Question

tìm x biết

(x-1)²-1+x²-(1-x)(x+3)=0

Answer 1

(x-1)²-1+x²-(1-x)(x+3)=0

⇒x²-2x+1-1+x²-x(1-x)+3(1-x)=0

⇒x²-2x+1-1+x²-x+x²+3-3x=0

⇒3x²-6x+3=0

⇒3(x²-2x+1)=0

⇒x²-2x+1=0

⇒(x-1)²=0

⇒x-1=0

⇒x=1

Answer 2

Lời giải:

$(x-1)^2-1+x^2-(1-x)(x+3)=0$

$\Leftrightarrow (x^2-2x+1)-1+x^2-(3-x^2-2x)=0$

$\Leftrightarrow x^2-2x+1-1+x^2-3+x^2+2x=0$

$\Leftrightarrow 3x^2-3=0$
$\Leftrightarrow x^2-1=0$

$\Leftrightarrow (x-1)(x+1)=0$

$\Leftrightarrow x=1$ hoặc $x=-1$

Answer 3

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

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

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

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

\(\Leftrightarrow3\left(x^2-1\right)=0\) \(\Leftrightarrow x^2-1=0\) \(\Leftrightarrow x^2=1\) \(\Leftrightarrow x=\pm1\)