ai nay dung kinh nghiem la chinh

cau a)

ta thay \(10+6\sqrt{3}=\left(1+\sqrt{3}\right)^3\)

\(6+2\sqrt{5}=\left(1+\sqrt{5}\right)^2\)

khi do \(x=\frac{\sqrt[3]{\left(\sqrt{3}+1\right)^3}\left(\sqrt{3}-1\right)}{\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{5}}\)

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

\(x=\frac{3-1}{1}=2\)

suy ra 

x^3-4x+1=1

A=1^2018

A=1

b)

ta thay

\(7+5\sqrt{2}=\left(1+\sqrt{2}\right)^3\)

khi do 

\(x=\sqrt[3]{\left(1+\sqrt{2}\right)^3}-\frac{1}{\sqrt[3]{\left(1+\sqrt{2}\right)^3}}\)

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

x=2

thay vao

x^3+3x-14=0

B=0^2018

B=0

\(\frac{2}{\sqrt{6}-2}+\frac{2}{\sqrt{6}+2}+\frac{5}{\sqrt{6}}.\)

\(=\frac{\sqrt{2}.\sqrt{2}}{\sqrt{2}\left(\sqrt{3}-1\right)}+\frac{\sqrt{2}.\sqrt{2}}{\sqrt{2}\left(\sqrt{3}+1\right)}+\frac{5}{\sqrt{6}}\)

\(=\frac{\sqrt{2}\left(\sqrt{3}+1\right)}{3-1}+\frac{\sqrt{2}\left(\sqrt{3}-1\right)}{3+1}+\frac{5}{\sqrt{6}}\)

\(=\frac{\left(\sqrt{3}+1\right)}{\sqrt{2}}+\frac{\sqrt{3}-1}{\sqrt{8}}+\frac{5}{\sqrt{6}}\)

\(=...\)

\(a,\frac{2}{\sqrt{6}-2}+\frac{2}{\sqrt{6}+2}+\frac{5}{\sqrt{6}}\)

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

\(=\frac{4\sqrt{6}}{6-2^2}+\frac{5\sqrt{6}}{6}=2\sqrt{6}+\frac{5\sqrt{6}}{6}\)

\(=\frac{17\sqrt{6}}{6}\)

\(b,\frac{1}{\sqrt{3}+\sqrt{2}-\sqrt{5}}-\frac{1}{\sqrt{3}+\sqrt{2}+\sqrt{5}}\)

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

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

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

a: Ta có: \((x - \sqrt{x^2+5})(x + \sqrt{x^2+5}) = x^2 - (x^2+5) = -5\)

=>\(x + \sqrt{x^2+5} = \frac{-5}{x - \sqrt{x^2+5}}\)

Ta có: \((x - \sqrt{x^2+5})(y - \sqrt{y^2+5}) = 5\)

=>\((x-\sqrt{x^2+5})(y-\sqrt{y^2+5})\left(x+\sqrt{x^2+5}\right)=5\left(x+\sqrt{x^2+5}\right)\)

=>\(-5 (y - \sqrt{y^2+5}) = 5 (x + \sqrt{x^2+5})\)

=>\(-(y-\sqrt{y^2+5})=x+\sqrt{x^2+5}\)

=>\(\sqrt{y^2+5}-y=x+\sqrt{x^2+5}\quad(1)\)

Ta có: \((x - \sqrt{x^2+5})(y - \sqrt{y^2+5}) = 5\)

=>\((x-\sqrt{x^2+5})(y-\sqrt{y^2+5})\left(y+\sqrt{y^2+5}\right)=5\left(y+\sqrt{y^2+5}\right)\)

=>\(-5 (x - \sqrt{x^2+5}) = 5 (y + \sqrt{y^2+5})\)

=>\(\sqrt{x^2+5}-x=y+\sqrt{y^2+5}\quad(2)\)

Từ (1),(2) suy ra -x-y=x+y

=>x+y=0

=>x=-y

\(M=x^{2015}+y^{2015}=\left(-y\right)^{2015}+y^{2015}=0\)

b: Đặt \(u=\sqrt[3]{5 + 2\sqrt{13}};v=\sqrt[3]{5-2\sqrt{13}}\)

=>\(uv = \sqrt[3]{(5 + 2\sqrt{13})(5 - 2\sqrt{13})} = \sqrt[3]{5^2 - (2\sqrt{13})^2} = \sqrt[3]{25 - 52} = \sqrt[3]{-27} = -3\)

x=u+v

=>\(x^3 = (u + v)^3 = u^3 + v^3 + 3uv(u + v)\)

=>\(x^3=(5+2\sqrt{13})+(5-2\sqrt{13})+3(-3)x\)

=>\(x^3=10-9x\)

=>\(x^3+9x-10=0\)

=>(x-1)(x^2+x+10)=0

=>x-1=0

=>x=1

\(A=x^{2015}-x^{2016}+2017\)

\(=1^{2015}-1^{2016}+2017\)

=2017

G = \(\sqrt{6}-2+5-\sqrt{6}+2^3=3+8=11\)

F= \(\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(2^5\right)^2}\)=\(2+\sqrt{3}-\sqrt{3}+1+2^5=3+32=35\)

H = \(\sqrt{6}-\frac{4\left(\sqrt{10}+\sqrt{6}\right)}{10-6}+\frac{\sqrt{10}\left(\sqrt{10}-1\right)}{\sqrt{10}-1}\)=\(\sqrt{6}-\sqrt{10}-\sqrt{6}+\sqrt{10}=0;\)

a) \(\left(\sqrt{125}+\sqrt{45}-2\sqrt{80}\right).\sqrt{5}=\left(5\sqrt{5}+3\sqrt{5}-8\sqrt{5}\right).\sqrt{5}\)

\(=0.\sqrt{5}=0\)

b) \(\frac{5-2\sqrt{6}}{\sqrt{2}-\sqrt{3}}=\frac{\left(5-2\sqrt{6}\right)\left(\sqrt{2}+\sqrt{3}\right)}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}=\frac{\left(5\sqrt{2}+5\sqrt{3}-4\sqrt{3}-6\sqrt{2}\right)}{-1}\)

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

a,\(\left(\sqrt{125}+\sqrt{45}-2\sqrt{80}\right).\sqrt{5}\)

\(=\left(5\sqrt{5}+3\sqrt{5}-8\sqrt{5}\right).\sqrt{5}\)

\(=0.\sqrt{5}\)

\(=0\)

b,\(\frac{5-2\sqrt{6}}{\sqrt{2}-\sqrt{3}}\)

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

\(=\frac{\sqrt{3}-\sqrt{2}}{-1}\)

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