\(\left(\sqrt{2}+\sqrt{3}\right)^2=5+2\sqrt{6}>2^2=4\left(5>4\right)\\ \Leftrightarrow\sqrt{2}+\sqrt{3}>2\)

\(\left(\sqrt{8}+\sqrt{5}\right)^2=13+2\sqrt{40};\left(\sqrt{7}-\sqrt{6}\right)^2=13-2\sqrt{42}\\ 2\sqrt{40}>0>-2\sqrt{42}\\ \Leftrightarrow13+2\sqrt{40}>13-2\sqrt{42}\\ \Leftrightarrow\left(\sqrt{8}+\sqrt{5}\right)^2>\left(\sqrt{7}-\sqrt{6}\right)^2\\ \Leftrightarrow\sqrt{8}+\sqrt{5}>\sqrt{7}-\sqrt{6}\)

\(\sqrt{2}\) + \(\sqrt{3}\)  > 2

Sửa đề: \(S=\frac{4+\sqrt7}{3\sqrt2+\sqrt{4+\sqrt7}}+\frac{4-\sqrt7}{3\sqrt2-\sqrt{4-\sqrt7}}\)

\(=\frac{\sqrt2\left(4+\sqrt7\right)}{6+\sqrt{8+2\sqrt7}}+\frac{\sqrt2\left(4-\sqrt7\right)}{6-\sqrt{8-2\sqrt7}}\)

\(=\frac{\sqrt2\left(4+\sqrt7\right)}{6+\sqrt{\left(\sqrt7+1\right)^2}}+\frac{\sqrt2\left(4-\sqrt7\right)}{6-\sqrt{\left(\sqrt7-1\right)^2}}\)

\(=\frac{\sqrt2\left(4+\sqrt7\right)}{6+\sqrt7+1}+\frac{\sqrt2\left(4-\sqrt7\right)}{6-\left(\sqrt7-1\right)}=\frac{\sqrt2\left(4+\sqrt7\right)}{7+\sqrt7}+\frac{\sqrt2\left(4-\sqrt7\right)}{7-\sqrt7}\)

\(=\frac{1}{\sqrt2}\cdot\left\lbrack\frac{2\left(4+\sqrt7\right)}{\sqrt7\left(\sqrt7+1\right)}+\frac{2\left(4-\sqrt7\right)}{\sqrt7\left(\sqrt7-1\right)}\right\rbrack\)

\(=\frac{1}{\sqrt2}\cdot\left\lbrack\frac{8+2\sqrt7}{\sqrt7\left(\sqrt7+1\right)}+\frac{8-2\sqrt7}{\sqrt7\left(\sqrt7-1\right)}\right\rbrack\)

\(=\frac{1}{\sqrt2}\cdot\left\lbrack\frac{\left(\sqrt7+1\right)^2}{\sqrt7\left(\sqrt7+1\right)}+\frac{\left(\sqrt7-1\right)^2}{\sqrt7\left(\sqrt7-1\right)}\right\rbrack=\frac{1}{\sqrt2}\cdot\frac{\sqrt7+1+\sqrt7-1}{\sqrt7}=\frac{2\sqrt7}{\sqrt2\cdot\sqrt7}=\sqrt2\)

Ta có: \(R=\frac{3+\sqrt5}{2\sqrt2+\sqrt{3+\sqrt5}}+\frac{3-\sqrt5}{2\sqrt2-\sqrt{3-\sqrt5}}\)

\(=\frac{\sqrt2\left(3+\sqrt5\right)}{4+\sqrt{6+2\sqrt5}}+\frac{\sqrt2\left(3-\sqrt5\right)}{4-\sqrt{6-2\sqrt5}}\)

\(=\frac{\sqrt2\left(3+\sqrt5\right)}{4+\sqrt5+1}+\frac{\sqrt2\left(3-\sqrt5\right)}{4-\sqrt5+1}\)

\(=\frac{\sqrt2\left(3+\sqrt5\right)}{5+\sqrt5}+\frac{\sqrt2\left(3-\sqrt5\right)}{5-\sqrt5}=\sqrt2\cdot\left\lbrack\frac{3+\sqrt5}{\sqrt5\left(\sqrt5+1\right)}+\frac{3-\sqrt5}{\sqrt5\left(\sqrt5-1\right)}\right\rbrack\)

\(=\frac{\sqrt2}{2}\cdot\left\lbrack\frac{6+2\sqrt5}{\sqrt5\left(\sqrt5+1\right)}+\frac{6-2\sqrt5}{\sqrt5\left(\sqrt5-1\right)}\right\rbrack\)

\(=\frac{\sqrt2}{2}\cdot\left\lbrack\frac{\left(\sqrt5+1\right)^2}{\sqrt5\left(\sqrt5+1\right)}+\frac{\left(\sqrt5-1\right)^2}{\sqrt5\left(\sqrt5-1\right)}\right\rbrack=\frac{\sqrt2}{2}\cdot\frac{\sqrt5+1+\sqrt5-1}{\sqrt5}=\frac{\sqrt2}{2}\cdot2=\sqrt2\)

Do đó: R=S

Ta có: 

\(R=\)\(\dfrac{3+\sqrt{5}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{3-\sqrt{5}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)

\(=\)\(\dfrac{\sqrt{10}+3\sqrt{2}}{5+\sqrt{5}}+\dfrac{\sqrt{10}-3\sqrt{2}}{5-\sqrt{5}}\)

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

\(=\dfrac{4\sqrt{2}}{4\sqrt{5}}=\sqrt{\dfrac{2}{5}}\)

Làm câu S tương tự như này rồi đối chiếu kết quả nha

1) So sánh:

N = \(\dfrac{5+\sqrt{5}}{\sqrt{5}+1}-\sqrt{6-2\sqrt{5}}\)

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

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

M = \(\sqrt{18}-\sqrt{8}\)

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

\(=\sqrt{2}\)

Ta có: \(1=\sqrt{1}\)

Mà 1 < 2

\(\Rightarrow\sqrt{1}< \sqrt{2}\)

Hay 1 \(< \sqrt{2}\)

Vậy N < M
 

2) Với \(x>0;x\ne4;x\ne9\), ta có:

A = \(\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{2x}{9-x}\right):\left(\dfrac{x-4}{x-3\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)

\(=\left[\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{2x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]:\left[\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\right]\)

\(=\dfrac{x-3\sqrt{x}-2x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{x-4-2\sqrt{x}+6}{\sqrt{x}\left(\sqrt{x-3}\right)}\)

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

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

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

1)  \(A^2=2+2.\frac{\sqrt{\left(8+\sqrt{15}\right)\left(8-\sqrt{15}\right)}}{2}\)

              \(2+\sqrt{64-15}=2+\sqrt{49}=2+7=9\) mà A>0

=> A=3

2) \(A=\sqrt{4-\sqrt{15}}\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right).\)

 \(A=\sqrt{\left(4-\sqrt{15}\right)\left(4+\sqrt{15}\right)}\sqrt{4+\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right).\)

​​\(A=\sqrt{4+\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right).\)

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

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

Mà A >0 

=> A=2

Mà 4>3

=> \(\sqrt{4}=2>\sqrt{3}\)

=> \(A>\sqrt{3}\)

Bài 1: 

Để M có nghĩa thì \(\left\{{}\begin{matrix}x+4\ge0\\2-x\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-4\\x\le2\end{matrix}\right.\Leftrightarrow-4\le x\le2\)

Số giá trị nguyên thỏa mãn điều kiện là:

\(\left(2+4\right)+1=7\)

giải giúp mình đi mai là mình đi học rồi

Ta có:

bla bla ........

vậy đáp số là... quên mất rồi

b: Ta có: \(4\sqrt{5}=\sqrt{4^2\cdot5}=\sqrt{80}\)

\(5\sqrt{3}=\sqrt{5^2\cdot3}=\sqrt{75}\)

mà 80>75

nên \(4\sqrt{5}>5\sqrt{3}\)