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Tô Thu Huyền

Tìm x, biết:

a.x3-x2=4x2-8x+4

b.(x-1)(x2+x+1)=7

c.2.(x+5)-x2-5x=0

d.x2-3x=-2

Đức Hiếu
9 tháng 9 2017 lúc 11:45

a, \(x^3-x^2=4x^2-8x+4\)

\(\Rightarrow x^3-x^2-4x^2+8x-4=0\)

\(\Rightarrow x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(x^2-2x-2x+4\right)=0\)

\(\Rightarrow\left(x-1\right)\left(x-2\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

b, \(\left(x-1\right)\left(x^2+x+1\right)=7\)

\(\Rightarrow x^3-1=7\Rightarrow x^3=2^3\Rightarrow x=2\)

c, \(2\left(x+5\right)-x^2-5x=0\)

\(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\)

\(\Rightarrow\left(x+5\right)\left(2-x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

d, \(x^2-3x=-2\)

\(\Rightarrow x^2-x-2x+2=0\)

\(\Rightarrow x\left(x-1\right)-2\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

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