ta có : x^5+2x^4+3x^3+3x^2+2x+1=0

\(\Leftrightarrow\)x^5+x^4+x^4+x^3+2x^3+2x^2+x^2+x+x+1=0

\(\Leftrightarrow\)(x^5+x^4)+(x^4+x^3)+(2x^3+2x^2)+(x^2+x)+(x+1)=0

\(\Leftrightarrow\)x^4(x+1)+x^3(x+1)+2x^2(x+1)+x(x+1)+(x+1)=0

\(\Leftrightarrow\)(x+1)(x^4+x^3+2x^2+x+1)=0

\(\Leftrightarrow\)(x+1)(x^4+x^3+x^2+x^2+x+1)=0

\(\Leftrightarrow\)(x+1)[x^2(x^2+x+1)+(x^2+x+1)]=0

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

VÌ x^2+x+1=(x+\(\dfrac{1}{2}\))^2+\(\dfrac{3}{4}\)\(\ne0\) và x^2+1\(\ne0\)

\(\Rightarrow\)x+1=0

\(\Rightarrow\)x=-1

CÒN CÂU B TỰ LÀM (02042006)

b: x^4+3x^3-2x^2+x-3=0

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

=>(x-1)(x^3+4x^2+2x+3)=0

=>x-1=0

=>x=1

Sau gõ latex.

\(\sqrt{x-5}-\left(x-\dfrac{14}{3}+\sqrt{x-5}\right)=3\\ \Leftrightarrow\sqrt{x-5}-x+\dfrac{14}{3}-\sqrt{x-5}=3\\ \Leftrightarrow-x=3-\dfrac{14}{3}=-\dfrac{5}{3}\\ \Rightarrow x=-\dfrac{5}{3}:\left(-1\right)=\dfrac{5}{3}\)

a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Ta có: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(x^2+3x+2-5x+10=12+x^2-4\)

\(\Leftrightarrow x^2-2x+12-8-x^2=0\)

\(\Leftrightarrow-2x+4=0\)

\(\Leftrightarrow-2x=-4\)

hay x=2(loại)

Vậy: \(S=\varnothing\)

b) Ta có: \(\left|2x+6\right|-x=3\)

\(\Leftrightarrow\left|2x+6\right|=x+3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+6=x+3\left(x\ge-3\right)\\-2x-6=x+3\left(x< -3\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3-6\\-2x-x=3+6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)

Vậy: S={-3}

(x-5)(x-1)+(x-3)(x-5)=0

<=> (x-5)(x-1+x-3)=0

<=> (x-5)(2x-4)=0

\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\2x-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\2x=4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=5\\x=2\end{cases}}}\)

Vậy x={5;2}

(x-5)(x-1)+(x-3)(x-5)=0

=>(x-5)(x-1+x-3)=0

=>(x-5)(2x-4)=0

\(\Rightarrow\hept{\begin{cases}x-5=0\\2x-4=0\end{cases}\Rightarrow\hept{\begin{cases}x=5\\2x=4\end{cases}\Rightarrow}}\).....

\(\left(x-5\right)\left(x-1\right)+\left(x-3\right)\left(x-5\right)=0\)

\(< =>\left(x-5\right).\left(x-1+x-3\right)=0\)

\(< =>\left(x-5\right).\left(2x-4\right)=0\)

\(< =>\orbr{\begin{cases}x-5=0\\2x-4=0\end{cases}}\)

\(< =>\orbr{\begin{cases}x=5\\2x=4\end{cases}< =>\orbr{\begin{cases}x=5\\x=2\end{cases}}}\)

\(\Leftrightarrow\dfrac{12\left(x-3\right)-2\left(x-3\right)\left(2x-5\right)-3\left(x-3\right)\left(3-x\right)}{12}=0\)

\(\Leftrightarrow12x-36-2\left(2x^2-5x-6x+15\right)-3\left(3x-x^2-9+3x\right)=0\)

\(\Leftrightarrow12x-36-4x^2+22x-30-18x+3x^2+27=0\)

\(\Leftrightarrow-x^2+16x-39=0\)

\(\Delta=b^2-4ac=16^2-4.\left(-1\right).\left(-39\right)=100>0\)

\(\Rightarrow PT\) có 2 nghiệm pb \(x_1,x_2\)

\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-16+10}{-2}=3\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-16-10}{-2}=13\end{matrix}\right.\)

Vậy \(S=\left\{3;13\right\}\)

ĐKXĐ: x+3>=0

=>x>=-3

\(x+\left(x+1\right)\sqrt{x+3}=5\)

=>\(x+\sqrt{\left(x+3\right)\left(x+1\right)^2}=5\)

=>\(x+\sqrt{\left(x+3\right)\left(x^2+2x+1\right)}=5\)

=>\(x+\sqrt{x^3+2x^2+x+3x^2+6x+3}=5\)

=>\(x+\sqrt{x^3+5x^2+7x+3}=5\)

=>\(x-1+\sqrt{x^3+5x^2+7x+3}-4=0\)

=>\(\left(x-1\right)+\dfrac{x^3+5x^2+7x+3-16}{\sqrt{x^3+5x^2+7x+3}+4}=0\)

=>\(\left(x-1\right)+\dfrac{x^3-x^2+6x^2-6x+13x-13}{\sqrt{x^3+5x^2+7x+3}+4}=0\)

=>\(\left(x-1\right)+\dfrac{\left(x-1\right)\left(x^2+6x+13\right)}{\sqrt{x^3+5x^2+7x+3}+4}=0\)

=>\(\left(x-1\right)\left(1+\dfrac{x^2+6x+13}{\sqrt{x^3+5x^2+7x+3}+4}\right)=0\)

=>x-1=0

=>x=1(nhận)

Câu 1: 

a: x+2=0

nên x=-2

b: (x-3)(2x+8)=0

=>x-3=0 hoặc 2x+8=0

=>x=3 hoặc x=-4

a . 

x + 2 = 0

=> x = 0 - 2 = -2 

b ) .

<=> x - 3 = 0 ; 2x + 8 = 0

= > x = 3 ; x = -8/2 = -4 

c ) .

ĐKXĐ của pt : x - 5 khác 0 = > ddk : x khác 5

1)

a) \(x+2=0\)

\(\Leftrightarrow x=-2\)

Vậy S = {\(-2\)}

b) \(\left(x-3\right)\left(2x+8\right)=0\)

\(\Leftrightarrow x-3=0\) hoặc \(2x+8=0\)

*) \(x-3=0\)

\(\Leftrightarrow x=3\)

*) \(2x+8=0\)

\(\Leftrightarrow2x=-8\)

\(\Leftrightarrow x=-4\)

Vậy S = \(\left\{-4;3\right\}\)

2) ĐKXĐ:

\(x-5\ne0\Leftrightarrow x\ne5\)

Lời giải:

$(x+5)(x-3)=(x-4)(3+x)$

$\Leftrightarrow x^2+2x-15=x^2-x-12$

$\Leftrightarrow 3x=3\Rightarrow x=1$