32 . (132 - 247) - 132 . (32 - 247)
= 32 . 132 - 32 . 247 - 132 . 32 + 132 . 247
= (32 . 132 - 132 . 32) + (132 . 247 - 32 . 247)
= [32 . (132 - 132)] + (132 . 247 - 32 . 247)
= [32 . 0] + (132 . 247 - 32 . 247)
= 0 + [247 . (132 - 32)]
= 0 + [247 . 100]
= 0 + 24700
= 24700
32 . (132 - 247) - 132 . (32 - 247)
= 32. 132 -32.247 - (132.32 -132.247)
= 32. 132 -32.247 - 132.32 +132.247
=(32. 132 -132.32 ) + (132.247 - 32.247)
= 0 + 247.(132-32)
=247.100
=24700
50[( 2 . 32 2 - 8 ): 2 +30 ]-215
4.55 - 32 : 24 -452 -(-67 + 75-452)
giup vs
tìm n thuộc N: 32-7n chia hết n
52. 63 + 52.37 - 250
b. 390- { 400 :[ 32.52 – (200 - 7. 52)]
c. (-37). 85 + 37.(-5) - 370
c. (-72): 32 – (- 18).2
-2-13+(-14)-19
221+4[(-5).8-4]
(-2)^3.(-2)^2+32
-15.12-8.(-12)
-2-13+(14)-19
221+4[(-5).8-4]
(-2)^3.(-2)^2+32
-15.12-8.(-12)
\(A=\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}>1\)
\(\dfrac{2n+5}{n+2}\)
\(S=\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+L+\dfrac{1}{29}\)
\(A=\dfrac{7}{4}.\left(\dfrac{3333}{1212}+\dfrac{3333}{2020}+\dfrac{3333}{3030}+\dfrac{3333}{4242}\right)\)
\(A=\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{48.49}+\dfrac{1}{49.50}\)
\(A=\dfrac{17}{1.3}+\dfrac{17}{3.5}+\dfrac{17}{5.7}+...+\dfrac{17}{49.51}\)
\(\dfrac{2n+5}{3n+1}\)
\(\left(\left|x\right|+\dfrac{2}{5}\right):\dfrac{2}{5}=1\)
\(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}< 1\)
\(A=\dfrac{4}{1.3}+\dfrac{4}{3.5}+\dfrac{4}{5.7}+...+\dfrac{4}{99.101}\)
\(5.\left(\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{10}\right)\le\dfrac{x}{20}\le-3\left(\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{5}\right)\left(x\in Z\right)\)
\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}< 1\)
Thay một thừa số bằng tổng để tính :
a) - 15.32
b)-75.18
f)95. (-28)
Thay một thừa số bằng hiệu để tính :
a) -43.99
b) -45.(-49)
