Question

[Haruno Sakura]

Chứng tỏ cấc phương trình sau vô nghiệm

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

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

Answer 1

[TNA Atula]

a) (x-1)²≥0 , x²≥0 => (x-1)²+x²≥0

Dau bang xay ra khi : \(\left\{{}\begin{matrix}x-1=0\\x=0\end{matrix}\right.\)

⇔ \(\left\{{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)

=> pt vo nghiem

b) (3x+1)²≥0 , (2x-1)²≥0 => (3x+1)²+(2x-1)²≥0

Dau bang xay ra khi : \(\left\{{}\begin{matrix}3x+1=0\\2x-1=0\end{matrix}\right.\)

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

=> pt vo nghiem

Answer 2

[Trần Quốc Lộc]

\(\text{a) }\left(x-1\right)^2+x^2=0\\ \Leftrightarrow x^2-2x+1+x^2=0\\ \Leftrightarrow2x^2-2x+\dfrac{1}{2}+\dfrac{1}{2}=0\\ \Leftrightarrow\left(2x^2-2x+\dfrac{1}{2}\right)+\dfrac{1}{2}=0\\ \Leftrightarrow2\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{2}=0\\ \Leftrightarrow2\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}=0\)

Do \(2\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow2\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}\ge\dfrac{1}{2}>0\forall x\left(\text{Trái với phương trình đã cho}\right)\)
Vậy phương trình vô nghiệm.

\(\text{ b) }\left(3x+1\right)^2+\left(2x-1\right)^2=0\\ \Leftrightarrow9x^2+6x+1+4x^2-4x+1=0\\ \Leftrightarrow13x^2+2x+2=0\\ \Leftrightarrow13x^2+2x+\dfrac{1}{13}+\dfrac{25}{13}=0\\ \Leftrightarrow\left(13x^2+2x+\dfrac{1}{13}\right)+\dfrac{25}{13}=0\\ \Leftrightarrow13\left(x^2+\dfrac{2}{13}x+\dfrac{1}{169}\right)+\dfrac{25}{13}=0\\ \Leftrightarrow13\left(x+\dfrac{1}{13}\right)^2+\dfrac{25}{13}=0\)

Do \(13\left(x+\dfrac{1}{13}\right)^2\ge0\forall x\)

\(\Rightarrow13\left(x+\dfrac{1}{13}\right)^2+\dfrac{25}{13}\ge\dfrac{25}{13}>0\forall x\left(\text{Trái với phương trình đã cho}\right)\)

Vậy phương trình vô nghiệm.