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
\(A^3=x^3-3x+3A\sqrt[3]{\frac{\left(x^3-3x\right)^2-\left(x^2-1\right)^2\left(x^2-4\right)}{4}}\)
\(A^3=x^3-3x+3A\sqrt[3]{\frac{x^6-6x^4+9x^2-\left(x^6-6x^4+9x^2-4\right)}{4}}\)
\(A^3=x^3-3x+3A\)
\(A^3-x^3-3\left(A-x\right)=0\)
\(\left(A-x\right)\left(A^2+x^2+Ax-3\right)=0\)
\(\Rightarrow A=x\) (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/ \(a+1=\sqrt{17}\Rightarrow a^2+2a+1=17\Rightarrow a^2+2a-17=-1\)
\(P=\left[a^3\left(a^2+2a-17\right)-a^2+18a-17\right]^{2018}\)
\(=\left(-a^3-a^2+18a-17\right)^{2018}\)
\(=\left(-a\left(a^2+2a-17\right)+a^2+a-17\right)^{2018}\)
\(=\left(a^2+2a-17\right)^{2018}\)
\(=\left(-1\right)^{2018}=1\)
Ta có \(x=\sqrt{\dfrac{1}{2\sqrt{3}-2}-\dfrac{3}{2\sqrt{3}+2}}=\sqrt{\dfrac{2\sqrt{3}+2}{\left(2\sqrt{3}-2\right)\left(2\sqrt{3}+2\right)}-\dfrac{3\left(2\sqrt{3}-2\right)}{\left(2\sqrt{3}-2\right)\left(2\sqrt{3}+2\right)}}=\sqrt{\dfrac{2\left(\sqrt{3}+1\right)}{12-4}-\dfrac{2\left(3\sqrt{3}-3\right)}{12-4}}=\sqrt{\dfrac{\sqrt{3}+1}{4}-\dfrac{3\sqrt{3}-3}{4}}=\sqrt{\dfrac{\sqrt{3}+1-3\sqrt{3}+3}{4}}=\sqrt{\dfrac{4-2\sqrt{3}}{4}}=\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{4}}=\dfrac{\sqrt{3-2\sqrt{3}+1}}{2}=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{2}=\dfrac{\left|\sqrt{3}-1\right|}{2}=\dfrac{\sqrt{3}-1}{2}\Leftrightarrow2x=\sqrt{3}-1\Leftrightarrow2x+1=\sqrt{3}\Leftrightarrow\left(2x+1\right)^2=3\Leftrightarrow4x^2+4x-2=0\Leftrightarrow2x^2+2x-1=0\)
Ta lại có \(P=\dfrac{4\left(x+1\right)x^{2018}-2x^{2017}+2x+1}{2x^2+3x}=\dfrac{2x^{2017}\left[2\left(x+1\right)x-1\right]+\sqrt{3}}{2x^2+2x-1+x+1}=\dfrac{2x^{2017}\left[2x^2+2x-1\right]+\sqrt{3}}{x+1}=\dfrac{\sqrt{3}}{x+1}=\sqrt{3}:\left(x+1\right)=\sqrt{3}:\left(\dfrac{\sqrt{3}-1}{2}+1\right)=\sqrt{3}:\dfrac{\sqrt{3}+1}{2}=\dfrac{2\sqrt{3}}{\sqrt{3}+1}=\dfrac{2\sqrt{3}\left(\sqrt{3}-1\right)}{3-1}=\dfrac{2\left(3-\sqrt{3}\right)}{2}=3-\sqrt{3}\)Vậy khi \(x=\sqrt{\dfrac{1}{2\sqrt{3}-2}-\dfrac{3}{2\sqrt{3}+2}}\) thì P=\(3-\sqrt{3}\)
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