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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
Bạn kiểm tra lại đề. Theo mình
\(H=5\left(\sqrt{2+\sqrt{3}}-\sqrt{3-\sqrt{5}}-\sqrt{\frac{5}{2}}\right)^2+\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}-\sqrt{\frac{3}{2}}\right)^2\)
a) \(A=\left|2-\sqrt{5}\right|+\left|2\sqrt{2}-\sqrt{5}\right|\)
\(=\sqrt{5}-2+2\sqrt{2}-\sqrt{5}=2\sqrt{2}-2\)
b) \(B=\left|\sqrt{7}-2\sqrt{2}\right|+\left|3-2\sqrt{2}\right|\)
\(=2\sqrt{2}-7+3-2\sqrt{2}=-4\)
c) \(C=\sqrt{9+6\sqrt{2}+2}-\sqrt{9-6\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{2}+3\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}=\left(3+\sqrt{2}\right)-\left(3-\sqrt{2}\right)=2\sqrt{2}\)
d) \(D=\sqrt{9+12\sqrt{2}+8}+\sqrt{9-12\sqrt{2}+8}\)
\(=\sqrt{\left(3+2\sqrt{2}\right)^2}+\sqrt{\left(3-2\sqrt{2}\right)^2}=\left(3+2\sqrt{2}\right)-\left(3-2\sqrt{2}\right)=4\sqrt{2}\)
Bài 3: \(3\left(\sqrt{2x^2+1}-1\right)=x\left(1+3x+8\sqrt{2x^2+1}\right)\)
\(\Leftrightarrow\left(3-8x\right)\sqrt{2x^2+1}=3x^2+x+3\)
\(\Rightarrow\left(3-8x\right)^2\left(2x^2+1\right)=\left(3x^2+x+3\right)^2\)
\(\Leftrightarrow119x^4-102x^3+63x^2-54x=0\)
\(\Leftrightarrow x\left(7x-6\right)\left(17x^2+9\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{6}{7}\end{cases}}\)
Thử lại, ta nhận được \(x=0\)là nghiệm duy nhất của phương trình
\(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)
\(=\left|2-\sqrt{5}\right|+\left|2\sqrt{2}-\sqrt{5}\right|\)
\(=\sqrt{5}-2+2\sqrt{2}-\sqrt{5}=2\sqrt{2}-2=2\left(\sqrt{2}-1\right)\)
Ta có: \(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)
\(=\sqrt{5}-2+2\sqrt{2}-\sqrt{5}\)
\(=2\sqrt{2}-2\)