b)CM: \(ab\sqrt{1+\dfrac{1}{a^2b^2}}-\sqrt{a^2b^2+1}=0\)

\(VT=ab\sqrt{\dfrac{a^2b^2+1}{\left(ab\right)^2}}-\sqrt{a^2b^2+1}\)

\(VT=ab\dfrac{\sqrt{a^2b^2+1}}{ab}-\sqrt{a^2b^2+1}\)

\(VT=\sqrt{a^2b^2+1}-\sqrt{a^2b^2+1}\)

\(VT=0=VP\)

\(a.\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}-\dfrac{3}{3-\sqrt{6}}=\dfrac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-\dfrac{\sqrt{3}.\sqrt{3}}{\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}=\sqrt{6}-\dfrac{\sqrt{3}}{\sqrt{3}-\sqrt{2}}=\dfrac{3\sqrt{2}-3\sqrt{3}}{\sqrt{3}-\sqrt{2}}=\dfrac{-3\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}=-3\) \(b.\left(2\sqrt{2}-\sqrt{3}\right)^2-2\sqrt{3}\left(\sqrt{3}-2\sqrt{2}\right)=\left(2\sqrt{2}-\sqrt{3}\right)\left(2\sqrt{2}+\sqrt{3}\right)=8-3=5\) \(c.\left(\dfrac{1}{3-\sqrt{5}}-\dfrac{1}{3+\sqrt{5}}\right):\dfrac{5-\sqrt{5}}{\sqrt{5}-1}=\dfrac{3+\sqrt{5}-3+\sqrt{5}}{9-5}:\sqrt{5}=\dfrac{2\sqrt{5}}{4}.\dfrac{1}{\sqrt{5}}=\dfrac{\sqrt{5}}{2}.\dfrac{1}{\sqrt{5}}=\dfrac{1}{2}\) \(d.\left(3-\dfrac{a-2\sqrt{a}}{\sqrt{a}-2}\right)\left(3+\dfrac{\sqrt{ab}-3\sqrt{a}}{\sqrt{b}-3}\right)=\left(3-\sqrt{a}\right)\left(3+\sqrt{a}\right)=9-a\)

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

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

\(b,\left(2-\frac{a-3\sqrt{a}}{\sqrt{a}-3}\right)\left(2-\frac{5\sqrt{a}-\sqrt{ab}}{\sqrt{b}-5}\right)=\left(2-\frac{\sqrt{a}\left(\sqrt{a}-3\right)}{\sqrt{a}-3}\right)\left(2-\frac{-\sqrt{a}\left(\sqrt{b}-5\right)}{\sqrt{b}-5}\right)\)

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

\(c,\left(3+\frac{a-2\sqrt{a}}{\sqrt{a}-2}\right)\left(3-\frac{3a+\sqrt{a}}{3\sqrt{a}+1}\right)=\left(3+\frac{\sqrt{a}\left(\sqrt{a}-2\right)}{\sqrt{a}-2}\right)\left(3-\frac{\sqrt{a}\left(3\sqrt{a}+1\right)}{3\sqrt{a}+1}\right)\)

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

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

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

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

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

1.

Đặt \(\sqrt{a^2+x^2}=m,\sqrt{a^2-x^2}=n\Rightarrow x^2=\frac{m^2-n^2}{2}\)

\(\frac{\sqrt{a^2+x^2}+\sqrt{a^2-x^2}}{\sqrt{a^2+x^2}-\sqrt{a^2-x^2}}-\sqrt{\frac{a^4}{x^4}-1}=\frac{\sqrt{a^2+x^2}+\sqrt{a^2-x^2}}{\sqrt{a^2+x^2}-\sqrt{a^2-x^2}}-\sqrt{\frac{(a^2+x^2)(a^2-x^2)}{x^4}}\)

\(=\frac{\sqrt{a^2+x^2}+\sqrt{a^2-x^2}}{\sqrt{a^2+x^2}-\sqrt{a^2-x^2}}-\frac{\sqrt{(a^2+x^2)(a^2-x^2)}}{x^2}\)

\(=\frac{m+n}{m-n}-\frac{mn}{\frac{m^2-n^2}{2}}=\frac{(m+n)^2}{m^2-n^2}-\frac{2mn}{m^2-n^2}=\frac{m^2+n^2}{m^2-n^2}\)

\(=\frac{2a^2}{2x^2}=\frac{a^2}{x^2}\)

2.

\(=\left[\frac{(1-\sqrt{a})(1+\sqrt{a}+a)}{1-\sqrt{a}}+\sqrt{a}\right].\left[\frac{(1+\sqrt{a})(1-\sqrt{a}+a)}{1+\sqrt{a}}-\sqrt{a}\right]\)

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

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

\(=(a-1)^2\)

3.

\(=\frac{3(1-x)}{\sqrt{1+x}.\sqrt{1-x}}:\frac{3+\sqrt{1-x^2}}{\sqrt{1-x^2}}=\frac{3(1-x)}{\sqrt{1-x^2}}.\frac{\sqrt{1-x^2}}{3+\sqrt{1-x^2}}=\frac{3(1-x)}{3+\sqrt{1-x^2}}\)

4. Bạn xem lại đề xem đã đúng chưa?

5.

\(=\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}.\frac{\sqrt{b}(a+\sqrt{ab})+\sqrt{b}(a-\sqrt{ab})}{(a-\sqrt{ab})(a+\sqrt{ab})}\)

\(=\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}.\frac{2a\sqrt{b}}{a^2-ab}\)

\(=\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}}.\frac{1}{a-b}\)

\(=\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})}\)

\(=\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{1}{a+\sqrt{ab}}=\frac{\sqrt{a}+\sqrt{b}}{a+\sqrt{ab}}=\frac{1}{\sqrt{a}}\)

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

\(=\dfrac{1.\left(2\sqrt{a}+3\right)}{2\sqrt{a}+3}=1\)

b) \(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right):\left(1+\sqrt{a}\right)^2\)

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

\(=\left(a+\sqrt{a}+1+\sqrt{a}\right).\dfrac{1}{\left(\sqrt{a}+1\right)^2}=\left(a+2\sqrt{a}+1\right).\dfrac{1}{\left(\sqrt{a}+1\right)^2}\)

\(=\left(\sqrt{a}+1\right)^2.\dfrac{1}{\left(\sqrt{a}+1\right)^2}=1\)

a, \(VT=\dfrac{\left(2+\sqrt{a}\right)^2-\left(\sqrt{a}+1\right)^2}{2\sqrt{a}+3}=\dfrac{a+4\sqrt{a}+4-a-2\sqrt{a}-1}{2\sqrt{a}+3}\)

\(=\dfrac{2\sqrt{a}+3}{2\sqrt{a}+3}=1=VP\)

Vậy ta có đpcm 

b, \(VT=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right):\left(1+\sqrt{a}\right)^2\)

\(=\left(1+\sqrt{a}+a+\sqrt{a}\right):\left(1+\sqrt{a}\right)^2=\dfrac{\left(1+\sqrt{a}\right)^2}{\left(1+\sqrt{a}\right)^2}=1=VP\)

Vậy ta có đpcm 

a) Ta có: \(\dfrac{\left(2+\sqrt{a}\right)^2-\left(\sqrt{a}+1\right)^2}{2\sqrt{a}+3}\)

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

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

b) Ta có: \(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right):\left(1+\sqrt{a}\right)^2\)

\(=\left(1+\sqrt{a}+\sqrt{a}+a\right):\left(1+\sqrt{a}\right)^2\)

\(=\left(1+\sqrt{a}\right)^2:\left(1+\sqrt{a}\right)^2=1\)

a) ĐKXĐ: x\(\ne\) 0;4

Ta có: Q= \(\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{2}{\sqrt{x}}\right)\)

                   = \(\frac{4\sqrt{x}\cdot\left(2-\sqrt{x}\right)+8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}:\frac{\sqrt{x}-1-2\cdot\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

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

b) Q=-1 => \(\frac{4}{\sqrt{x}-3}=-1\)

<=> \(4=3-\sqrt{x}\)

<=> \(\sqrt{x}=-1\) (vô lí)

Vậy ko tìm được x.

\(\frac{\left(5\sqrt{3}+\sqrt{50}\right)\left(5-\sqrt{24}\right)}{\sqrt{75}-5\sqrt{2}}\)

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

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

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

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

DÀI QUÁ MK KO GHI ĐƯỢC NÊN VIẾT KQ LUÔN NHA !!!

ĐẲNG THỨC ĐÓ = 1 NHA  Hatsune Miku !

Nguyễn Thị Thanh Trúc: Vậy nêu bước giải bằng lời giúp mk được ko T^T

a) \(\Leftrightarrow\left(\sqrt{3+\sqrt{5}}\right)^2\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}=8\)

\(\Leftrightarrow\sqrt{3+\sqrt{5}}\cdot\left(\sqrt{10}-\sqrt{2}\right)\sqrt{\left(3-\sqrt{5}\right)\cdot\left(3+\sqrt{5}\right)}=8\)

\(\Leftrightarrow\sqrt{\frac{6+2\sqrt{5}}{2}}\cdot\left(\sqrt{5}\sqrt{2}-\sqrt{2}\right)\sqrt{3^2-5}=8\).

\(\Leftrightarrow\sqrt{\frac{5+2\sqrt{5}+1}{2}}\cdot\sqrt{2}\cdot\left(\sqrt{5}-1\right)\cdot\sqrt{4}=8\)

\(\Leftrightarrow\frac{\sqrt{\left(\sqrt{5}+1\right)^2}}{\sqrt{2}}\cdot\sqrt{2}\cdot\left(\sqrt{5}-1\right)\cdot2=8\)

\(\Leftrightarrow\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)=4\Leftrightarrow\left(\sqrt{5}\right)^2-1=4\Leftrightarrow5-1=4\)Đúng -ĐPCM.

Cậu giải dùm mình câu b luôn nhé! cảm ơn c! :)))))))

b) Rõ ràng: \(\sqrt{\sqrt{2}+1}>\sqrt{\sqrt{2}-1}\)

Bình phương 2 vế ta được:

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

\(\Leftrightarrow2\sqrt{2}-2\sqrt{\left(\sqrt{2}\right)^2-1}=2\sqrt{2}-2\)

\(\Leftrightarrow2\sqrt{2}-2=2\sqrt{2}-2\)Hiển nhiên đúng. ĐPCM