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\(xy\le\frac{\left(x+y\right)^2}{4}=\frac{1}{4}.\)
\(A=x^2y^2+\frac{1}{x^2y^2}+\frac{x^2}{y^2}+\frac{y^2}{x^2}=\left(x^2y^2+\frac{1}{256x^2y^2}\right)+\left(\frac{x^2}{y^2}+\frac{y^2}{x^2}\right)+\frac{255}{256x^2y^2}\)
\(\ge2\sqrt{x^2y^2.\frac{1}{256x^2y^2}}+2\sqrt{\frac{x^2}{y^2}.\frac{y^2}{x^2}}+\frac{255}{256.\left(\frac{1}{4}\right)^2}=\frac{289}{16}.\)
\("="\text{ }for\text{ }x=y=\frac{1}{2}.\)
\(x=\frac{\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5-1}\right)^2}}{\sqrt{20}}=\frac{2\sqrt{5}}{\sqrt{20}}=1\)
=>P=(1+1-1)2016=1
Điều kiện có 2 nghiệm phân biệt tự làm nha
Theo vi-et ta có:
\(\hept{\begin{cases}x_1+x_2=5\\x_1.x_2=m-2\end{cases}}\)
\(2\left(\frac{1}{\sqrt{x_1}}+\frac{1}{\sqrt{x_2}}\right)=3\)
\(\Leftrightarrow4\left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{2}{\sqrt{x_1.x_2}}\right)=9\)
\(\Leftrightarrow4\left(\frac{5}{m-2}+\frac{2}{\sqrt{m-2}}\right)=9\)
Làm nốt nhé
Câu 1:
M=\(\left(x^2+2xy+y^2\right)+\left(2x+2y\right)+1+\left(4x^2-4x+1\right)+2014\)
=\(\left(\left(x+y\right)^2+2\left(x+y\right)+1\right)+\left(2x-1\right)^2+2014\)
=\(\left(x+y+1\right)^2+\left(2x-1\right)^2+2014\ge2014\)
\(\Rightarrow M\ge2014\Leftrightarrow minM=2014\)
\(\Leftrightarrow\hept{\begin{cases}x+y+1=0\\2x-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0,5\\y=1,5\end{cases}}\)
đkxđ là \(x\ne1;x>0\)
\(Q=\frac{\sqrt{x}\left(\left(\sqrt{x}\right)^3-1\right)}{x+\sqrt{x}+1}-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(Q=\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-2\sqrt{x}-1+\frac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(Q=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2=x-\sqrt{x}+1\)
gtnn \(x-\sqrt{x}+1=x-\frac{1}{2}.2.\sqrt{x}+\frac{1}{4}+\frac{3}{4}=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
gtnn 3/4
ý c bạn tự làm nha mk chịu
Ta có : \(A=\frac{x^2-2x+2018}{x^2}=\frac{2018x^2-4036x+2018^2}{x^2}=\frac{2017x^2}{x^2}+\frac{x^2-4036x+2018^2}{x^2}\)
\(=2017+\frac{\left(x-2018\right)^2}{x^2}\)
Vì \(\frac{\left(x-2018\right)^2}{x^2}\ge0\forall x\)
Nên : \(A=2017+\frac{\left(x-2018\right)^2}{x^2}\ge2017\)
Vậy \(A_{max}=2017\) khi x = 2018
Cảm ơn bn nha