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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{99}{100}\)
\(=\frac{1\cdot2\cdot3\cdot...\cdot99}{2\cdot3\cdot4\cdot...\cdot100}\)
\(=\frac{1}{100}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1:
\(4\sqrt[4]{\left(a+1\right)\left(b+4\right)\left(c-2\right)\left(d-3\right)}\le a+1+b+4+c-2+d-3=a+b+c+d\)
Dấu = xảy ra khi a = -1; b = -4; c = 2; d= 3
\(\frac{a^2}{b^5}+\frac{1}{a^2b}\ge\frac{2}{b^3}\)\(\Leftrightarrow\)\(\frac{a^2}{b^5}\ge\frac{2}{b^3}-\frac{1}{a^2b}\)
\(\frac{2}{a^3}+\frac{1}{b^3}\ge\frac{3}{a^2b}\)\(\Leftrightarrow\)\(\frac{1}{a^2b}\le\frac{2}{3a^3}+\frac{1}{3b^3}\)
\(\Rightarrow\)\(\Sigma\frac{a^2}{b^5}\ge\Sigma\left(\frac{5}{3b^3}-\frac{2}{3a^3}\right)=\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{1}{d^3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 3/4x16/9-7/5:(-21/20)
=4/3-(-4/3)
=8/3
b)=7/3-1/3x[-3/2+(2/3+2)]
=7/3-1/3x[-3/2+8/3]
=7/3-1/3x7/6
=7/3-7/18
=35/18
c)=(20+37/4):9/4
=117/4:9/4
=13
d)=6-14/5x25/8-8/5:1/4
=6-35/4-32/5
=-11/4-32/5
=-183/20
Hoàng Lê Bảo Ngọc alibaba nguyễn Thắng Nguyễn giup e vs