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1/h=1/2(1/a+1/b)=1/2a+1/2b=(a+b)/2ab
=>(a+b/)2ab-1/h=0
quy dong len ta co
(a+b)h/2abh-2ab/2abh=0=> (ah+bh-2ab)/2abh=0 =>ah+bh-2ab=0
=>ah+bh-ab-ab=0
=>a(h-b)-b(a-h)=0
=>a(h-b)=b(a-h)
=>a/b=(a-h)(h-b)
\(M=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{37.38.39}=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+..+\frac{1}{37.38}-\frac{1}{38.39}\)
\(=\frac{1}{1.2}-\frac{1}{38.39}=\frac{1}{1}-\frac{1}{2}-\frac{1}{38}+\frac{1}{39}=\frac{370}{741}\)
\(A=\frac{\left(1+2+3+...+100\right)\left(\frac{1}{4}+\frac{1}{6}-\frac{1}{2}\right)\left(63.1,2-21.3,6+1\right)}{1-2+3-4+....+99-100}\)
\(=\frac{\frac{100\left(100+1\right)}{2}\left(\frac{3+2-6}{12}\right)\left[63\left(1,2-1,2\right)+1\right]}{\left(1-2\right)+\left(3-4\right)+....+\left(99-100\right)}\)
\(=\frac{5050.\left(-\frac{1}{12}\right).1}{-1+\left(-1\right)+\left(-1\right)+...+\left(-1\right)}\)
\(=\frac{2525.\left(-\frac{1}{6}\right)}{-50}=\frac{101}{12}\)
A=(1-1/1)+(1-1/4)+(1-1/9)+(1/16)+..........+(1-1/100)
=>1-99/100
A=\(\frac{1-2^2}{2^2}.\frac{1-3^2}{3^2}...\frac{1-100^2}{100^2}\)
trong biểu thức trên có 99 số âm nên tích sẽ âm nên ta có thể viết lại như sau:
A=-\(\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}...\frac{100^2-1}{100^2}\),
Chú ý: \(a^2-b^2=\left(a-b\right)\left(a+b\right)\)
do vậy: A=-\(\frac{1.3}{2^2}.\frac{2.4}{3^2}...\frac{99.101}{100^2}=\frac{1.2.3...100.101}{2^2.3^2...100^2}=\frac{-101}{100!}>\frac{-101}{2.101}=\frac{-1}{2}\)
Vậy A>\(-\frac{1}{2}\)
A = \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^6}\)
3A = \(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\)
2A = 3A - A = \(1-\frac{1}{3^6}\)
=> A = \(\frac{1-\frac{1}{3^6}}{2}\)
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^6}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\)
\(\Rightarrow3A-A=1-\frac{1}{3^6}=\frac{3^6}{3^6}-\frac{1}{3^6}=\frac{728}{729}\)
\(\Rightarrow2A=\frac{728}{729}\)
\(\Rightarrow A=\frac{\frac{728}{729}}{2}=\frac{364}{729}\)
Ta có: \(\hept{\begin{cases}\left|x^2-1\right|+2\ge2\\\frac{6}{\left(y+1\right)^2+3}\le\frac{6}{3}=2\end{cases}}\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}x=\pm1\\y=-1\end{cases}}\)
