Question

ai giúp mình bài 4 với ạ

 

Answer 1

Bài 4: 

a: \(\Leftrightarrow\left\{{}\begin{matrix}x>=2\\x^2+4=x^2-4x+4\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

b: \(\Leftrightarrow\left|x-5\right|=3-19x\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{19}\\\left(3-19x-x+5\right)\left(3-19x+x-5\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{19}\\\left(-20x+8\right)\left(-18x-2\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{1}{9}\right\}\)

c: \(\Leftrightarrow\sqrt{x^2-9}+\left|x-3\right|=0\)

=>x-3=0

hay x=3

Answer 2

\(d;pt\Leftrightarrow\sqrt{2x-3+2\sqrt{2x-3}+1}+\sqrt{2x-3+2.4\sqrt{2x-3}+16}=5\Leftrightarrow\sqrt{\left(\sqrt{2x-3}+1\right)^2}+\sqrt{\left(\sqrt{2x-3}+4\right)^2}=5\Leftrightarrow2\sqrt{2x-3}+1+4=5\Leftrightarrow\sqrt{2x-3}=0\Leftrightarrow x=\dfrac{3}{2}\)

\(\)