Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
mk nghĩ bạn chép sai đề hình như đề bài phải là \(A=\sqrt[3]{\frac{x^3-3x+\left(x^2-1\right)\sqrt{x^2-4}}{2}}+\sqrt[3]{\frac{x^3-3x-\left(x^2-1\right)\sqrt{x^2-4}}{2}}\)
ta xét \(A^3=\left(\sqrt[3]{\frac{x^3-3x+\left(x^2-1\right)\sqrt{x^2-4}}{2}}+\sqrt[3]{\frac{x^3-3x-\left(x^2-1\right)\sqrt{x^2-4}}{2}}\right)^3\)
<=> \(A^3=x^3-3x+3A\cdot\sqrt[3]{\frac{4}{4}}\)
<=> \(A^3=x^3-3x+3A\)
<=> \(A^3-3A-x^3+3x=0\)
<=>\(\left(A^3-x^3\right)-3A+3x=0\)
<=> \(\left(A-x\right)\left(A^2+Ax+x^2\right)-3\left(A-x\right)=0\)
<=> \(\left(A-x\right)\left(A^2+Ax+x^2-3\right)=0\)
<=> \(\orbr{\begin{cases}A=x\\A^2+Ax+x^2-3=0\end{cases}}\)(vô lí )
vậy \(A=x\)
\(A^3=x^3-3x+3A\sqrt[3]{\frac{\left(x^3-3x\right)^2-\left(x^2-1\right)^2\left(x^2-4\right)}{4}}\)
\(A^3=x^3-3x+3A\sqrt[3]{\frac{x^6-6x^4+9x^2-\left(x^6-6x^4+9x^2-4\right)}{4}}\)
\(A^3=x^3-3x+3A\)
\(A^3-x^3-3\left(A-x\right)=0\)
\(\left(A-x\right)\left(A^2+x^2+Ax-3\right)=0\)
\(\Rightarrow A=x\) (do \(\left\{{}\begin{matrix}A>0\\x\ge2\end{matrix}\right.\) \(\Rightarrow x^2-3>0\Rightarrow A^2+x^2+Ax-3>0\))
2/ \(a+1=\sqrt{17}\Rightarrow a^2+2a+1=17\Rightarrow a^2+2a-17=-1\)
\(P=\left[a^3\left(a^2+2a-17\right)-a^2+18a-17\right]^{2018}\)
\(=\left(-a^3-a^2+18a-17\right)^{2018}\)
\(=\left(-a\left(a^2+2a-17\right)+a^2+a-17\right)^{2018}\)
\(=\left(a^2+2a-17\right)^{2018}\)
\(=\left(-1\right)^{2018}=1\)