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

a) Thay m=2 vào hệ phương trình, ta được: 

\(\left\{{}\begin{matrix}x-2y=5\\2x-y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-4y=10\\2x-y=7\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-3y=3\\x-2y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=5+2y=5+2\cdot\left(-1\right)=3\end{matrix}\right.\)

Vậy: Khi m=2 thì hệ phương trình có nghiệm duy nhất là (x,y)=(3;-1)

ĐKXĐ: \(x\notin\left\{-3;1\right\}\)

Ta có: \(\frac{4}{x^2+2x-3}=\frac{2x-5}{x+3}-\frac{2x}{x-1}\)

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

Suy ra: \(\left(2x-5\right)\left(x-1\right)-2x\left(x+3\right)=4\)

\(\Leftrightarrow2x^2-2x-5x+5-2x^2-6x=4\)

\(\Leftrightarrow-13x+5=4\)

\(\Leftrightarrow-13x=4-5=-1\)

hay \(x=\frac{1}{13}\)(nhận)

Vậy: \(S=\left\{\frac{1}{13}\right\}\)

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

<=>\(12\left(x^2+2x+5\right)-5\left(x^2+2x+4\right)=2\left(x^2+2x+5\right)\left(x^2+2x+4\right)\)

\(\Leftrightarrow12x^2+24x+60-5x^2-10x-20=2x^4+8x^3+26x^2+36x+40\)

\(\Leftrightarrow7x^2+14x+40=2x^4+8x^3+26x^2+36x+40\)

\(\Leftrightarrow2x^4+8x^3+19x^2+22x=0\)

\(\Leftrightarrow x\left(2x^3+8x^2+19x+22\right)=0\)

\(\Leftrightarrow x\left(2x^3+4x^2+4x^2+8x+11x+22\right)=0\)

\(\Leftrightarrow x\left[2x^2\left(x+2\right)+4x\left(x+2\right)+11\left(x+2\right)\right]=0\)

\(\Leftrightarrow x\left(x+2\right)\left(2x^2+4x+11\right)=0\)

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

Vậy PT có nghiệm duy nhất S ={0 ; -2 }  vì(  \(2x^2+4x+11\ne0\))

Bài làm

( 2x + 5 )( x - 4 ) = ( x - 5 )( 4 - x )

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

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

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

<=> ( x - 4 ) . 3x = 0

<=> \(\orbr{\begin{cases}x-4=0\\3x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=0\end{cases}}}\)

Vậy tập nghiệm phương trình S = { 4; 0 }

# Học tốt #

( 2x + 5 )( x - 4 ) = ( x - 5 )( 4 - x )
=> ( 2x + 5 )( x - 4 ) - ( x - 5 )( 4 - x ) = 0
=> ( 2x + 5 )( x - 4 ) + ( x - 5 )( x - 4 ) = 0
=> ( x - 4 )( 2x + 5 + x - 5 ) = 0
=> ( x - 4 ) . 3x = 0
=> \(\orbr{\begin{cases}x-4=0\\3x=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=4\\x=0\end{cases}}\)

Đặt 3-x = a ; 2-x = b

=> 5-2x = a+b

pt <=> a^4+b^4 = (a+b)^4 = a^4+4a^3b+6a^2b^2+4ab^3+b^4

<=> a^4+4a^3b+6a^2b^2+4ab^3+b^4-a^4-b^4 = 0

<=> 4a^3b+6a^2b^2+4ab^3 = 0

<=> 2a^3b+3a^2b^2+2ab^3 = 0

<=> ab.(2a^2+3ab+2b^2) = 0

<=> ab=0 ( vì 2a^2+3ab+2b^2 > 0 )

<=> a=0 hoặc b=0

<=> 3-x=0 hoặc 2-x=0

<=> x=3 hoặc x=2

Vậy .............

Tk mk nha

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}