Tính
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\)\(\frac{1}{100\sqrt{99}+99\sqrt{100}}\)
K
Khách
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18 tháng 9 2019
Tớ giải bừa
\(\left(a-b\right)^2\left(\sqrt{\frac{a+b}{a-b}}+1\right)\left(\sqrt{\frac{a+b}{a-b}}-1\right)\)
\(=\left(a-b\right)^2\left(\sqrt{\frac{a+b}{a-b}}\right)^2-1^2\)
\(=\left(a-b\right)^2\left(\frac{a+b}{a-b}-1\right)\)
\(=2ab-2b^2\)
18 tháng 9 2019
E=\(|x-2005|+|2006-x|\ge|x-2005+2006-x|=1\)
Dấu = xảy ra khi \(\orbr{\begin{cases}x\ge2006\\x\le2005\end{cases}}\)
LT
tính giá trị của biểu thức \(\left(\sqrt{\frac{3}{4}}-\sqrt{3}+5\sqrt{\frac{4}{3}}\right)\sqrt{12}\)
1
A
18 tháng 9 2019
\(\left(\sqrt{\frac{3}{4}}-\sqrt{3}+5\cdot\sqrt{\frac{4}{3}}\right)\sqrt{12}.\)
\(=\left(\frac{\sqrt{3}}{\sqrt{4}}-\sqrt{3}+5\cdot\frac{\sqrt{4}}{\sqrt{3}}\right)\sqrt{12}.\)
\(=\left(\frac{\sqrt{3}}{2}-\sqrt{3}+5\cdot\frac{2}{\sqrt{3}}\right)\sqrt{12}.\)
\(=\left(\frac{1}{2}\cdot\sqrt{3}-\sqrt{3}+5\cdot\frac{2}{\sqrt{3}}\right)\sqrt{12}.\)
\(=\left(-\frac{1}{2}\sqrt{3}+\frac{10}{\sqrt{3}}\right)\sqrt{12}\)
\(=\left(-\frac{1}{2}\sqrt{3}+\frac{10}{3}\sqrt{3}\right)\sqrt{12}\)
\(=\frac{17}{6}\sqrt{3}\sqrt{12}=\frac{17}{6}\sqrt{36}=\frac{17}{6}\cdot6=17\)
\(\Rightarrow\left(\sqrt{\frac{3}{4}}-\sqrt{3}+5\cdot\sqrt{\frac{4}{3}}\right)\sqrt{12}=17\)
DU
0
N
18 tháng 9 2019
Ta co:
\(P\ge21\left(a^2+b^2+c^2\right)+12\left(a+b+c\right)^2+\frac{2017.9}{2}\)
\(=21\left(a^2+b^2+c^2\right)+12\left(a+b+c\right)^2+\frac{18153}{2}\)
\(\Leftrightarrow\frac{P}{\left(a+b+c\right)^2}\ge21\left[\left(\frac{a}{a+b+c}\right)^2+\left(\frac{b}{a+b+c}\right)^2+\left(\frac{c}{a+b+c}\right)^2\right]+12+\frac{\frac{18153}{2}}{\left(a+b+c\right)^2}\)
Dat \(\left(\frac{a}{a+b+c};\frac{b}{a+b+c};\frac{c}{a+b+c}\right)\rightarrow\left(x;y;z\right)\)
\(\Rightarrow x+y+z=1\)
\(\Rightarrow\left(a+b+c\right)^2=\frac{a^2}{x^2}\)
BDT tro thanh:
\(\frac{P}{\left(a+b+c\right)^2}\ge21\left(x^2+y^2+z^2\right)+12+\frac{18153}{2\left(a+b+c\right)^2}\)
\(\Leftrightarrow\frac{P}{\frac{a^2}{x^2}}\ge21\left(x^2+y^2+z^2\right)+12+\frac{18153}{2\left(a+b+c\right)^2}\ge21.\frac{\left(x+y+z\right)^2}{3}+12+\frac{18153}{8}\)
\(\Leftrightarrow\frac{x^2P}{a^2}\ge7+12+\frac{18153}{8}\)
Ta lai co:\(x=\frac{a}{a+b+c}\ge\frac{a}{2}\Rightarrow a^2\le4x^2\)
Suy ra:\(\frac{x^2P}{a^2}\ge\frac{x^2P}{4x^2}=\frac{P}{4}\)
\(\Rightarrow\frac{P}{4}\ge\frac{18503}{8}\)
\(\Leftrightarrow P\ge\frac{18503}{2}\)
Dau '=' xay ra khi \(a=b=c=\frac{2}{3}\)
Vay \(P_{min}=\frac{18503}{2}\)khi \(a=b=c=\frac{2}{3}\)