Question

[Nguyễn Hân]

mong mọi người giúp đỡ

Answer 1

[Trúc Giang]

a)

\(\left(x+1\right)\left(x-3\right)\left(x^2-2x\right)=-2\)

<=> (x + 1).(x - 3).x.(x - 2) = -2

<=> [ (x + 1). (x - 3) ]. [ x. (x - 2) ] = -2

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

Đặt \(x^2-2x=a\)

PT (1) <=> (a - 3).a + 2 = 0

\(\Leftrightarrow a^2-3a+2=0\)

\(\Leftrightarrow a^2-a-2a+2=0\)

<=> a. (a - 1) - 2. (a - 1) = 0

<=> (a - 1). (a - 2) = 0

<=> a - 1 = 0 hoặc a - 2 = 0

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1-\sqrt{2}\right).\left(x-1+\sqrt{2}\right)=0\\\left(x-1-\sqrt{3}\right).\left(x-1+\sqrt{3}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1+\sqrt{2}\\x=1-\sqrt{2}\\x=1+\sqrt{3}\\x=1-\sqrt{3}\end{matrix}\right.\)

Answer 2

[Lê Thị Thục Hiền]

b) \(\left\{{}\begin{matrix}x^2+x-y^2-y=0\left(1\right)\\x^2+y^2-2\left(x+y\right)=0\left(2\right)\end{matrix}\right.\)

PT (1)\(\Leftrightarrow\left(x-y\right)\left(x+y\right)+\left(x-y\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x+y=-1\end{matrix}\right.\)

TH1: x=y thay vào Pt (2) ta được: \(2x^2-4x=0\Leftrightarrow\left[{}\begin{matrix}x=0\Rightarrow y=0\\x=2\Rightarrow y=2\end{matrix}\right.\)

TH2: Thay x+y=-1 vào Pt (2) ta được: \(x^2+y^2+2=0\left(vn\right)\)

Vậy hẹ pt có nghiệm (x;y)=(0;0) ; (2;2)