`110-(10x+20)=50`

`=> 10x+20=110-50`

`=> 10x+20=60`

`=> 10x=60-20`

`=>10x=40`

`=>x=40:10`

`=>x=4`

Vậy `x=4`

`@ gvy`

k quá thì còn lm đc;-;

kiến thức vừa vừa thôi với những người kém toán như tui;-;

Bài `13`

\(a,\sqrt{27}+\sqrt{48}-\sqrt{108}-\sqrt{12}\\ =\sqrt{9\cdot3}+\sqrt{16\cdot3}-\sqrt{36\cdot3}-\sqrt{4\cdot3}\\ =3\sqrt{3}+4\sqrt{3}-6\sqrt{3}-2\sqrt{3}\\ =\left(3+4-6-2\right)\sqrt{3}\\ =-\sqrt{3}\\ b,\left(\sqrt{28}+\sqrt{12}-\sqrt{7}\right)\cdot\sqrt{7}+\sqrt{84}\\ =\left(\sqrt{4\cdot7}+\sqrt{4\cdot3}-\sqrt{7}\right)\cdot\sqrt{7}+\sqrt{4\cdot21}\\ =\left(2\sqrt{7}+2\sqrt{3}-\sqrt{7}\right)\cdot\sqrt{7}+2\sqrt{21}\\ =2\cdot7+2\sqrt{21}-7+2\sqrt{21}\\ =14+2\sqrt{21}-7+2\sqrt{21}\\ =7+4\sqrt{21}\)

số 2 gần số 3 quá viết nhanh nên lú @@

\(\sqrt{x^2-6x+9}-4=3x\left(đkxđ:x\ge-\dfrac{4}{3}\right)\\ \Leftrightarrow\sqrt{\left(x-3\right)^2}=3x+4\\ \Leftrightarrow\left|x-3\right|=3x+4\\ \Leftrightarrow\left[{}\begin{matrix}x-3=3x+4\\x-3=-3x-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x-3x=4+3\\x+3x=-4+3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-2x=7\\4x=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\left(ktm\right)\\x=-\dfrac{1}{4}\left(tm\right)\end{matrix}\right.\)