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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
\(\sqrt[3]{2-\sqrt{5}}\left(\sqrt[6]{9+4\sqrt{5}}+\sqrt[3]{2+\sqrt{5}}\right)\)
\(=\sqrt[3]{2-\sqrt{5}}\left(\sqrt[6]{\left(2^2+2.2\sqrt{5}+\sqrt{5^2}\right)}+\sqrt[3]{2+\sqrt{5}}\right)\)
\(=\sqrt[3]{2-\sqrt{5}}\left(\sqrt[6]{\left(2+\sqrt{5}\right)^2}+\sqrt[3]{2+\sqrt{5}}\right)\)
\(=2\sqrt[3]{2-\sqrt{5}}.\sqrt[3]{2+\sqrt{5}}=2\sqrt[3]{\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)}=2\sqrt[3]{4-5}=2\sqrt[3]{-1}=-1.2=-2\)
a/ \(\sqrt{4a^4-12a^2+9}-\sqrt{a^4-8a^2+16}\)
= \(\sqrt{\left(2a^2-3\right)^2}-\sqrt{\left(a^2-4\right)^2}\)
= \(|2a^2-3|-|a^2-4|\)
= \(2a^2-3+a^2-4\)
= \(3a^2-7\)
Thay a=\(\sqrt{3}\).Ta có:
\(3.\left(\sqrt{3}\right)^2-7\)
= 3.3-7=2
b/ \(\sqrt{10a^2-12a\sqrt{10}+36}\)
= \(\sqrt{\left(a\sqrt{10}\right)^2-2.a\sqrt{10}.6+6^2}\)
= \(\sqrt{\left(a\sqrt{10}-6\right)^2}\)
= \(|a\sqrt{10}-6|\)
= \(-a\sqrt{10}+6\)
Thay a= \(\sqrt{\frac{5}{2}}-\sqrt{\frac{2}{5}}\)=\(\frac{3}{\sqrt{10}}\),Ta có:
\(-\frac{3}{\sqrt{10}}.\sqrt{10}+6\)
= -3+6 =3
\(A=\left(\sqrt{8}-3\sqrt{2}+10\right)\left(\sqrt{2}-3\sqrt{0.4}\right)=\sqrt{16}-\frac{12\sqrt{5}}{5}+\sqrt{20}-6\sqrt{10}-6+\frac{18\sqrt{5}}{5}\)
\(A=-2+\frac{16\sqrt{5}}{5}-6\sqrt{10}\)
b)\(B=\frac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}-\frac{\sqrt{5}-1}{2}=\frac{\sqrt{6+2\sqrt{5}}}{2}-\frac{\sqrt{5}-1}{2}=\frac{\sqrt{\left(\sqrt{5}+1\right)^2}}{2}-\frac{\sqrt{5}-1}{2}=\frac{\sqrt{5}+1}{2}-\frac{\sqrt{5}-1}{2}=1\)
Bạn kiểm tra lại đề. Theo mình
\(H=5\left(\sqrt{2+\sqrt{3}}-\sqrt{3-\sqrt{5}}-\sqrt{\frac{5}{2}}\right)^2+\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}-\sqrt{\frac{3}{2}}\right)^2\)
\(\frac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)
\(\frac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{2\sqrt{5}+2}\)
\(=\frac{\sqrt{\left(\sqrt{5}-1\right)^2}\left(3+\sqrt{5}\right)}{2\sqrt{5}+2}\)
\(=\frac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{2\sqrt{5}+2}\)
\(=\frac{3\sqrt{5}-3+5-\sqrt{5}}{2\sqrt{5}+2}\)
\(=\frac{2\sqrt{5}+2}{2\sqrt{5}+2}=1\)