\(75\%+1,1:\left(\frac{2}{5}-1\frac{1}{2}\right)-\left(\frac{1}{3}\right)^2\)

\(=\frac{3}{4}+\frac{11}{10}:\left(\frac{2}{5}-\frac{3}{2}\right)-\frac{1}{9}\)

=\(\frac{3}{4}+\frac{11}{10}:\frac{-11}{10}-\frac{1}{9}\)

\(=\frac{3}{4}+\left(-1\right)-\frac{1}{9}\)

\(=\frac{27}{36}+\left(\frac{-36}{36}\right)-\frac{4}{36}\)

\(=\frac{-13}{36}\)

chúc bạn học tốt

\(=\left(1+3+...+2021\right)\cdot\left[1001\cdot135\cdot137-1001\cdot135\cdot137\right]\)

\(=\left(1+3+...+2021\right)\cdot0\)

=0

= (1 + 3 + ... + 2021) x (135 x 1001 x 137 - 135 x 137 x 1001)

= (1 + 3 + ... + 2021) x 0

= 0

Tính:

(-2)².3 -(1¹⁰+8):(-3)²

=4.3-(1+8):9

=12-9:9

=12-1

=11

-(+15)-12=-27

0-(+7)=-7

0-(-5)=5

Tk cho mk nha

1. 

-15 - 12 = -15 + (-12) = -27 

0+ (-7)= 0-7 = -7 

0-(-5) = 0 + 5 = 5 

đap số 1 . - 27 

2. -7

3. 5

KHO THE

\(A=\frac{\left[\left(25-1\right):1+1\right]\left(25+1\right)}{2}=325.\)

\(B=\frac{\left[\left(51-3\right):2+1\right]\left(51+3\right)}{2}=675\)

\(C=\frac{\left[\left(81-1\right):4+1\right]\left(81+1\right)}{2}=861\)

Ta có:

$S=\dfrac{5}{1\cdot2\cdot3}+\dfrac{8}{2\cdot3\cdot4}+\dfrac{11}{3\cdot4\cdot5}+\cdots+\dfrac{6026}{2008\cdot2009\cdot2010}$

Nhận thấy tử số có dạng $3n+2$, nên:

$S=\sum_{n=1}^{2008}\dfrac{3n+2}{n(n+1)(n+2)}$

Ta có: $\dfrac{3n+2}{n(n+1)(n+2)}=\dfrac{1}{n(n+1)}+\dfrac{2}{(n+1)(n+2)}$

Do đó:

$S=\left(\dfrac1{1\cdot2}+\dfrac1{2\cdot3}+\cdots+\dfrac1{2008\cdot2009}\right)$

$+2\left(\dfrac1{2\cdot3}+\dfrac1{3\cdot4}+\cdots+\dfrac1{2009\cdot2010}\right)$

Mà: $\dfrac1{n(n+1)}=\dfrac1n-\dfrac1{n+1}$

Nên: $S=\left(1-\dfrac1{2009}\right)+2\left(\dfrac12-\dfrac1{2010}\right)$

$=1-\dfrac1{2009}+1-\dfrac1{1005}$

$=2-\dfrac1{2009}-\dfrac1{1005}$

Vì: $\dfrac1{2009}+\dfrac1{1005}>0$

Nên $S<2$