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ta có \(25x^2-20ax+5a^2=25x^2-20ax+4a^2+a^2=\left(5x-2a\right)^2+a^2\ge a^2\)
=>\(\frac{a^2}{25x^2-20ax+5a^2}\le\frac{a^2}{a^2}=1\Rightarrow P\le1\)
dấu = xảy ra <=> x=2/5.a
em gửi bài qua fb thầy chữa cho nhé, tìm fb của thầy bằng sđt: 0975705122 nhé.
Ta có: \(2x^2+\frac{y^2}{4}+\frac{1}{x^2}=4\)
=> \(\left(x^2+\frac{y^2}{4}\right)+\left(x^2+\frac{1}{x^2}\right)=4\)
Lại có: \(x^2+\frac{y^2}{4}\ge2.x.\frac{y}{2}=xy\) Và \(x^2+\frac{1}{x^2}\ge2.x.\frac{1}{x}=2\)
=> \(4\ge xy+2\)=> \(2\ge xy\)
=> \(A=2016+xy\le2016+2=2018\)
=> Amin=2018
\(\sqrt[]{\sqrt{ }\frac{ }{ }\sqrt[]{}3\hept{\begin{cases}\\\\\end{cases}}3\frac{ }{ }\sqrt{ }\cos\hept{\begin{cases}\\\\\end{cases}}\Omega3\cong}\)
Bài giải:
a) 2 – 25x2 = 0 => (√2)2 – (5x)2 = 0
=> (√2 – 5x)( √2 + 5x) = 0
Hoặc √2 – 5x = 0 => 5x = √2 => x = √2525
Hoặc √2 + 5x = 0 => 5x = -√2 => x = - √2525
b) x2 - x + 1414 = 0 => x2 – 2 . x . 1212 + (1212)2 = 0
=> (x - 1212)2 = 0 => x - 1212 = 0 => x = 12
\(a,x^3-13x=0\)
\(x.\left(x^2-13\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^2=13\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\sqrt{13}\end{cases}}}\)
\(b,2-25x^2=0\)
\(\Rightarrow25x^2=2\Rightarrow x^2=\frac{2}{25}\Rightarrow x=\sqrt{\frac{2}{25}}\)
\(c,x^2-x+\frac{1}{4}=0\)
\(\left(x-\frac{1}{2}\right)^2=0\Rightarrow x=\frac{1}{2}\)
a, x 3 - 13 x = 0
=> x ( x 2 - 13 ) = 0
=> \(\orbr{\begin{cases}x=0\\x^2=13\end{cases}\Rightarrow[\begin{cases}x=0\\x=\sqrt{13}\\x=-\sqrt{13}\end{cases}}\)
b, 2 - 25 x 2 = 0
=> 25 x 2 = 2
=> x 2 = 0,08
=> \(\orbr{\begin{cases}x=\frac{\sqrt{2}}{5}\\x=\frac{-\sqrt{2}}{5}\end{cases}}\)
x, x 2 - x + \(\frac{1}{4}\)= 0
=> \(\left(x-\frac{1}{2}\right)^2=0\)
=> \(x-\frac{1}{2}=0\)
=> \(x=\frac{1}{2}\)
ta có \(4=2a^2+\frac{b^2}{4}+\frac{1}{a^2}=a^2+a^2+\frac{b^2}{4}+\frac{1}{a^2}\ge4\sqrt[4]{\frac{a^2.a^2.b^2}{4a^2}}\)
Vậy\(\sqrt[4]{\frac{a^2b^2}{4}}\le1\Leftrightarrow a^2b^2\le4\Leftrightarrow-2\le ab\le2\)
Vậy \(2007\le ab+2009\le2011\)
a: Ta có: \(A=\left(\dfrac{4x}{\left(x-2\right)\left(x+2\right)}+\dfrac{2x-4}{x+2}\right)\cdot\dfrac{x+2}{2x}-\dfrac{2}{x-2}\)
\(=\dfrac{4x+2\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{2x}-\dfrac{2}{x-2}\)
\(=\dfrac{4x+2x^2-8x+8}{x-2}\cdot\dfrac{1}{2x}-\dfrac{2}{x-2}\)
\(=\dfrac{2x^2-12x+8}{2x\left(x-2\right)}-\dfrac{2}{x-2}\)
\(=\dfrac{2x^2-12x+8-4x}{2x\left(x-2\right)}=\dfrac{2x^2-16x+8}{2x\left(x-2\right)}\)
\(=\dfrac{x^2-8x+4}{x\left(x-2\right)}\)
b: Thay x=4 vào A, ta được:
\(A=\dfrac{4^2-8\cdot4+4}{4\cdot\left(4-2\right)}=\dfrac{-12}{4\cdot2}=\dfrac{-12}{8}=-\dfrac{3}{2}\)
\(A=\dfrac{a^2}{25x^2-20ax+5a^2}\\ =\dfrac{a^2}{\left(5x\right)^2-2.5x.2a+4a^2+a^2}\\ =\dfrac{a^2}{\left(5x-2a\right)^2+a^2}\)
rồi sao bn, tìm GTLN mak