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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
\(\dfrac{1}{\sqrt[3]{7+5\sqrt{2}}}=\dfrac{\sqrt[3]{7-5\sqrt{2}}}{\sqrt[3]{\left(7+5\sqrt{2}\right)\left(7-5\sqrt{2}\right)}}=-\sqrt[3]{7-5\sqrt{2}}\)
\(\Rightarrow x=\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\)
\(\Rightarrow x^3=\left(\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\right)^3=14-3\left(\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\right)\)
\(\Rightarrow x^3=14-3x\Rightarrow x^3+3x-14=0\)
Vậy F=0
Lời giải:
Đặt \(\sqrt[3]{\sqrt{5}+2}=a; \sqrt[3]{\sqrt{5}-2}=b\)
\(\Rightarrow a^3-b^3=4; ab=1\)
Ta có:
$x=a-b$
$\Rightarrow x^3=(a-b)^3=a^3-b^3-3ab(a-b)=4-3x$
$\Rightarrow x^3+3x=4$
$\Rightarrow f(x)=4$
Lời giải:
Đặt $\sqrt[3]{5\sqrt{2}+7}=a; \sqrt[3]{5\sqrt{2}-7}=b$
Ta có:
$a^3-b^3=14$
$ab=\sqrt[3]{(5\sqrt{2}+7)(5\sqrt{2}-7)}=1$
$x=a-b$
$\Rightarrow x^3=(a-b)^3=a^3-b^3-3ab(a-b)=14-3.1.x$
$\Leftrightarrow x^3+3x-14=0$
$\Leftrightarrow (x-2)(x^2+2x+7)=0$
Dễ thấy $x^2+2x+7>0$ nên $x-2=0$
$\Rightarrow x=2$
$\Rightarrow f(x)=x^3+2x=2^3+2.2=12$
b) Ta có: \(x+\sqrt{3}=2\Leftrightarrow x-2=-\sqrt{3}\Leftrightarrow\left(x-2\right)^2=3\Leftrightarrow x^2-4x+1=0\)
\(B=x^5-3x^4-3x^3+6x^2-20x+2021\)
\(B=\left(x^5-4x^4+x^3\right)+\left(x^4-4x^3+x^2\right)+\left(5x^2-20x+5\right)+2016\)
\(B=x^3\left(x^2-4x+1\right)+x^2\left(x^2-4x+1\right)+5\left(x^2-4x+1\right)+2016\)
Thế \(x^2-4x+1=0\)\(\Rightarrow B=2016.\)
\(x=\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\Rightarrow x^3=5\sqrt{2}+7-\left(5\sqrt{2}-7\right)-3\sqrt[3]{\left(5\sqrt{2}\right)^2-7^2}.x\)
\(=14-3.\sqrt[3]{50-49}.x=14-3x\)
\(\Rightarrow x^3=14-3x\Rightarrow x^3+3x=14\)