a) 

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

\(=a+\sqrt{b}\ne2\sqrt{\left(a+\sqrt{b}\right)\left(a-\sqrt{b}\right)}+a-\sqrt{b}\)

\(=2a\ne2\sqrt{a^2-b}=2\left(a\ne\sqrt{a^2}-b\right)\)

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

\(\Rightarrowđpcm\)

b)

\(\left(\sqrt{\frac{a+\sqrt{a^2-b}}{2}\ne}\sqrt{\frac{a-\sqrt{a^2-b}}{2}}\right)^2\)

\(=\frac{a+\sqrt{a^2-b}}{2}\ne\sqrt[2]{\frac{a+\sqrt{a^2-b}}{2}.\frac{a-\sqrt{a^2-b}}{2}}+\frac{a-\sqrt{a^2-b}}{2}\)

\(=\frac{a}{2}+\frac{\sqrt{a^2-b}}{2}\ne\sqrt[2]{\frac{a^2-a^2+b}{2.2}}+\frac{a}{2}-\frac{\sqrt{a^2-b}}{2}\)

\(=a\ne2\frac{\sqrt{b}}{2}=a\ne\sqrt{b}\)

\(\Rightarrow\sqrt{\frac{a+\sqrt{a^2-b}}{2}}\ne\sqrt{\frac{a-\sqrt{a^2-b}}{2}}=\sqrt{a\ne\sqrt{b}}\)

\(\Rightarrowđpcm\)

Á,nhầm dấu,đổi mấy cái dấu \(\ne\)thành \(\pm\)giúp mình nhé :)

Với b\(\ge\)0, a\(\ge\)\(\sqrt{b}\) ta bình phương 2 vế lên có:

\(\sqrt{a\pm \sqrt{b}}^2\)=\((\sqrt{\dfrac{\sqrt{a+\sqrt{a^2-b}}}{2}}\)\pm \(\sqrt{\dfrac{\sqrt{a-\sqrt{a^2-b}}}{2}})^2\)

Xét vế trái ta có:

\(\sqrt{(a\pm \sqrt{b})^2}\)=\(a\pm \sqrt{b})

Bình phương vế phải của đẳng thức ta đc :

\(\frac{a+\sqrt{a^2-b}}{2}+\frac{a-\sqrt{a^2-b}}{2}\pm2\sqrt{\frac{a+\sqrt{a^2-b}}{2}\cdot\frac{a-\sqrt{a^2-b}}{2}}\)

\(=a\pm2\sqrt{\frac{a^2-\left(a^2-b\right)}{4}}\)

\(=a\pm2\sqrt{\frac{b}{4}}=a\pm\sqrt{b}\)

=> đpcm

a) \(\left(4\sqrt{2}+\sqrt{30}\right)\left(\sqrt{5}-\sqrt{3}\right).\sqrt{4-\sqrt{15}}\)

\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}\right).\sqrt{\dfrac{8-2\sqrt{15}}{2}}\)

\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{25.6}-\sqrt{9.10}\right).\sqrt{\dfrac{\left(\sqrt{5}\right)^2-2\sqrt{5}.\sqrt{3}+\left(\sqrt{3}\right)^2}{2}}\)

\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right).\sqrt{\dfrac{\left(\sqrt{5}-\sqrt{3}\right)^2}{2}}\)

\(=\left(\sqrt{10}+\sqrt{6}\right).\dfrac{\left|\sqrt{5}-\sqrt{3}\right|}{\sqrt{2}}=\sqrt{2}.\left(\sqrt{5}+\sqrt{3}\right).\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}\)

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

a) Ta có: \(\left(4\sqrt{2}+\sqrt{30}\right)\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{4-\sqrt{15}}\)

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

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

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

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

=2

Ta có : \(\sqrt{\text{a}-\sqrt{\text{b}}}\text{=}\sqrt{\frac{a+\sqrt{a^2-b}}{2}}-\sqrt{\frac{a-\sqrt{a^2-b}}{2}}\) \(\left(b\ge0,a\ge\sqrt{b}\right)\)

Đặt \(x=\sqrt{a-\sqrt{b}}+\sqrt{a+\sqrt{b}}\) => \(x>0\Rightarrow x=\sqrt{x^2}\)

Ta có  : \(x^2=2a+2\sqrt{a^2-b}=4\left(\frac{a+\sqrt{a^2-b}}{2}\right)\)\(\Rightarrow x=2\sqrt{\frac{a+\sqrt{a^2-b}}{2}}\)

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

Đặt \(y=\sqrt{a+\sqrt{b}}-\sqrt{a-\sqrt{b}}\Rightarrow y>0\Rightarrow y=\sqrt{y^2}\)

Ta có  ; \(y^2=2a-2\sqrt{a^2-b}=4\left(\frac{a-\sqrt{a^2-b}}{2}\right)\Rightarrow y=2\sqrt{\frac{a-\sqrt{a^2-b}}{2}}\)

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

Trử (1) và (2) theo vế ta được : 

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

có cách nào ngắn hơn không mấy bn

Mình thấy cách này là chuẩn rồi bạn nhé :))

Cách khác thì mình chưa nghĩ ra ^.^

\(\Rightarrow\left(\sqrt{a+\sqrt{b}}\mp\sqrt{a-\sqrt{b}}\right)^2=\left(\sqrt{2\left(a\mp\sqrt{a^2-b}\right)}\right)^2\Leftrightarrow a+\sqrt{b}+a-\sqrt{b}\mp2\sqrt{\left(a+\sqrt{b}\right)\cdot\left(a-\sqrt{b}\right)}=2a\mp2\sqrt{a^2-b}\Leftrightarrow2a\mp2\sqrt{a^2-b}=2a\mp2\sqrt{a^2-b}\) (luôn đúng) \(\Rightarrowđpcm\)

hộ mình câu c ạ :(((

GIÚP MK NHANH NHÉ