Bài 1: Căn bậc hai

a) ta có : \(a>1\Leftrightarrow a-1>0\Leftrightarrow\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)>0\)

\(\Leftrightarrow\sqrt{a}-1>0\Leftrightarrow\sqrt{a}>1\Leftrightarrow a>\sqrt{a}\left(đpcm\right)\)

b) ta có : \(a< 1\Leftrightarrow a-1< 0\Leftrightarrow\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)< 0\)

\(\Leftrightarrow\sqrt{a}-1< 0\Leftrightarrow\sqrt{a}< 1\Leftrightarrow a< \sqrt{a}\left(đpcm\right)\)

a , \(\sqrt{3\sqrt{2}}>\sqrt{2\sqrt{3}}\)

a,Ta có:\(\sqrt{3\sqrt{2}}=\sqrt{\sqrt{18}}\)
\(\sqrt{2\sqrt{3}}=\sqrt{\sqrt{12}}\)
Vì 12<18\(\Rightarrow\sqrt{12}< \sqrt{18}\)
\(\Rightarrow\sqrt{\sqrt{12}}< \sqrt{\sqrt{18}}\) hay \(\sqrt{2\sqrt{3}}< \sqrt{3\sqrt{2}}\)
b,Vì \(\sqrt{10}+\sqrt{17}+1>\sqrt{9}+\sqrt{16}+1\)
\(\Rightarrow\sqrt{10}+\sqrt{17}+1>3+4+1 \)
\(\Rightarrow\sqrt{10}+\sqrt{17}+1>8\)
\(\Rightarrow\sqrt{10}+\sqrt{17}+1>\sqrt{64}\)
Mà \(\sqrt{64}>\sqrt{61}\)
\(\Rightarrow\sqrt{10}+\sqrt{17}+1>\sqrt{61}\)

ta có : \(a+b+\dfrac{1}{2}=a+\dfrac{1}{4}+b+\dfrac{1}{4}\ge\dfrac{2\sqrt{a}}{2}+\dfrac{2\sqrt{b}}{2}=\sqrt{a}+\sqrt{b}\left(đpcm\right)\)

A < B

\(A=\sqrt{2018-2017}=1=\sqrt{2019-2018}\)

\(\Rightarrow A=B\) nhé EDOGAWA CONAN :3

\(A=\sqrt{2018-2017}=\sqrt{1}=1\)

\(B=\sqrt{2019-2018}=\sqrt{1}=1\)

Vì \(1=1\Rightarrow A=B\)

\(\sqrt{3}\left(\sqrt{27}-\sqrt{2}+1\right)+\sqrt{6}\)

\(=\sqrt{81}-\sqrt{6}+\sqrt{3}+\sqrt{6}\)

\(=9+\sqrt{3}\)

giúp mình với ạ

\(a,\dfrac{-3}{5}.\sqrt{\left(-0.5\right)^2}\\ =\dfrac{-3}{5}.0,5\\ =\dfrac{-3}{5}.\dfrac{1}{2}\\ =-\dfrac{3}{10}\)

Câu (b) nhìn hơi lạ lạ á :v

\(c,\sqrt{\left(1-\sqrt{7}\right)^2}+\sqrt{7}\\ =\sqrt{7}-1+\sqrt{7}\\ =2\sqrt{7}-1\)

\(d,\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\\ =\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\\ =3+\sqrt{2}-\left(3-\sqrt{2}\right)\\ =3+\sqrt{2}-3+\sqrt{2}\\ =2\sqrt{2}\)

b , 128

c , \(2\sqrt{7}-1\)

d , \(2\sqrt{2}\)

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

\(=4\sqrt{3}-2\sqrt{3}=2\sqrt{3}\)

\(\sqrt{8^2+6^2}+2\sqrt{\sqrt{625}}=\sqrt{100}+2\sqrt{25}=10+2\cdot5=10+10=20\)

Đặt: \(A=\dfrac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}=\dfrac{\sqrt{2}-\sqrt{2+\sqrt{3}}}{\left(\sqrt{2}+\sqrt{2+\sqrt{3}}\right)\left(\sqrt{2}-\sqrt{2+\sqrt{3}}\right)}+\dfrac{\sqrt{2}+\sqrt{2-\sqrt{3}}}{\left(\sqrt{2}-\sqrt{2-\sqrt{3}}\right)\left(\sqrt{2}+\sqrt{2-\sqrt{3}}\right)}=\dfrac{2-\sqrt{2+\sqrt{3}}}{-\sqrt{3}}+\dfrac{2+\sqrt{2-\sqrt{3}}}{\sqrt{3}}=\dfrac{\sqrt{2+\sqrt{3}}-2+2+\sqrt{2-\sqrt{3}}}{\sqrt{3}}=\dfrac{\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}{\sqrt{3}}\)

\(\Rightarrow A^2=\dfrac{2+\sqrt{3}+2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+2-\sqrt{3}}{3}=\dfrac{4+2}{3}=\dfrac{6}{3}=2\)

\(\Rightarrow A=\sqrt{2}\)

Đặt A = \(\dfrac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)

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

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