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Lời giải:
$A=(99-97)+(95-93)+(91-89)+...+(7-5)+(8-1)$
$=\underbrace{2+2+2+...+2}_{24}+7$
$=2\times 24+7$
$=55$
Ta có: 99–97 +95–93+...+7–5+3−1
=(99−97)+(95−93)+...+(7−5)+(3−1)
=2+2+...+2+2 (có 25 số 2)
=2.25
=50
\(A=\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\\ A=\dfrac{35}{6}-\dfrac{31}{6}-\dfrac{19}{6}=-\dfrac{15}{6}\)
a) \(29.31=\left(30-1\right)\left(30+1\right)=30^2-1=900-1=899\)
b) \(33,3^2-2.33,3.3,3+3,3^2=\left(33,3-3,3\right)^2=30^2=900\)
c) \(20,1.19,9=\left(20+0,1\right)\left(20-0,1\right)=20^2-0,1^2=400-0.01=399,99\)
d) \(37^2+2.37.13+13^2=\left(37+13\right)^2=50^2=2500\)
\(=\left(300-52\right)\left(300+52\right)=300^2-52^2\\ =90000-2704=87296\)
1
a)48,124
b)-28,538
c)56,96
d)111,5
Câu 2
a)132,1
b)21,35
Câu 3
a)4,8
b)71,5
a)
\(=\dfrac{4.4.25+4.11.4.25}{29.2.48+71.2.48}=\dfrac{400+400.11}{96.29+71.96}=\dfrac{400\left(1+11\right)}{96\left(29+71\right)}=\dfrac{400.12}{96.100}=\dfrac{2.2.100.12}{12.2.2.2.100}=\dfrac{1}{2}\)
\(\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{3}{7}\right)\times\left(1-\dfrac{3}{10}\right)\times\left(1-\dfrac{3}{97}\right)\times\left(1-\dfrac{3}{100}\right)\)
\(=\dfrac{2}{3}\times\dfrac{4}{7}\times\dfrac{7}{10}\times...\times\dfrac{94}{97}\times\dfrac{97}{100}\)
\(=\dfrac{2\times4\times7\times...\times94\times97}{3\times7\times10\times...\times97\times100}\)
\(=\dfrac{2\times4}{3\times100}=\dfrac{8}{300}=\dfrac{2}{75}\)
\(A=\dfrac{2}{3}\cdot\dfrac{4}{7}\cdot\dfrac{7}{10}\cdot...\cdot\dfrac{94}{97}\cdot\dfrac{97}{100}=\dfrac{2}{3}\cdot\dfrac{1}{25}=\dfrac{2}{75}\)
a) 48 . 19 + 48 . 115 + 134 . 52
= 48 . (19 + 115) + 134 . 52
= 48 . 134 + 134 . 52
= 134 . (48 + 52)
= 134 . 100
= 13400
b) 43 . 37 + 43 . 73 + 57 . 66 + 57 . 44
= 43 . (37 + 73) + 57 . (66 + 44)
= 43 . 110 + 57 . 110
= 110 . (43 + 57)
= 110 . 100
= 11000
k cho mình đã rồi mình giải tiếp nha!
nhưng nếu bạn không giải thì