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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
Đầu tiên là rút gọn P
P
\(=\frac{1}{\sqrt{x}+1}+\frac{2}{x-\sqrt{x}+1}-\frac{3}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\frac{1x+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\frac{\sqrt{x}}{x-\sqrt{x}+1}\)
\(\Rightarrow\frac{1}{P}=\frac{x-\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}-1+\frac{1}{\sqrt{x}}=-1+2008=2007\)
\(\Rightarrow P=\frac{1}{2007}\)
x= ...... - ....... = a -b
P=(a-b)^3 + 3(a-b) +2018 = a^3-3a^2b+3ab^2-b^3 +3a-3b+2018
=a^3-b^3 -3a(ab-1) -3b(ab -1) +2018 = a^3-b^3 - 3(ab-1)(a+b) +2018
a.b = 1 => ab-1 =0 => P =a^3 -b^3 +2018=\(\sqrt{2}\)-1 -\(\frac{1}{\sqrt{2}-1}\)+2018
=\(\frac{2+1-2\sqrt{2}-1+2018\sqrt{2}-2018}{\sqrt{2}-1}\)=\(\frac{2016\sqrt{2}-2016}{\sqrt{2}-1}\)=2016
Vậy P=2016