Câu hỏi của pham lan phuong

Question

Bài 1: Tìm x

f) x3 - 6x2 + 11x - 6 = 0

g) x + / 2x-1/ = 5

h) 2x3 + 3x2 - 32x= 48

Chú ý: /......./  là dấu giá trị tuyệt đối nhá.........

Hắc Hường · Accepted Answer

Giải:

a) $x^3-6x^2+11x-6=0$

$\Leftrightarrow x^3-3.x^2.2+3.x.2^2-x-2^3+2=0$

$\Leftrightarrow x^3-3.x^2.2+3.x.2^2-2^3+2-x=0$

$\Leftrightarrow\left(x-2\right)^3+2-x=0$

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

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

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

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

$\Leftrightarrow\left[{}\begin{matrix}x-2=0\x-3=0\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\x=3\x=1\end{matrix}\right.$

Vậy ...

b) $x+\left|2x-1\right|=5$

$\Leftrightarrow\left|2x-1\right|=5-x$

$\Leftrightarrow\left[{}\begin{matrix}2x-1=5-x\2x-1=x-5\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}3x=6\x=-4\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=2\x=-4\end{matrix}\right.$

Vậy ...Câu trả lời: Giải:

a) $x^3-6x^2+11x-6=0$

$\Leftrightarrow x^3-3.x^2.2+3.x.2^2-x-2^3+2=0$

$\Leftrightarrow x^3-3.x^2.2+3.x.2^2-2^3+2-x=0$

$\Leftrightarrow\left(x-2\right)^3+2-x=0$

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

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

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

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

$\Leftrightarrow\left[{}\begin{matrix}x-2=0\x-3=0\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\x=3\x=1\end{matrix}\right.$

Vậy ...

b) $x+\left|2x-1\right|=5$

$\Leftrightarrow\left|2x-1\right|=5-x$

$\Leftrightarrow\left[{}\begin{matrix}2x-1=5-x\2x-1=x-5\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}3x=6\x=-4\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=2\x=-4\end{matrix}\right.$

Vậy ...

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