a)\(\sqrt{400.0,81}=\sqrt{4.81}=\sqrt{2^2.9^2}=2.9=18\)

b)\(\sqrt{\dfrac{5}{27}.\dfrac{3}{20}}=\sqrt{\dfrac{5}{3^3}.\dfrac{3}{2^2.5}}=\sqrt{\dfrac{1}{3^2.2^2}}=\dfrac{1}{3.2}=\dfrac{1}{6}\)

c)\(\sqrt{\left(-5\right)^2.3^2}=\sqrt{5^2.3^2}=5.3=15\)

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

\(1.\\ A=\sqrt{\left(2+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\\ =\left|2+\sqrt{3}\right|+\left|2-\sqrt{3}\right|\\ =2+\sqrt{3}+2-\sqrt{3}=4\)

\(2.\\a.\\ P=3x-\sqrt{\left(x-5\right)^2}=3x-\left|x-5\right|\\ b.\\ x=2\Rightarrow P=3\)

\(3.\\ M=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{\left|x-1\right|}{x-1}\)

\(\cdot x>1\Rightarrow M=1\\ \cdot x=1\Rightarrow M=0\\\cdot x< 1\Rightarrow M=-1\)

B1.

Ta có:A\(=\sqrt{3+4\sqrt{3}+4}+\sqrt{3-4\sqrt{3}+4}\)

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

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

Bài 1 : 

\(A=\sqrt{\left(\sqrt{3}+2\right)^2}+\sqrt{\left(\sqrt{3}-2\right)^2}\\ =\sqrt{3}+2+2-\sqrt{3}=4\)

Bài 2 : 

a) \(P=3x-\sqrt{\left(x-5\right)^2}=3x-\left|x-5\right|\)

b) khi x = 2 thì \(P=3.2-\left|2-5\right|=3\)

Bài 3 : 

\(M=\dfrac{\sqrt{\left(\sqrt{x}-1\right)^2}}{x-1}=\dfrac{\left|\sqrt{x}-1\right|}{x-1}\)

Bài 1:

a)\(Q=2x-\sqrt{x^2+2x+1}=2x-\sqrt{\left(x+1\right)^2}=2x-\left|x+1\right|\)

b)Tại x=7 thay vào Q ta được:

\(Q=2.7-\left|7+1\right|=14-8=6\)

Bài 2:

\(\sqrt{x^2-6x}+7x=13\)\(\Leftrightarrow\sqrt{x^2-6x}=13-7x\)

\(\Leftrightarrow\left\{{}\begin{matrix}13-7x\ge0\\x^2-6x=\left(13-7x\right)^2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{13}{7}\\0=48x^2-85x+169\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{13}{7}\\\Delta=\left(-85\right)^2-4.48.169=-25223< 0\end{matrix}\right.\)

\(\Rightarrow x\in\varnothing\)

Vậy pt vô nghiệm.

a,\(\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{5}\)

\(=|^{ }_{ }2-\sqrt{5}|^{ }_{ }-\sqrt{5}\)

\(=\sqrt{5}-2-\sqrt{5}\)(vì \(2-\sqrt{5}< 0\))

=-2

b,\(\sqrt{16}\cdot\sqrt{25}+\sqrt{256}\cdot\sqrt{64}\)

\(=4\cdot5-16\cdot8=20+128=148\)

c,\(\sqrt{\left(\sqrt{2}-3\right)^2}-\sqrt{\left(5-\sqrt{2}\right)^2}\)

\(=|^{ }_{ }\sqrt{2}-3|^{ }_{ }-|^{ }_{ }5-\sqrt{2}|^{ }_{ }\)

\(=3-\sqrt{2}-5+\sqrt{2}\)(vì \(\sqrt{2}-3< 0;5-\sqrt{2}>0\))

\(=-2\)

cảm ơn

à bạn ơi , mình không hiểu câu c gì à , bạn làm ơn viết lại mình xem, 

a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)

Ta có: \(A=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)

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

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

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

b: Thay x=16 vào A, ta được:

\(A=\dfrac{3}{4+3}=\dfrac{3}{7}\)

c)\(A=\dfrac{3}{\sqrt{x}+3}=\dfrac{1}{3}\)

\(\Rightarrow\sqrt{x}+3=9\\ \Rightarrow\sqrt{x}=6\\ \Rightarrow x=36\)

d) \(A=\dfrac{3}{\sqrt{x}+3}\)

Vì \(3>0;\sqrt{x}+3>0\Rightarrow\dfrac{3}{\sqrt{x}+3}>0\)

e) \(2A\in Z\Rightarrow\dfrac{6}{\sqrt{x}+3}\in Z \Rightarrow6⋮x+3\\\Rightarrow\sqrt{x}+3\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\Rightarrow x=\left\{0;9\right\}\)

Bài 1: 

\(D=\dfrac{1}{2}\sqrt{48}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}=\dfrac{1}{2}.4\sqrt{3}-\sqrt{3}+5.\dfrac{2\sqrt{3}}{3}=2\sqrt{3}-\sqrt{3}+\dfrac{10\sqrt{3}}{3}=\dfrac{3\sqrt{3}+10\sqrt{3}}{3}=\dfrac{13\sqrt{3}}{3}\)

\(E=\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}-\sqrt{\dfrac{3+\sqrt{5}}{3-\sqrt{5}}}=\sqrt{\dfrac{\left(3-\sqrt{5}\right)^2}{9-5}}-\sqrt{\dfrac{\left(3+\sqrt{5}\right)^2}{9-5}}=\dfrac{3-\sqrt{5}}{2}-\dfrac{3+\sqrt{5}}{2}=-\sqrt{5}\)

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

Bài 2: 

Ta có: G-1

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

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

\(=\dfrac{-\left(\sqrt{x}-1\right)^2}{x-\sqrt{x}+1}\le0\forall x\) thỏa mãn ĐKXĐ

hay \(G\le1\)

B1 

a,\(\frac{5}{3\sqrt{8}}\)\(=\frac{5.\sqrt{8}}{3\sqrt{8}.\sqrt{8}}=\frac{5.\sqrt{8}}{3}\)

b,\(\frac{2a}{1-\sqrt{a}}=\frac{2a\left(1+\sqrt{a}\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}=\frac{2a\left(1+\sqrt{a}\right)}{1-a}\)

c,\(\frac{2+\sqrt{2}}{1+\sqrt{2}}=\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{1+\sqrt{2}}=\sqrt{2}\)

d,\(\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}=\frac{\sqrt{5}\left(\sqrt{3-1}\right)}{1-\sqrt{3}}=\sqrt{-5}\)

a) \(a+b\ge2\sqrt{a}\cdot\sqrt{b}\)

\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\)( luôn đúng )

Dấu "=" xảy ra \(\Leftrightarrow a=b\)

b) \(a+b+c\ge\sqrt{a}\cdot\sqrt{b}+\sqrt{a}\cdot\sqrt{c}+\sqrt{b}\cdot\sqrt{c}\)

\(\Leftrightarrow2a+2b+2c-2\sqrt{a}\cdot\sqrt{b}-2\sqrt{a}\cdot\sqrt{c}-2\sqrt{b}\cdot\sqrt{c}\ge0\)

\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2+\left(\sqrt{b}-\sqrt{c}\right)^2+\left(\sqrt{c}-\sqrt{a}\right)^2\ge0\)( luôn đúng )

Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)

a)

\(a+b\ge2\sqrt{a}.\sqrt{b}\)

\(\Leftrightarrow\) \(a+b\ge2\sqrt{ab}\)

\(\Leftrightarrow\) \(a-2\sqrt{ab}+b\ge0\)

\(\Leftrightarrow\) \(\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\) ( vì a, b > 0) luôn đúng

=> Bất đẳng thức đã cho luôn đúng với ∀ a, b dương (đpcm)