a) \(\sqrt{x-5}=3\)

\(x-5=9\)

\(x=14\)

b) Vì \(\sqrt{x-10}\) ≥0

⇒không có x thỏa mãn

c) \(\sqrt{2x-1}=\sqrt{7}\)

\(2x-1=7\)

\(2x=8\)

\(x=4\)

Bài 3

a) \(\sqrt{x-5}=3\)

\(\Rightarrow x-5=9\)

\(\Rightarrow x=14\)

b) \(\sqrt{x-10}=-21\)

\(\Rightarrow x\in\varnothing\)

c) \(\sqrt{2x-1}=\sqrt{7}\)

\(\Rightarrow2x-1=7\)

\(\Rightarrow2x=8\)

\(\Rightarrow x=4\)

Bài `13`

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

giải hết giùm em luôn được không ạ, em cảm ơn.

\(\sqrt{\dfrac{x^2+2x+1}{16x^2}}=\sqrt{\dfrac{\left(x+1\right)^2}{16x^2}}=\dfrac{\left|x+1\right|}{4\left|x\right|}=\dfrac{1-x}{-4x}=\dfrac{x-1}{4x}\left(do.x\le-1\right)\)

\(\dfrac{x-2\sqrt{x}}{x-4}=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}}{\sqrt{x}+2}\)

b: OH=1,8cm

Câu 2: 

Ta có: \(\sqrt{x^2-4x+4}=x-1\)

\(\Leftrightarrow2-x=x-1\left(x< 2\right)\)

\(\Leftrightarrow-2x=-3\)

hay \(x=\dfrac{3}{2}\left(tm\right)\)