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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
\(\left(\frac{x+1}{x-2}\right)^2+\frac{x+1}{x-4}-3\left(\frac{2x-4}{x-4}\right)^2=0\)
<=> \(\left(x+1\right)^2.\left(x-2\right)^2.\left(x-4\right)^2+\frac{x+1}{x-4}.\left(x-2\right)^2.\left(x-4\right)^2-\frac{3\left(2x-4\right)^2}{\left(x-4\right)^2}.\left(x-2\right)^2.\left(x-4\right)^2\)\(=0.\left(x-2\right)^2.\left(x-4\right)^2\)
<=> \(\left(x+1\right)^2.\left(x-4\right)^2+\left(x+1\right).\left(x-2\right)^2.\left(x-4\right)^2-3\left(2x-4\right)^2.\left(x-2\right)^2=0\)
<=> \(-\left(x-3\right)\left(5x-4\right)\left(2x^2-9x+16\right)=0\)
<=> \(\orbr{\begin{cases}x-3=0\\5x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{4}{5}\end{cases}}\)
Mà vì: \(2x^2-9x+16\ne0\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{4}{5}\end{cases}}\)
\(\left(x^2+x-2\right)^2=3\left(x^4+x^2+1\right)\)
\(\Leftrightarrow\left[\left(x-1\right)\left(x+2\right)\right]^2=3\left(x^4+x^2+1\right)\)
\(\Leftrightarrow\left(x-1\right)^2\left(x+2\right)^2=3\left(x^4+x^2+1\right)\)
\(\Leftrightarrow x^4+4x^3+4x^2-2x^3-8x^2-8x+x^2+4x+4=3x^4+3x^2+3\)
\(\Leftrightarrow x^4+2x^3-3x^2-4x+4-3x^4-3x^2-3=0\)
\(\Leftrightarrow-2x^4+2x^3-6x^2-4x+1=0\)
\(\left(x+1\right)\left(x+4\right)=\left(2-x\right)\left(x+2\right)\)
\(\Leftrightarrow x^2+5x+4=4-x^2\)
\(\Leftrightarrow2x^2+5x=0\)
\(\Leftrightarrow x\left(2x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{5}{2}\end{cases}}}\)
\(x\left(x+2\right)=x\left(x+3\right)\)
\(\Leftrightarrow x\left(x+2\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow-x=0\)
\(\Leftrightarrow x=0\)
Ta có :
\(\left(x+1\right)\left(x+4\right)=\left(2-x\right)\left(x+2\right)\)
\(\Leftrightarrow\)\(x\left(x+1\right)+4\left(x+1\right)=x\left(2-x\right)+2\left(2-x\right)\)
\(\Leftrightarrow\)\(x^2+x+4x+4=2x-x^2+4-2x\)
\(\Leftrightarrow\)\(2x^2+5x=4-4\)
\(\Leftrightarrow\)\(2x^2+5x=0\)
Suy ra \(2x^2=-\left(5x\right)\) hoặc \(2x^2=0\)và \(5x=0\)
\(+)\)Nếu \(2x^2=-\left(5x\right)\)ta có :
\(-\left(5x\right)+5x=0\)
\(\Leftrightarrow\)\(5x\left(-1+1\right)=0\)
\(\Leftrightarrow\)\(5x=0\)
\(\Leftrightarrow\)\(x=0\)
\(+)\)Nếu \(2x^2=0\) và \(5x=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x^2=0\\x=0\end{cases}=\Leftrightarrow\orbr{\begin{cases}x=0\\x=0\end{cases}}}\)
Vậy \(x=0\)
\(\left(x-2\right)\left(x+4\right)-\left(x+2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[\left(x^2+4x\right)-\left(2x+8\right)\right]-\left(x^2+3x+2x+6\right)=0\)
\(\Leftrightarrow x^2+4x-2x-8-x^2-3x-2x-6=0\)
\(\Leftrightarrow\left(x^2-x^2\right)+\left[4x-2x-2x-3x\right]+\left(-8-6\right)=0\)
\(\Leftrightarrow0+0+\left(-3x\right)+\left(-14\right)=0\)
\(\Leftrightarrow-3x-14=0\)
\(\Leftrightarrow-3x=0+14\)
\(\Leftrightarrow-3x=14\)
\(\Leftrightarrow x=14\div\left(-3\right)\)
\(\Leftrightarrow x=-\frac{14}{3}\)
Vậy S = { \(-\frac{14}{3}\) }
\(x^2-4=2\left(x-2\right)\left(x+3\right)\)
\(\Leftrightarrow x^2-4=2\left(x^2+3x-2x-6\right)\)
\(\Leftrightarrow x^2-4=2x^2+2x-12\)
\(\Leftrightarrow x^2-2x^2-2x=-12+4\)
\(\Leftrightarrow-x^2-2x=-8\)
\(\Leftrightarrow-x^2-2x+8=0\)
\(\Leftrightarrow-x^2+2x-4x+8=0\)
\(\Leftrightarrow-x\left(x-2\right)-4\left(x-2\right)=0\)
\(\Leftrightarrow\left(-x-4\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x-4=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=2\end{matrix}\right.\)
Vậy \(S=\left\{-4;2\right\}\)
\(x^2-4=2\left(x-2\right)\left(x+3\right)\)
\(\Leftrightarrow\left(x+2\right)\left(x-2\right)=2\left(x-2\right)\left(x+3\right)\)
\(\Leftrightarrow\left(x+2\right)\left(x-2\right)-2\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[\left(x+2\right)-2\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2-2x-6\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(-x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)