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Bài 2:
Gọi tử là x
Mẫu là x+6
Theo đề, ta có:
\(\dfrac{x+3}{x+5}=\dfrac{4}{5}\)
=>5x+15=4x+20
=>x=5
(3) \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\)
\(\Leftrightarrow\) \(\dfrac{a^2+b^2}{ab}\ge2\)
\(\Leftrightarrow a^2+b^2\ge2ab\)
\(\Leftrightarrow a^2-2ab+b^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\left(luôn đúng\right)\)
(4)\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a +b+c}\)
\(\Leftrightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(a+b+c\right)\ge9\)
\(\Leftrightarrow1+\dfrac{b}{a}+\dfrac{c}{a}+\dfrac{a}{b}+1+\dfrac{c}{b}+\dfrac{a}{c}+\dfrac{b}{c}+1\ge9\)
\(\Leftrightarrow3+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)\ge3+2+2+2\ge9\) (đpcm)
\(56^2+44^2+2.44.56\)
\(=\left(56+44\right)^2\)
\(=100^2=10000\)
\(\Leftrightarrow\left[{}\begin{matrix}x\left(x+1\right)=x+1\\x\left(x+1\right)=-\left(x+1\right)\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}\left(x+1\right)\left(x-1\right)=0\\\left(x+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)