Ta có: \(\frac{1}{1+\sqrt2}+\frac{1}{\sqrt2+\sqrt3}+\cdots+\frac{1}{\sqrt{99}+\sqrt{100}}\)

\(=\frac{-1+\sqrt2}{\left(\sqrt2+1\right)\left(\sqrt2-1\right)}+\frac{-\sqrt2+\sqrt3}{\left(\sqrt3-\sqrt2\right)\left(\sqrt3+\sqrt2\right)}+\cdots+\frac{-\sqrt{99}+\sqrt{100}}{\left(\sqrt{100}+\sqrt{99}\right)\left(\sqrt{100}-\sqrt{99}\right)}\)

\(=-1+\sqrt2-\sqrt2+\sqrt3-\cdots-\sqrt{99}+\sqrt{100}\)

\(=-1+\sqrt{100}\)

=-1+10

=9

?

\(\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}-\sqrt{\left(1-\sqrt{5}\right)^2}+\dfrac{3}{\sqrt{3}}\)

\(=\left|\sqrt{3}-\sqrt{5}\right|-\left|1-\sqrt{5}\right|+\dfrac{\left(\sqrt{3}\right)^2}{\sqrt{3}}\)

\(=\left(\sqrt{5}-\sqrt{3}\right)-\left(\sqrt{5}-1\right)+\sqrt{3}\)

\(=\sqrt{5}-\sqrt{3}-\sqrt{5}+1+\sqrt{3}\)

\(=1\)

Cảm ơn bạn nha

\(\sqrt{1}+\sqrt{4}+\sqrt{9}+...+\sqrt{100}=1+2+3+...+10\)

=(1+10)+(2+9)+(3+8)+(4+7)+(5+6)=11+11+11+11+11=11.5=55

a, \(A=\left(\sqrt{12}-2\sqrt{5}\right)\sqrt{3}+\sqrt{60}\)

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

\(=2\sqrt{9}-2\sqrt{15}+2\sqrt{15}=2\sqrt{9}\)

b, \(B=\frac{\sqrt{4x}}{x-3}\sqrt{\frac{x^2-6x+9}{x}}=\frac{2\sqrt{x}}{x-3}.\sqrt{\frac{\left(x-3\right)^2}{x}}\)

\(=\frac{2\sqrt{x}}{x-3}.\frac{x-3}{\sqrt{x}}=2\)

em thiếu, giờ mới nhìn lại \(2\sqrt{9}=2.3=6\)

\(\left(3\sqrt{12}-4\sqrt{3}+\sqrt{15}\right)\cdot\sqrt{3}-2\sqrt{5}\)

\(=\left(6\sqrt{3}-4\sqrt{3}+\sqrt{15}\right)\cdot\sqrt{3}-2\sqrt{5}\)

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

(3\(\sqrt{12}\)-4\(\sqrt{3}\)+\(\sqrt{15}\)).\(\sqrt{3}\)-2\(\sqrt{5}\)

=\(\left(6\sqrt{3}-4\sqrt{3}+\sqrt{15}\right).\sqrt{3}-2\sqrt{5}\)

=\(\left(2\sqrt{3}+\sqrt{15}\right).\sqrt{3}-2\sqrt{5}\)

=\(6+\sqrt{45}-2\sqrt{5}\)

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

=\(6+\sqrt{5}\)

a: \(\sqrt{6-4\sqrt2}+\sqrt{22-12\sqrt2}\)

\(=\sqrt{4-2\cdot2\cdot\sqrt2+2}+\sqrt{18-2\cdot3\sqrt2\cdot2+4}\)

\(=\sqrt{\left(2-\sqrt2\right)^2}+\sqrt{\left(3\sqrt2-2\right)^2}\)

\(=2-\sqrt2+3\sqrt2-2=2\sqrt2\)

b: \(\sqrt{\left(\sqrt3-\sqrt2\right)^2}+\sqrt2=\sqrt3-\sqrt2+\sqrt2=\sqrt3\)

c: \(3\sqrt5-\sqrt{\left(1-\sqrt5\right)^2}\)

\(=3\sqrt5-\left|1-\sqrt5\right|\)

\(=3\sqrt5-\left(\sqrt5-1\right)=2\sqrt5+1\)

d:Sửa đề: \(\sqrt{17-12\sqrt2}+\sqrt{6+4\sqrt2}\)

\(=\sqrt{9-2\cdot3\cdot2\sqrt2+8}+\sqrt{4+2\cdot2\cdot\sqrt2+2}\)

\(=\sqrt{\left(3-2\sqrt2\right)^2}+\sqrt{\left(2+\sqrt2\right)^2}=3-2\sqrt2+2+\sqrt2=5-\sqrt2\)

\(\sqrt{5+2\sqrt{6}}+\sqrt{8-2\sqrt{15}}=\sqrt{\left(\sqrt{3}\right)^2+2\sqrt{3}\sqrt{2}+\left(\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\sqrt{3}+\sqrt{2}+\sqrt{5}-\sqrt{3}=\sqrt{2}+\sqrt{5}\)

\(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}-\dfrac{5}{\sqrt{3}-2\sqrt{2}}-\dfrac{5}{\sqrt{3}+\sqrt{8}}=\sqrt{\sqrt{3}^2+2\sqrt{3}.1+1^2}+\sqrt{\sqrt{3}^2-2\sqrt{3}.1+1^2}-\dfrac{5\left(\sqrt{3}+2\sqrt{2}\right)}{\left(\sqrt{3}-2\sqrt{2}\right)\left(\sqrt{3}+2\sqrt{2}\right)}-\dfrac{5\left(\sqrt{3}-2\sqrt{2}\right)}{\left(\sqrt{3}+2\sqrt{2}\right)\left(\sqrt{3}-2\sqrt{2}\right)}=\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}-\dfrac{5\sqrt{3}+10\sqrt{2}}{9-8}-\dfrac{5\sqrt{3}-10\sqrt{2}}{9-8}=\sqrt{3}+1+\sqrt{3}-1-5\sqrt{3}-10\sqrt{2}-5\sqrt{3}+10\sqrt{2}=-8\sqrt{3}\)\(\sqrt{8+2\sqrt{15}}-\sqrt{8-2\sqrt{15}}=\sqrt{\left(\sqrt{5}\right)^2+2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\sqrt{5}+\sqrt{3}-\sqrt{5}+\sqrt{3}=2\sqrt{3}\)

  Chúc Bạn Học Tốt.