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Đây này: chứng minh $(a^m-1, a^n-1)=a^{(m,n)}-1$ - Số học - Diễn đàn Toán học
Chưa lên mạng thì hỏi làm gì?
\(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=a+b+c+2\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)=4\)
\(\Leftrightarrow\sqrt{ab}+\sqrt{ac}+\sqrt{bc}=1\)
\(\Rightarrow a+1=a+\sqrt{ab}+\sqrt{ac}+\sqrt{bc}=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\)
Tương tự: \(b+1=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+\sqrt{c}\right)\)
\(c+1=\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)\)
\(VT=\sum\dfrac{\sqrt{a}}{a+1}=\sum\dfrac{\sqrt{a}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)}\)
\(=\dfrac{\sqrt{a}\left(\sqrt{b}+\sqrt{c}\right)+\sqrt{b}\left(\sqrt{a}+\sqrt{c}\right)+\sqrt{c}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}\)
\(=\dfrac{2\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}=\dfrac{2}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}\)
\(VP=\dfrac{2}{\sqrt{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}=\dfrac{2}{\sqrt{\left(\sqrt{a}+\sqrt{b}\right)^2\left(\sqrt{a}+\sqrt{c}\right)^2\left(\sqrt{b}+\sqrt{c}\right)^2}}\)
\(=\dfrac{2}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}\)
\(\Rightarrow VT=VP\) (đpcm)
a) \(\left(a-b\right)^2\ge0\forall a,b\)
\(\Rightarrow a^2+b^2\ge2ab\)
\(\Rightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)
\(\Rightarrow\sqrt{2\left(a^2+b^2\right)}\ge\sqrt{\left(a+b\right)^2}=\left|a+b\right|\)
Dấu "=" \(\Leftrightarrow a=b\)
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Bài 1:
a: \(=\dfrac{1}{mn^2}\cdot\dfrac{n^2\cdot\left(-m\right)}{\sqrt{5}}=\dfrac{-\sqrt{5}}{5}\)
b: \(=\dfrac{m^2}{\left|2m-3\right|}=\dfrac{m^2}{3-2m}\)
c: \(=\left(\sqrt{a}+1\right):\dfrac{\left(a-1\right)^2}{\left(1-\sqrt{a}\right)}=\dfrac{-\left(a-1\right)}{\left(a-1\right)^2}=\dfrac{-1}{a-1}\)
Đặt \(\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)=\left(x,y,z\right)\)
\(x+y+z\ge\frac{x^2+2xy}{2x+y}+\frac{y^2+2yz}{2y+z}+\frac{z^2+2zx}{2z+x}\)
\(\Leftrightarrow x+y+z\ge\frac{3xy}{2x+y}+\frac{3yz}{2y+z}+\frac{3zx}{2z+x}\)
\(\frac{3xy}{2x+y}\le\frac{3}{9}xy\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}\right)=\frac{1}{3}\left(x+2y\right)\)
\(\Rightarrow\Sigma_{cyc}\frac{3xy}{2x+y}\le\frac{1}{3}\left[\left(x+2y\right)+\left(y+2z\right)+\left(z+2x\right)\right]=x+y+z\)
Dấu "=" xảy ra khi x=y=z
\(b^2+c^2\le a^2\Leftrightarrow\left(\frac{b}{a}\right)^2+\left(\frac{c}{a}\right)^2\le1\)
Đặt \(\left\{{}\begin{matrix}\left(\frac{b}{a}\right)^2=x\\\left(\frac{c}{a}\right)^2=y\end{matrix}\right.\) \(\Rightarrow x+y\le1\)
\(P=\left(\frac{b}{a}\right)^2+\left(\frac{c}{a}\right)^2+\left(\frac{a}{b}\right)^2+\left(\frac{a}{c}\right)^2=x+y+\frac{1}{x}+\frac{1}{y}\)
\(P=x+\frac{1}{4x}+y+\frac{1}{4y}+\frac{3}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\ge2\sqrt{\frac{x}{4x}}+2\sqrt{\frac{y}{4y}}+\frac{3}{4}.\frac{4}{\left(x+y\right)}\)
\(P\ge2+\frac{3}{\left(x+y\right)}\ge2+\frac{3}{1}=5\) (đpcm)
Dấu "=" xảy ra khi \(x=y=\frac{1}{2}\) hay \(\left(\frac{b}{a}\right)^2=\left(\frac{c}{a}\right)^2=\frac{1}{2}\Rightarrow b=c=\frac{a}{\sqrt{2}}\)