Ta có : x + y = 1 => y = 1 - x

Do đó: \(0\le x\le1\)

\(A=x^2+\left(1-x\right)^2=2x^2-2x+1\)

\(=2\left(x-\frac{1}{2}\right)^2+\frac{1}{2}\ge\frac{1}{2}\)

Min A = 1/2

Dấu = xảy ra khi: \(x=y=\frac{1}{2}\)

Do \(0\le x\le1\) nên \(x\left(x-1\right)\le0\)

\(\Rightarrow A=2x\left(x-1\right)+1\le1\)

Max A =1

Dấu = xảy ra khi: \(\orbr{\begin{cases}x=1\Rightarrow y=0\\x=0\Rightarrow y=1\end{cases}}\)

=.= hok tốt!!

nguyên tín??

\(\sqrt{4x^2-4x+1}< 5-x\)

\(\Leftrightarrow\sqrt{\left(2x-1\right)^2}< 5-x\)

\(\Leftrightarrow\left|2x-1\right|< 5-x\)⁽¹⁾

Đk : \(5-x\ge0\Leftrightarrow x\le5\)

(1)\(\Rightarrow\orbr{\begin{cases}2x-1=5-x\\2x-1=x-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\left(n\right)\\x=-4\left(n\right)\end{cases}}}\)

Vậy \(x\in\left\{-4;2\right\}\)

\(\sqrt{\frac{27}{25}}.\sqrt{\frac{44}{189}}.\sqrt{\frac{700}{99}}\)

\(=\sqrt{\frac{27}{25}.\frac{44}{189}.\frac{700}{99}}\)

\(=\sqrt{\frac{16}{9}}\)

\(=\frac{4}{3}\)

học tốt

b,Ap dung bdt cauchy schwarz dang engel ta co

\(B=\frac{x^2}{1}+\frac{y^2}{1}+\frac{z^2}{1}>=\frac{\left(x+y+z\right)^2}{3}=\frac{a^2}{3}\)

xay ra dau = khi x=y=z=a/3

Sorry nha nhưng em mới học lớp 7 thôi à ~~

\(1=x+y=\frac{x}{2}+\frac{x}{2}+\frac{y}{3}+\frac{y}{3}+\frac{y}{3}\ge5\sqrt[5]{\left(\frac{x}{2}\right)^2\left(\frac{y}{3}\right)^3}\)

\(\Leftrightarrow1\ge5\sqrt[5]{\frac{x^2y^3}{108}}\Rightarrow\frac{1}{5}\ge\sqrt[5]{\frac{x^2y^3}{108}}\Rightarrow\frac{x^2y^3}{108}\le\frac{1}{3125}\)

\(\Rightarrow x^2y^3\le\frac{108}{3125}\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}\frac{x}{2}=\frac{y}{3}\\x+y=1\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{2}{5}\\y=\frac{3}{5}\end{cases}}}\)
Vậy...