Ta có: \(a=\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}=\frac{\sqrt{3}}{\sqrt{5}}+\frac{\sqrt{5}}{\sqrt{3}}=\frac{8\sqrt{15}}{15}\)

=> \(a^2=\frac{64}{15}\)

=> \(M=\sqrt{15a^2-8a\sqrt{15}+16}=\sqrt{15.\frac{64}{15}-8.\frac{8\sqrt{15}}{15}.\sqrt{15}+16}\)

\(M=\sqrt{64-64+16}=4\)

Đây mà là toán lớp 2 á ???

Giải

Ta có: \(\sqrt{\dfrac{5}{3}}+\sqrt{\dfrac{3}{5}}=\dfrac{\sqrt{5}}{\sqrt{3}}+\dfrac{\sqrt{3}}{\sqrt{5}}=\dfrac{8}{\sqrt{15}}\)

Vậy M = \(\sqrt{15\left(\dfrac{8}{15}\right)^2-8.\dfrac{8}{\sqrt{15}}.\sqrt{15}+16}\)

= \(\sqrt{8^2-8^2+16}=\sqrt{16}=4\)

2) \(A=\sqrt{15a^2-8a\sqrt{15}+16}\\ =\sqrt{\left(a\sqrt{15}-4\right)^2}\)

b) Khi a=\(\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}\)  thì 

     \(A=\sqrt{\left[\left(\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}\right)\sqrt{15}-4\right]^2}\)

         \(=\sqrt{\left[\left(3+5\right)-4\right]^2}\)

        \(=\sqrt{4^2}\)

         \(=4\)

\(C=\sqrt{x^2+2\sqrt{x^2-1}}-\sqrt{x^2-2\sqrt{x^2-1}}\)

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

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

\(C^2=2x^2-2\sqrt{x^4-2x^2\sqrt{x^2-1}+2x^2\sqrt{x^2-1}-\left(2\sqrt{x^2-1}\right)^2}\)

\(C^2=2x^2-2\sqrt{x^4-4\left(x^2-1\right)}\)

\(C^2=2x^2-2\sqrt{x^4-4x^2+4}\)

\(C=\sqrt{2x^2-2\sqrt{x^4-4x^2+4}}\) 

Thay: \(x=\sqrt{5}\) vào C, ta có:

\(C=\sqrt{2\sqrt{5}^2-2\sqrt{\sqrt{5}^4-4\sqrt{5}^2+4}}\)

\(C=\sqrt{10-2\sqrt{25-20+4}}\)

\(C=\sqrt{10-2\sqrt{9}}\)

\(C=\sqrt{10-6}\)

\(C=\orbr{\begin{cases}-2\\2\end{cases}}\)

Mà theo bài ra: \(\sqrt{x^2+2\sqrt{x^2-1}}>\sqrt{x^2-2\sqrt{x^2-1}}\)

\(\Rightarrow\sqrt{x^2+2\sqrt{x^2-1}}-\sqrt{x^2-2\sqrt{x^2-1}}>0\)

\(\Rightarrow C=2\)

Đề câu a là \(4\sqrt{5}a\) hay \(4\sqrt{5a}\) . Thấy \(4\sqrt{5}a\) đúng hơn
 

\(A=\sqrt{5a^2-4\sqrt{5}a+4}\) 

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

\(A=\sqrt{5}a-2\)

Thay \(a=\sqrt{5}+\frac{1}{\sqrt{5}}\) vào A, có:

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

\(A=5+1-2\) 

\(A=4\)

Nhớ không nhầm thì gọi là trục căn thức ở mẫu thì phải, cậu dở lại lý thuyết coi nha :v

\(A=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\frac{\sqrt{5}-3}{\sqrt{5}+\sqrt{3}}\)

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

\(=\frac{5+2\sqrt{18}+3}{5-3}+\frac{5-2\sqrt{18}+3}{5-3}\)

\(=\frac{8+6\sqrt{2}}{2}+\frac{8-6\sqrt{2}}{2}\)

\(=\frac{16}{2}\)

\(=8\)

Vậy...

Bài 1: Tính

a) Ta có: \(\frac{\sqrt{6+\sqrt{11}}-\sqrt{7-\sqrt{33}}}{\sqrt{6}+\sqrt{2}}\)

\(=\frac{\sqrt{12+2\sqrt{11}}-\sqrt{14-2\sqrt{33}}}{\sqrt{12}+2}\)

\(=\frac{\sqrt{11+2\cdot\sqrt{11}\cdot1+1}-\sqrt{11-2\cdot\sqrt{11}\cdot\sqrt{3}+3}}{2\sqrt{3}+2}\)

\(=\frac{\sqrt{\left(\sqrt{11}+1\right)^2}-\sqrt{\left(\sqrt{11}-\sqrt{3}\right)^2}}{2\sqrt{3}+2}\)

\(=\frac{\left|\sqrt{11}+1\right|-\left|\sqrt{11}-\sqrt{3}\right|}{2\left(\sqrt{3}+1\right)}\)

\(=\frac{\sqrt{11}+1-\left(\sqrt{11}-\sqrt{3}\right)}{2\left(1+\sqrt{3}\right)}\)(Vì \(\left\{{}\begin{matrix}\sqrt{11}>1>0\\\sqrt{11}>\sqrt{3}\end{matrix}\right.\))

\(=\frac{\sqrt{11}+1-\sqrt{11}+\sqrt{3}}{2\left(1+\sqrt{3}\right)}\)

\(=\frac{1+\sqrt{3}}{2\left(1+\sqrt{3}\right)}=\frac{1}{2}\)

b) Ta có: \(\frac{5\sqrt{3}-3\sqrt{5}}{\sqrt{5}-\sqrt{3}}+\frac{2}{4+\sqrt{15}}-\frac{5\sqrt{5}+3\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)

\(=\frac{\sqrt{15}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}+\frac{2}{4+\sqrt{15}}-\frac{\left(\sqrt{5}+\sqrt{3}\right)\left(8-\sqrt{15}\right)}{\sqrt{5}+\sqrt{3}}\)

\(=\sqrt{15}+\frac{2}{4+\sqrt{15}}-\left(8-\sqrt{15}\right)\)

\(=\sqrt{15}+\frac{2}{4+\sqrt{15}}-8+\sqrt{15}\)

\(=2\sqrt{15}-8+\frac{2}{4+\sqrt{15}}\)

\(=\frac{2\sqrt{15}\left(4+\sqrt{15}\right)}{4+\sqrt{15}}-\frac{8\left(4+\sqrt{15}\right)}{4+\sqrt{15}}+\frac{2}{4+\sqrt{15}}\)

\(=\frac{8\sqrt{15}+30-32-8\sqrt{15}+2}{4+\sqrt{15}}\)

\(=\frac{0}{4+\sqrt{15}}=0\)

Bài 2: Rút gọn

Ta có: \(B=\left(\frac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\left(\frac{1+\sqrt{a}}{a-1}\right)^2\)

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

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

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

\(=\frac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)^2}=1\)

Bài 3:

a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\notin\left\{9;4\right\}\end{matrix}\right.\)

b) Ta có: \(A=\frac{\sqrt{x}+2}{\sqrt{x}-3}-\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{3-3\sqrt{x}}{x-5\sqrt{x}+6}\)

\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{3-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

\(=\frac{x-4-\left(x-2\sqrt{x}-3\right)+3-3\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\frac{x-3\sqrt{x}-1-x+2\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

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

\(=\frac{-\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{1}{3-\sqrt{x}}\)

c) Để A<-1 thì A+1<0

\(\Leftrightarrow\frac{1}{3-\sqrt{x}}+1< 0\)

\(\Leftrightarrow\frac{-1}{\sqrt{x}-3}+\frac{\sqrt{x}-3}{\sqrt{x}-3}< 0\)

\(\Leftrightarrow\frac{-1+\sqrt{x}-3}{\sqrt{x}-3}< 0\)

\(\Leftrightarrow\frac{\sqrt{x}-4}{\sqrt{x}-3}< 0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}-4>0\\\sqrt{x}-3< 0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}-4< 0\\\sqrt{x}-3>0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}>4\\\sqrt{x}< 3\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}< 4\\\sqrt{x}>3\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< 16\\x>9\end{matrix}\right.\Leftrightarrow9< x< 16\)