Question

[Nguyễn Ngọc Nhã Hân]

Giải các pt sau:

a)\(\left|3x+1\right|=\left|x+1\right|\)

b)\(\left|x^2-3\right|=\left|x-\sqrt{3}\right|\)

c)\(\sqrt{9x^2-12x+4}=\sqrt{x^2}\)

d)\(\sqrt{x^2+4x+4}=\sqrt{4x^2-12x+9}\)

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

f)\(\sqrt{x^2-8x+16}+\left|x+2\right|=0\)

g) \(\sqrt{1-x^2}+\sqrt{x+1}=0\)

h) \(\sqrt{x^2-4}+\sqrt{x^2+4x+4}=0\)

Mọi người giúp em gấp với!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 1

[Nhã Doanh]

a) \(\left|3x+1\right|=\left|x+1\right|\)

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

c) \(\sqrt{9x^2-12x+4}=\sqrt{x^2}\)

\(\Leftrightarrow\sqrt{\left(3x-2\right)^2}=\sqrt{x^2}\)

\(\Leftrightarrow\left|3x-2\right|=\left|x\right|\)

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

d) \(\sqrt{x^2+4x+4}=\sqrt{4x^2-12x+9}\)

\(\Leftrightarrow\sqrt{\left(x+2\right)^2}=\sqrt{\left(2x-3\right)^2}\)

\(\Leftrightarrow\left|x+2\right|=\left|2x-3\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=2x-3\\x+2=-2x+3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{1}{3}\end{matrix}\right.\)

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

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

\(\Leftrightarrow x=-1\)

f) \(\sqrt{x^2-8x+16}+\left|x+2\right|=0\)

\(\Leftrightarrow\sqrt{\left(x-4\right)^2}+\left|x+2\right|=0\)

\(\Leftrightarrow\left|x-4\right|+\left|x+2\right|=0\)

⇒ vô nghiệm