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

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(  \(2x^2+4x+11\ne0\))

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

ĐKXĐ: \(x\notin\left\{0;5;-5\right\}\)

Ta có: \(\frac{x+25}{2x^2-50}-\frac{x+5}{x^2-5x}=\frac{5-x}{2x^2+10x}\)

\(\Leftrightarrow\frac{x\left(x+25\right)}{2x\left(x+5\right)\left(x-5\right)}-\frac{2\left(x+5\right)^2}{2x\left(x-5\right)\left(x+5\right)}+\frac{\left(x-5\right)^2}{2x\left(x+5\right)\left(x-5\right)}=0\)

Suy ra: \(x^2+25x-2\left(x^2+10x+25\right)+x^2-10x+25=0\)

\(\Leftrightarrow2x^2+15x+25-2x^2-20x-50=0\)

\(\Leftrightarrow-5x-25=0\)

\(\Leftrightarrow-5x=25\)

hay x=-5(loại)

Vậy: \(S=\varnothing\)

a, đk : x >= 1

\(\left[{}\begin{matrix}3x+5=2x-2\\3x+5=2-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=-\dfrac{3}{5}\end{matrix}\right.\left(ktm\right)\)

vậy pt vô nghiệm 

b, đk >= 0 

\(\left[{}\begin{matrix}x^2+1=2x\\x^2+1=-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(x+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)

c, \(\left[{}\begin{matrix}2x^2+2x=0\\2x^2+4x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x\left(x+1\right)=0\\x^2+2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0;x=-1\\x=-1\end{matrix}\right.\)

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}

a, \(5\left|2x-1\right|-3=7\Leftrightarrow5\left|2x-1\right|=10\Leftrightarrow\left|2x-1\right|=2\)

TH1 : \(2x-1=2\Leftrightarrow x=\frac{3}{2}\)

TH2 : \(2x-1=-2\Leftrightarrow x=-\frac{1}{2}\)

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

\(\Leftrightarrow\left(x-2\right)\left(2x+3-x-2\right)=0\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow x=-1;x=2\)

c, \(\frac{2x-3}{2}< \frac{1-3x}{-5}\Leftrightarrow\frac{2x-3}{2}+\frac{1-3x}{5}< 0\)

\(\Leftrightarrow\frac{10x-15+2-6x}{10}< 0\Rightarrow4x-13< 0\Leftrightarrow x< \frac{13}{4}\)

ĐKXĐ: \(x\notin\left\{0;\frac{3}{2}\right\}\)

Ta có: \(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)

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

\(\Leftrightarrow x-3-5\left(2x-3\right)=0\)

\(\Leftrightarrow x-3-10x+15=0\)

\(\Leftrightarrow-9x+12=0\)

\(\Leftrightarrow-9x=-12\)

hay \(x=\frac{4}{3}\)(tm)

Vậy: \(x=\frac{4}{3}\)

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

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

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

\(x=\frac{1}{2}\)

ĐKXĐ: \(x\ge\dfrac{5}{2}\)

\(\sqrt{2x-4+2\sqrt{2x-5}}+\sqrt{2x+4+6\sqrt{2x-5}}=14\)

\(\Leftrightarrow\sqrt{\left(\sqrt{2x-5}+1\right)^2}+\sqrt{\left(\sqrt{2x-5}+3\right)^2}=14\)

\(\Leftrightarrow\left|\sqrt{2x-5}+1\right|+\left|\sqrt{2x-3}+3\right|=14\)

\(\Leftrightarrow2\sqrt{2x-5}=10\)

\(\Leftrightarrow\sqrt{2x-5}=5\)

\(\Leftrightarrow2x-5=25\)

\(\Leftrightarrow x=15\)

\(\sqrt{5-x}=9-2x\left(đk:x\le5\right)\)

\(\Leftrightarrow5-x=\left(9-2x\right)^2\)

\(\Leftrightarrow5-x=81-36x+4x^2\)

\(\Leftrightarrow4x^2-35x+76=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x_1=\dfrac{19}{4}\\x_2=4\end{matrix}\right.\) (đoạn này e tự giải đc nha)