Question

 (x-1)² +2(x-1)(x+2) + (x+2)²= 0         

Answer 1

⇔x²-2x+1+2x²+4x-2x-4=0

⇔3x²-3=0

⇔3(x²-1)=0

⇔x²-1=0

⇔x²=1

⇔x=1

Vậy S={1}

Answer 2

\(\left(x-1\right)^2+2\left(x-1\right)\left(x+2\right)+\left(x+2\right)^2=0\\ \Leftrightarrow\left(x-1+x+2\right)^2=0\\ \Leftrightarrow\left(2x+1\right)^2=0\\ \Leftrightarrow2x+1=0\\ \Leftrightarrow x=\dfrac{-1}{2}\)

Vậy \(x=\dfrac{-1}{2}\) là nghiệm của pt.

Answer 3

(x-1)² +2(x-1)(x+2) + (x+2)²=0

<=> (2x+1)²=0

<=> 2x+1=0

<=> 2x= -1

<=> \(x=-\dfrac{1}{2}\)

Vậy .....

Answer 4

 (x-1)² +2(x-1)(x+2) + (x+2)²= 0       

\(< =>\left[\left(x-1\right)^2+2\left(x-1\right)\left(x+2\right)\right]+\left(x+2\right)^2=0\)

\(< =>\left[x-1\left(x-1+2\left(x+2\right)\right)\right]+\left(x+2\right)^2=0\)

\(< =>\left[x-1\left(x-1+2x+4\right)\right]+\left(x+2\right)^2=0\)

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

\(< =>3x^2-3x+3x-3+x^2+4x+4=0\)

\(< =>4x^2+4x+1=0\)

\(< =>\left(2x+1\right)^2=0\)

\(=>2x+1=0\)

\(< =>2x=-1\)

\(< =>x=-\dfrac{1}{2}\)

Vậy tập nghiệm của phương trình là \(S=\left\{-\dfrac{1}{2}\right\}\)

Answer 5

(x-1)²+2(x-1)(x+2)+(x+2)²=0

⇔(x-1+x+2)²=0

⇔(2x+1)²=0

⇔x=-1/2

S={-1/2}

 

Answer 6

Ta có: \(\left(x-1\right)^2+2\left(x-1\right)\left(x+2\right)+\left(x+2\right)^2=0\)

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

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

\(\Leftrightarrow2x+1=0\)

\(\Leftrightarrow2x=-1\)

\(\Leftrightarrow x=-\dfrac{1}{2}\)Vậy: \(S=\left\{-\dfrac{1}{2}\right\}\)