\(\left(\sqrt{4,5}-\frac{1}{2}.\sqrt{72}+5\sqrt{\frac{1}{2}}\right).\left(42\sqrt{\frac{25}{6}}-10\sqrt{\frac{3}{2}}-12\sqrt{\frac{98}{3}}\right)\)

=\(\left(\frac{3\sqrt{2}}{2}-3\sqrt{2}+\frac{5\sqrt{2}}{2}\right).\left(35\sqrt{6}-5\sqrt{6}-28\sqrt{6}\right)\)

=\(\left(\frac{3\sqrt{2}-6\sqrt{2}+5\sqrt{2}}{2}\right).2\sqrt{6}\)

=\(2\sqrt{2}.\sqrt{6}=4\sqrt{3}\)

\(\left(20.\sqrt{0.03}+12.\sqrt{3}-\frac{1}{5}.\sqrt{75}\right).\sqrt{6}\)

\(=20.\sqrt{0,03.6}+12.\sqrt{3.6}-\frac{1}{5}.\sqrt{75.6}\)

\(=20.\sqrt{\frac{9}{50}}+12.\sqrt{3^2.2}-\frac{1}{5}.\sqrt{15^2.2}\)

\(=6\sqrt{2}+36\sqrt{2}-3\sqrt{2}\)

\(=39\sqrt{2}\)

\(=\sqrt{6}\left(2\sqrt{3}+12\sqrt{3}-\sqrt{3}\right)\)

\(=\sqrt{6}.13\sqrt{3}=13\sqrt{18}=39\sqrt{2}\)

Bài 2:

a)\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}}=-4\) (đk: \(x\ge2\))

\(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9\left(x-2\right)}+\dfrac{6}{\sqrt{81}}\sqrt{x-2}=-4\)

\(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-2\sqrt{x-2}+\dfrac{2}{3}\sqrt{x-2}=-4\)

\(\Leftrightarrow-\sqrt{x-2}=-4\) \(\Leftrightarrow x-2=16\)

\(\Leftrightarrow x=18\) (thỏa)

Vậy...

b)\(\sqrt{9x^2+12x+4}=4x\)(Đk:\(9x^2+12x+4\ge0\))

\(\Leftrightarrow\left\{{}\begin{matrix}4x\ge0\\9x^2+12x+4=16x^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\-7x^2+12x+4=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\-7x^2+14x-2x+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left(x-2\right)\left(-7x-2\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left[{}\begin{matrix}x=2\\x=-\dfrac{2}{7}\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow x=2\) (tm đk)

Vậy...

c) \(\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1}\) (đk: \(x\ge1\))

\(\Leftrightarrow x-2\sqrt{x-1}=x-1\)

\(\Leftrightarrow\sqrt{x-1}=\dfrac{1}{2}\) \(\Leftrightarrow x=\dfrac{5}{4}\) (tm)

Vậy...

Bài 1:

\(\sqrt{\left(4-\sqrt{5}\right)^2}+\sqrt{5+2\sqrt{5}+1}\)

\(=\left|4-\sqrt{5}\right|+\sqrt{\left(\sqrt{5}+1\right)^2}\)

\(=4-\sqrt{5}+\sqrt{5}+1=5\)

Bài 2:

a: ĐKXĐ: x>=3

\(\sqrt{x-3}=6\)

=>x-3=36

=>x=36+3=39(nhận)

b: ĐKXĐ: \(x\in R\)

\(\sqrt{\left(x-3\right)^2}=12\)

=>\(\left|x-3\right|=12\)

=>\(\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)

Bài 3:

a: \(P=\left(\dfrac{3-x\sqrt{x}}{3-\sqrt{x}}+\sqrt{x}\right)\cdot\left(\dfrac{3-\sqrt{x}}{3-x}\right)\)

\(=\dfrac{3-x\sqrt{x}+\sqrt{x}\left(3-\sqrt{x}\right)}{3-\sqrt{x}}\cdot\dfrac{3-\sqrt{x}}{3-x}\)

\(=\dfrac{3-x\sqrt{x}+3\sqrt{x}-x}{3-x}\)

\(=\dfrac{-\sqrt{x}\left(x-3\right)-\left(x-3\right)}{-\left(x-3\right)}=\dfrac{\left(x-3\right)\left(\sqrt{x}+1\right)}{x-3}=\sqrt{x}+1\)

b: \(P=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{x+\sqrt{x}}\right):\dfrac{x-\sqrt{x}+1}{x\sqrt{x}+1}\)

\(=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right):\dfrac{x-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)

c: \(A=\sqrt{3x-1}+3\cdot\sqrt{12x-4}-\sqrt{6^2\left(3x-1\right)}+\sqrt{5}\)

\(=\sqrt{3x-1}+6\sqrt{3x-1}-6\sqrt{3x-1}+\sqrt{5}\)

\(=\sqrt{3x-1}+\sqrt{5}\)

d: \(A=\left(\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\dfrac{a+2}{a-2}\)

\(=\left(\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\right)\cdot\dfrac{a-2}{a+2}\)

\(=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}\cdot\dfrac{a-2}{a+2}\)

\(=\dfrac{2\left(a-2\right)}{a+2}\)

\(B=\left(\dfrac{4}{1-\sqrt{5}}+\dfrac{1}{2+\sqrt{5}}-\dfrac{4}{3-\sqrt{5}}\right)\left(\sqrt{5}-6\right)\)

\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}+\dfrac{2-\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}-\dfrac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right]\left(\sqrt{5}-6\right)\)

\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{1-5}+\dfrac{2-\sqrt{5}}{4-5}-\dfrac{4\left(3+\sqrt{5}\right)}{9-5}\right]\left(\sqrt{5}-6\right)\)

\(B=\left[-\dfrac{4\left(1+\sqrt{5}\right)}{4}-\dfrac{2-\sqrt{5}}{1}-\dfrac{4\left(3+\sqrt{5}\right)}{4}\right]\left(\sqrt{5}-6\right)\)

\(B=\left(-1-\sqrt{5}-2+\sqrt{5}-3-\sqrt{5}\right)\left(\sqrt{5}-6\right)\)

\(B=\left(-\sqrt{5}-6\right)\left(\sqrt{5}-6\right)\)

\(B=-\left(\sqrt{5}+6\right)\left(\sqrt{5}-6\right)\)

\(B=-\left(5-36\right)\)

\(B=-\left(-31\right)\)

\(B=31\)

_____________________________

\(\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}\)

\(=4\sqrt{3}-\dfrac{\sqrt{3}\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{7}-\sqrt{5}}+\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

\(=4\sqrt{3}-\sqrt{3}-\dfrac{2\left(\sqrt{3}-1\right)}{2}\)

\(=3\sqrt{3}-\sqrt{3}+1\)

\(=2\sqrt{3}+1\)

\(\left(2\sqrt{2}-\sqrt{5}+3\sqrt{2}\right)\left(\sqrt{18}-\sqrt{20}+2\sqrt{2}\right)\)

\(=\left(2\sqrt{2}-\sqrt{5}+3\sqrt{2}\right)\left(3\sqrt{2}-2\sqrt{5}+2\sqrt{2}\right)\)

\(=\left(5\sqrt{2}-\sqrt{5}\right)\left(5\sqrt{2}-2\sqrt{5}\right)\)

\(=50-10\sqrt{10}-5\sqrt{10}+10\)

\(=60-15\sqrt{10}\)

\(\left(1+\sqrt{2}-\sqrt{5}\right)\left(1+\sqrt{2}+\sqrt{5}\right)\)

\(=\left(1+\sqrt{2}\right)^2-5\)

\(=1+2\sqrt{2}+2-5\)

\(2\sqrt{2}-2\)

\(a,=\sqrt{5}\left(2\sqrt{5}-3\right)+3\sqrt{5}=10-3\sqrt{5}+3\sqrt{5}=10\\ b,=5-\sqrt{3}-\left(2-\sqrt{3}\right)=3\\ c,=\dfrac{2\left(\sqrt{5}-1\right)}{4}-\dfrac{2\left(3+\sqrt{5}\right)}{4}=\dfrac{2\sqrt{5}-2-6-2\sqrt{5}}{4}=\dfrac{-8}{4}=-2\)

a: \(=12\sqrt{80}=48\sqrt{5}\)

b: \(=2\sqrt{5}\cdot2\sqrt{3}-10=4\sqrt{15}-10\)

c: =20-9=11