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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
Ta có:
\(x=1+\sqrt[3]{5}+\sqrt[3]{25}\)
\(\Rightarrow x^3=\left(1+\sqrt[3]{5}+\sqrt[3]{25}\right)^3=61+33\sqrt[3]{5}+21\sqrt[3]{25}\)
\(=\left(33+21\sqrt[3]{5}+9\sqrt[3]{25}\right)+\left(12+12\sqrt[3]{5}+12\sqrt[3]{25}\right)+16=3x^2+12x+16\)
\(\Rightarrow P=\left(x^3-3x^2-12x-15\right)^{10}+2018\)
\(=\left(3x^2+12x+16-3x^2-12x-15\right)^{10}+2018=2019\)
\(x^3=4\left(\sqrt{5}+1\right)-4\left(\sqrt{5}-1\right)-3\sqrt[3]{4\left(\sqrt{5}+1\right).4\left(\sqrt{5}-1\right)}.\left(\sqrt[3]{4\left(\sqrt{5}+1\right)}-\sqrt[3]{4\left(\sqrt{5}-1\right)}\right)\)\(\Rightarrow x^3=8-12x\)
\(\Rightarrow x^3+12x-9=-1\)
\(\Rightarrow P=\left(-1\right)^{2015}=-1\)