Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x=\frac{1}{2}\sqrt{\sqrt{2}+\frac{1}{8}}-\frac{1}{8}\sqrt{2}\)
\(\Leftrightarrow x+\frac{\sqrt{2}}{8}=\frac{1}{2}\sqrt{\sqrt{2}+\frac{1}{8}}\)
\(\Leftrightarrow\left(x+\frac{\sqrt{2}}{8}\right)^2=\frac{1}{4}\left(\sqrt{2}+\frac{1}{8}\right)\)
\(\Leftrightarrow x^2+\frac{x\sqrt{2}}{4}+\frac{1}{32}=\frac{\sqrt{2}}{4}+\frac{1}{32}\)
\(\Leftrightarrow x^2+\frac{x\sqrt{2}}{4}-\frac{\sqrt{2}}{4}=0\)
\(\Leftrightarrow4x^2+x\sqrt{2}-\sqrt{2}=0\)(1)
\(\Leftrightarrow x\sqrt{2}=\sqrt{2}-4x^2\)
\(\Leftrightarrow x=1-2x^2\sqrt{2}\)
Thay vào M ta sẽ được
\(M=x^2+\sqrt{x^4+1-2x^2\sqrt{2}+1}\)
\(=x^2+\sqrt{\left(x^2-\sqrt{2}\right)^2}\)
\(=x^2+\left|x^2-\sqrt{2}\right|\)
Từ \(\left(1\right)\Rightarrow\sqrt{2}-x\sqrt{2}=4x^2\ge0\)
\(\Leftrightarrow\sqrt{2}\left(1-x\right)\ge0\)
\(\Leftrightarrow x\le1\)
\(\Leftrightarrow x^2\le1< \sqrt{2}\)
\(\Rightarrow\left|x^2-\sqrt{2}\right|=\sqrt{2}-x^2\)
Khi đó \(M=x^2+\left|x^2-\sqrt{2}\right|=x^2-\sqrt{2}+x^2=\sqrt{2}\)
|N|
\(x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y\right)\left(x^2-xy+y^2\right)+3xy\left(x+y\right)+z^3-3xyz-3xy\left(x+y\right)=0\)
\(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\x^2+y^2+z^2-xy-yz-xz=0\end{cases}}\)
Mà \(x,y,z>0\Rightarrow x+y+z\ne0\)
\(\Rightarrow x^2+y^2+z^2-xy-xz-yz=0\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}\Leftrightarrow}x=y=z\)
Thay vào biểu thức A ta được :
\(A=\frac{2018x-2019x+2020x}{\sqrt[3]{x^3}}=\frac{2019x}{x}=2019\)
Vậy ...
Thay 2020=x+y+z vao mẫu đc
\(\frac{xy}{\sqrt{xy+zx+zy+z^2}}=\frac{xy}{\sqrt{\left(x+z\right)\left(y+z\right)}}\le\frac{xy}{2}\left(\frac{1}{x+z}+\frac{1}{y+z}\right)\)(Cauchy)
Làm tương tự mấy cái kia sau đó ghép mấy cái cũng mẫu lại là ra
Lời giải:
Đặt mẫu số của $B$ là $M$.
Từ \(2018x^3=2019y^3=2020z^3\)
\(\Rightarrow \sqrt[3]{2018}x=\sqrt[3]{2019}y=\sqrt[3]{2020}z=\frac{\sqrt[3]{2018}}{\frac{1}{x}}=\frac{\sqrt[3]{2019}}{\frac{1}{y}}=\frac{\sqrt[3]{2020}}{\frac{1}{z}}=\frac{\sqrt[3]{2018}+\sqrt[3]{2019}+\sqrt[3]{2020}}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}\)
\(=\frac{\sqrt[3]{2018}+\sqrt[3]{2019}+\sqrt[3]{2020}}{8}=\frac{M}{8}\)
\(\Rightarrow \left\{\begin{matrix} x=\frac{M}{8\sqrt[3]{2018}}\\ y=\frac{M}{8\sqrt[3]{2019}}\\ z=\frac{M}{8\sqrt[3]{2020}}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} 2018x^2=\frac{\sqrt[3]{2018}M^2}{64}\\ 2019y^2=\frac{\sqrt[3]{2019}M^2}{64}\\ 2020z^2=\frac{\sqrt[3]{2020}M^2}{64}\end{matrix}\right.\)
\(\Rightarrow 2018x^2+2019y^2+2020z^2=\frac{M^2(\sqrt[3]{2018}+\sqrt[3]{2019}+\sqrt[3]{2020})}{64}=\frac{M^3}{64}\)
\(\Rightarrow B=\frac{\sqrt[3]{\frac{M^3}{64}}}{M}=\frac{M}{4M}=\frac{1}{4}\)