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\(\sqrt{8-4\sqrt{3}}-\dfrac{4}{\sqrt{2}+\sqrt{6}}=\dfrac{\left(\sqrt{2}+\sqrt{6}\right)\left(\sqrt{8-4\sqrt{3}}\right)-4}{\sqrt{2}+\sqrt{6}}=\dfrac{\sqrt{16-8\sqrt{3}}+\sqrt{3}.\sqrt{16-8\sqrt{3}}-4}{\sqrt{2}+\sqrt{6}}=\dfrac{2\sqrt{3}-2+\sqrt{3}\left(2\sqrt{3}-2\right)-4}{\sqrt{2}+\sqrt{6}}=0\)\(\Rightarrow a^3+b^3=0\Rightarrow a^3=-b^3\Rightarrow a=-b\Rightarrow a^5=-b^5\Rightarrow a^5+b^5=0\)
a, Ta có : \(x=\sqrt{3+2\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(3-\sqrt{2}\right)^2}=4\)
Thay x = 4 => \(\sqrt{x}=2\) vào B ta được :
\(B=\frac{2+5}{2-3}=-7\)
b, Ta có : Với \(x\ge0;x\ne9\)
\(A=\frac{4}{\sqrt{x}+3}+\frac{2x-\sqrt{x}-13}{x-9}-\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(=\frac{4\left(\sqrt{x}-3\right)+2x-\sqrt{x}-13-\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}\)
\(=\frac{4\sqrt{x}-12+2x-\sqrt{x}-13-x-3\sqrt{x}}{x-9}=\frac{x-25}{x-9}\)
Lại có \(P=\frac{A}{B}\Rightarrow P=\frac{\frac{x-25}{x-9}}{\frac{\sqrt{x}+5}{\sqrt{x}-3}}=\frac{\sqrt{x}-5}{\sqrt{x}+3}\)
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
\(a^3+b^3=\sqrt{6-2\sqrt{6.2}+2}-\dfrac{4\left(\sqrt{6}-\sqrt{2}\right)}{\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{6}-\sqrt{2}\right)}\)
\(a^3+b^3=\sqrt{\left(\sqrt{6}-\sqrt{2}\right)^2}-\dfrac{4\left(\sqrt{6}-\sqrt{2}\right)}{4}=\sqrt{6}-\sqrt{2}-\sqrt{6}+\sqrt{2}=0\)
\(\Rightarrow a^3+b^3=0\Rightarrow a=-b\)
\(\Rightarrow M=a^5+b^5=a^5+\left(-a\right)^5=a^5-a^5=0\)