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C/m nó nhỏ hơn 3/4 hả bạn ?
Có \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{100}< \frac{3}{4}\)
\(3^6:3^2+2^3.2^2-3^3.3\)
\(=3^4+2^5-3^4\)
\(=3^4-3^4+2^5\)
\(=0+2^5=2^5\)
\(3^6:3^2+2^3.2^2-3^3.3\\ =3^4+2-3^4\\ =\left(3^4-3^4\right)+2\\ =0+2\\ =2.\)
Ta có: \(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)
\(\frac{1}{4^2}<\frac{1}{3\cdot4}=\frac13-\frac14\)
...
\(\frac{1}{100^2}<\frac{1}{99\cdot100}=\frac{1}{99}-\frac{1}{100}\)
Do đó: \(\frac{1}{3^2}+\frac{1}{4^2}+\cdots+\frac{1}{100^2}<\frac12-\frac13+\frac13-\frac14+\cdots+\frac{1}{99}-\frac{1}{100}=\frac12-\frac{1}{100}<\frac12\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{100^2}<\frac14+\frac12=\frac34\)
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2020^2}\)
\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(=1-\frac{1}{2020}< 1\)
a)31x32x33x........x3100
=31+2+3+4+...+100
=3(100+1)x(100-1+1):2
=3101x100:2
=35050
Bài b mình không biết làm
A = 1.2 + 2.3 + 3.4 + ...+ 59.60
3A = 1.2.3 + 2.3.3 + 3.4.3 + ...+ 59.60.3
3A = 1.2.(3-0) + 2.3.(4-1) + 3.4.(5-2) +...+ 59.60.(61-58)
3A = 1.2.3 - 0 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 +...+ 59.60.61 - 58.59.60
3A = 58.59.60 => A = 58.59.60 : 3 = 68 440
B = 12 + 22 + 32 + 592
B = 1.2 + 2.2 + 3.3 + 59.59
B = 2 + 4 + 9 + 3481
B = 3496
vậy A - B = 68 440 - 3 496 = 64 944
( bấm nhé )
a: \(=\dfrac{6\left(12-7\right)}{60}=\dfrac{6\cdot5}{6\cdot10}=\dfrac{5}{10}=\dfrac{1}{2}\)
b: \(=\dfrac{35\cdot\left(18-1\right)}{34\cdot7}=5\cdot\dfrac{1}{2}=\dfrac{5}{2}\)
c: \(=\dfrac{42\left(1-11\right)}{21\cdot\left(-15\right)}=2\cdot\dfrac{-10}{-15}=2\cdot\dfrac{2}{3}=\dfrac{4}{3}\)
d: \(=\dfrac{2^5\cdot3^2}{2^5\cdot3^3}=\dfrac{1}{3}\)
e: \(=\dfrac{-36\cdot14}{27\left(14+7\right)}=\dfrac{-36}{27}\cdot\dfrac{14}{21}=\dfrac{-4}{3}\cdot\dfrac{2}{3}=-\dfrac{8}{9}\)
Sửa đề: \(\frac{5\cdot\left(2^2\cdot3^2\right)^9\cdot\left(2^2\right)^6-2\cdot\left(2^2\cdot3\right)^{14}\cdot3^6}{5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)
Ta có: \(5\cdot\left(2^2\cdot3^2\right)^9\cdot\left(2^2\right)^6-2\cdot\left(2^2\cdot3\right)^{14}\cdot3^6\)
\(=5\cdot2^{18}\cdot3^{18}\cdot2^{12}-2\cdot2^{28}\cdot3^{14}\cdot3^6\)
\(=5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}=2^{29}\cdot3^{18}\left(5\cdot2-3^2\right)=2^{29}\cdot3^{18}\)
Ta có: \(5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}\)
\(=2^{28}\cdot3^{18}\left(5\cdot3-7\cdot2\right)=2^{28}\cdot3^{18}\)
Ta có: \(\frac{5\cdot\left(2^2\cdot3^2\right)^9\cdot\left(2^2\right)^6-2\cdot\left(2^2\cdot3\right)^{14}\cdot3^6}{5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)
\(=\frac{2^{29}\cdot3^{18}}{2^{28}\cdot3^{18}}=2\)