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\(a,=\dfrac{x^3-\left(x-1\right)\left(x^2+x+1\right)}{1-x}=\dfrac{x^3-x^3+1}{1-x}=\dfrac{1}{1-x}\\ b,=\dfrac{2x+x^2+3x+2+2-x}{\left(x+2\right)^2}=\dfrac{\left(x+2\right)^2}{\left(x+2\right)^2}=1\)
ta có:\(x^3+x^2+2x^2+2x+2x+2=0\)0
\(\Leftrightarrow x^2\left(x+1\right)+2x\left(x+1\right)+2\left(x+1\right)=0\)
\(\Leftrightarrow\left(x^2+2x+2\right)\left(x+1\right)=0\)
Do \(x^2+2x+2\ne0\)
\(\Rightarrow x+1=0\)
\(\Rightarrow x=-1\)
vậy phương trình trên có tập nghiệm là :S=(-1)
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bạn viết rõ đề ra nhé
b, \(\left|4x-8\right|=1-x\)ĐK : \(x\le1\)
TH1 : \(4x-8=1-x\Leftrightarrow5x=9\Leftrightarrow x=\dfrac{9}{5}\)( ktm )
TH2 : \(4x-8=x-1\Leftrightarrow3x=7\Leftrightarrow x=\dfrac{7}{3}\)( ktm )
b) Ta có: \(\left|4x-8\right|=1-x\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-8=1-x\left(x\ge2\right)\\4x-8=x-1\left(x< 2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x+x=1+8\\4x-x=-1+8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=9\\3x=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{5}\left(loại\right)\\x=\dfrac{7}{3}\left(loại\right)\end{matrix}\right.\)
(3x-4-x-1)(3x-4+x+1)=0
(2x-5)(4x-3)=0
2x-5 = 0 hoặc 4x-3=0
2x=5 hoặc 4x=3
x=5/2 hoặc x=3/4
`(x+3)(x^2-5x+8)=(x+3).x^2`
`<=>(x+3)(x^2-5x+8-x^2)=0`
`<=>(x+3)(8-5x)=0`
`<=>` \(\left[ \begin{array}{l}x+3=0\\8-5x=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac85\\x=-3\end{array} \right.\)
Vậy `S={-3,8/5}`
`(x+3)(x^2-5x+8)=(x+3).x^2`
`<=>(x+3)(x^2-5x+8-x^2)=0`
`<=>(x+3)(-5x+8)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\-5x+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{8}{5}\end{matrix}\right.\)
Vậy `S={-3;8/5}`.
= x^2(X-1) - 4(x^2-2x+1)
=x^2(x-1)-4(x-1)^2
=(x-1)(x^2-4x+4)
=(x-1)(x-2)^2
a) x3-x2-21x+45=0
<=> x3+5x2-6x2-30x+9x+45=0
<=> (x+5)(x2-6x+9)=0
<=> (x+5)(x2-3x-3x+9)=0
<=> (x+5)(x-3)2=0
Vậy S={-5;3}
b) X3+3X2+4X+2=0
<=> X3+X2+2X2+2X+2X+2=0
<=> (X+1)(X2+2X+2)=0
VÌ X2+2X+2 >=0
NÊN S={-1}
C) X4+7X-8=0
<=> X4-X3+X3-X2+X2-X+8X-8=0
<=> (X-1)(X3+X2+X+8)=0
VÌ X3+X2+X+8>=0
NÊN S={1}
D) 6X4-X3-7X2+X+1=0
<=> 6X4-6X3+5X3-5X2-2X2+2X-X+1=0
<=> (X-1)(6X3+5X2-2X-1)=0
<=> (X-1)(6X3-3X2+8X2-4X+2X-1)=0
<=> (X-1)(2X-1)(3X2_4X+1)=0
<=> (X-1)(2X-1)(3X2-3x-x+1)=0
<=> (X-1)2(2X-1)(3x-1)=0
vậy S={1/3;1/2;1}
