\(\sqrt{x}-2=5\\ \Leftrightarrow\sqrt{x}=7\\ \Leftrightarrow x=49\\ \sqrt{\left(x-3\right)^2}=9\\ \Leftrightarrow\left|x-3\right|=9\\ \Leftrightarrow\left[{}\begin{matrix}x-3=9\\x-3=-9\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=12\\x=-6\end{matrix}\right.\\ \sqrt{x^2+1}=2\\ \Leftrightarrow\sqrt{\left(x^2+1\right)^2}=2^2\\ \Leftrightarrow x^2+1=4\\ \Leftrightarrow x^2=3\\ \Leftrightarrow x=\sqrt{3}\\ \sqrt{2x+1}=3\\ \Leftrightarrow\sqrt{\left(2x+1\right)^2}=3^2\\ \Leftrightarrow2x+1=9\\ \Leftrightarrow2x=8\\ \Leftrightarrow x=4\\ \sqrt{x^2+6x+9}=3\\ \Leftrightarrow\sqrt{\left(x+3\right)^2}=3\\ \Leftrightarrow\left|x+3\right|=3\\ \Leftrightarrow\left[{}\begin{matrix}x+3=3\\x+3=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)

\(\dfrac{\sqrt{x-3}}{\sqrt{x}+3}\cdot1=4\) đề là vậy ạ?

\(\sqrt{0,36}-\left|-0,75\right|:\left(-1\dfrac{3}{4}\right)\\ =0,6-0,75:\dfrac{-7}{4}\\ =\dfrac{3}{5}-\dfrac{3}{4}\cdot\dfrac{-4}{7}\\ =\dfrac{3}{5}-\dfrac{-12}{28}\\ =\dfrac{3}{5}+\dfrac{3}{7}\\ =\dfrac{36}{35}\)

\(1-\left(\dfrac{5}{9}-\dfrac{2}{3}\right)^2:\dfrac{4}{27}\\ =1-\left(\dfrac{5}{9}-\dfrac{6}{9}\right)^2\cdot\dfrac{27}{4}\\ =1-\left(-\dfrac{1}{9}\right)^2\cdot\dfrac{27}{4}\\ =1-\dfrac{1}{81}\cdot\dfrac{27}{4}\\ =1-\dfrac{1}{12}\\ =\dfrac{11}{12}\)

`(x^2-4)(2x+x+3)=0`

`=>(x-2)(x+2)(3x+3)=0`

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

Vậy \(x\in\left\{-1;-2;2\right\}\)

Bài `2`

\(A=2^4\cdot5-\left[131-\left(13-4\right)^2\right]\\ =16\cdot5-\left(131-9^2\right)\\ =16\cdot5-\left(131-81\right)\\ =16\cdot5-50\\ =80-50=30\\ B=2^3+3\left(\dfrac{1}{2}\right)^0-1+\left[\left(-2\right)^2:\dfrac{1}{2}\right]-8\\ =8+3\cdot1-1+\left(4\cdot2\right)-8\\ =8+3-1+8-8\\ =10\)

(x+2)-2 chứ không phải -4 nha.

Ta có : `(x+2)^2 -4(x+2)+4`

`=(x+2)^2 - 2*(x+2)*2 +4^2` 

`= [(x+2) -4)]^2`

`= (x+2-2)^2`

`= x^2`

`->D`