rightarrow a=-b$

$1=a^3+b^3+c^3=(-b)^3+b^3+c^3=c^3\Rightarrow c=1$

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{-1}{b}+\frac{1}{b}+\frac{1}{1}=1$

Vậy..........

Giải

a) Ta có : 2.x² -2.x = 5.x 

<=> 2.x² -3.x-5=0 : a = 2 ; b = 3 ; c = -5 

b) Ta có : x² +2.x = m. x + m 

<=> x² + ( 2-m ) .x - m = 0 : a = 1 ; b=2-m ; c=-m

c) Ta có : 2.x² \(+\sqrt{2}.\left(3.x-1\right)=1+\sqrt{2}\)

<=>  2.x²  + 3.\(\sqrt{2}.x-2.\sqrt{2}-1=0\): a = 2 ; b= 3\(\sqrt{2};c=-2\sqrt{2}-1\)

a) \(2x^2-2x=5+x\)

\(\Leftrightarrow2x^2-x-5=0\)với \(\hept{\begin{cases}a=2\\b=-3\\c=-5\end{cases}}\)

b) \(x^2+2x=mx+m\)

\(\Leftrightarrow x^2+\left(2-m\right)x-m=0\)với \(\hept{\begin{cases}z=1\\b=3-m\\c=-m\end{cases}}\)

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

\(\Leftrightarrow2x^2+3\sqrt{2}\cdot x-2\sqrt{2}-1=0\)

với \(\hept{\begin{cases}a=2\\b=3\sqrt{2}\\c=-2\sqrt{2}-1\end{cases}}\)

\(=2x.x-2x.3+x-3\\ =2x^2-6x+x-3\\ =2x^2-5x-3\)

=> Chọn B

a: ĐKXĐ của A là x<>1; x<>-3

ĐKXĐ của B là x<>4

ĐKXĐ của C là x<>0; x<>2

ĐKXĐ của D là x<>3

ĐKXĐ của E là x<>0; x<>2

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

Để A=0 thì 2x=0

=>x=0

\(B=\dfrac{\left(x-4\right)\left(x+4\right)}{\left(x-4\right)^2}=\dfrac{x+4}{x-4}\)

Để B=0 thì x+4=0

=>x=-4

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

Để C=0 thì x+2=0

=>x=-2

\(D=\dfrac{\left(x+4\right)\left(x-3\right)}{\left(x-3\right)\left(x^2+3x+9\right)}=\dfrac{x+4}{x^2+3x+9}\)

Để D=0 thi x+4=0

=>x=-4
\(E=\dfrac{2x\left(x^2+2x+1\right)}{2x\left(x-2\right)}=\dfrac{\left(x+1\right)^2}{x-2}\)

Để E=0 thì (x+1)^2=0

=>x=-1

a) \(x^2-2x-4y^2-4y=\left(x^2-4y^2\right)-\left(2x+4y\right)=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)=\left(x+2y\right)\left(x-2y-2\right)\)

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

c) \(x^3-4x^2+12x-27=x^3-3x^2-x^2+3x+9x-27=x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)=\left(x-3\right)\left(x^2-x+9\right)\)

d) \(a^6-a^4+2a^3+2a^2=a^2\left(a^4-a^2+2a+2\right)=a^2\left[a^2\left(a-1\right)\left(a+1\right)+2\left(a+1\right)\right]=a^2\left(a+1\right)\left(a^3-a^2+2\right)=a^2\left(a+1\right)\left[a^3+a^2-2a^2+2\right]=a^2\left(a+1\right)\left[a^2\left(a+1\right)-2\left(a-1\right)\left(a+1\right)\right]=a^2\left(a+1\right)^2\left(a^2-2a+2\right)\)

a) Ta có: \(x^2-2x-4y^2-4y\)

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

\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)

\(=\left(x+2y\right)\left(x-2y-2\right)\)

b) Ta có: \(x^3+2x^2+2x+1\)

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

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

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

d) Ta có: \(a^6-a^4+2a^3+2a^2\)

\(=a^2\left(a^4-a^2+2a+2\right)\)

\(=a^2\left[a^2\left(a^2-1\right)+\left(2a+2\right)\right]\)

\(=a^2\left[a^2\left(a-1\right)\left(a+1\right)+2\left(a+1\right)\right]\)

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

c) Ta có: \(x^3-4x^2+12x-27\)

\(=\left(x^3-27\right)-\left(4x^2-12x\right)\)

\(=\left(x-3\right)\left(x^2+3x+9\right)-4x\left(x-3\right)\)

\(=\left(x-3\right)\left(x^2-x+9\right)\)