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\(5+\frac{3}{5}=\frac{25}{5}+\frac{3}{5}=\frac{28}{5}\)
\(10-\frac{9}{16}=\frac{160}{16}-\frac{9}{16}=\frac{151}{16}\)
\(\frac{2}{3}-\left(\frac{1}{6}+\frac{1}{8}\right)=\frac{2}{3}-\frac{1}{6}-\frac{1}{8}=\frac{16}{24}-\frac{4}{24}-\frac{3}{24}=\frac{9}{24}=\frac{3}{8}\)
1-2+3-4+5-6+...+99-100+101
= (1+3+5+...+101) - (2+4+6+...+100)
tu 1 den 101 co : (101-1):2+1=51
1+..+101 = (1+101)x 51:2= 2601
tu 2 den 100 co : (100-2);2+1=50
2+...+100 = (100 +2) x 50:2=2550
=> A= 2601-2550=51
học tốt
\(\dfrac{4}{3}:\dfrac{2}{5}=\dfrac{4}{3}\cdot\dfrac{5}{2}=\dfrac{20}{6}=\dfrac{10}{3}\)
\(\dfrac{1}{8}:7=\dfrac{1}{8\cdot7}=\dfrac{1}{56}\)
\(3:\dfrac{4}{5}=3\cdot\dfrac{5}{4}=\dfrac{15}{4}\)
\(\dfrac{4}{5}:3=\dfrac{4}{5\cdot3}=\dfrac{4}{15}\)
1-2+3-4+5-6+.....+99-100+101.
Ta viết lại tổng như sau:
101 - 100 + 99 - 98 + ... + 5 - 4 + 3 - 2 + 1
1 + 1 + ... + 1 + 1 + 1
Số phép trừ trong dãy tính là:
( 101 - 1 ) : 2 = 50 ( phép trừ )
Kết quả dãy số là:
1 x 50 + 1 = 51
Vậy: 1-2+3-4+5-6+.....+99-100+101.
= 51
a) y + 3/7 x 21/9 = 5
y + 3/7 x 7/3 = 5
y + 1 = 5
y = 4
b) y x 3/5 - 1/2 = 4/5
y x 3/5 = 4/5 + 1/2
y x 3/5 = 13/10
y = 13/6
c) 9/8 - y + 2/5 = 1/4
y - 2/5 = 9 /8 - 1/4
y - 2/5 = 7/8
y = 51/40
d) y - 6/5 : 2/3 = 5/6
y - 9/5 = 5/6
y = 79/30
e) y : 7/8 + 3/4 = 7/4
y : 7/8 = 1
y = 7/8
\(a.\dfrac{3}{4}\times\dfrac{5}{6}-\dfrac{1}{6}=\dfrac{5}{8}-\dfrac{1}{6}=\dfrac{11}{24}\)
\(b.\left(\dfrac{8}{9}+\dfrac{1}{3}\right):\dfrac{2}{5}=\dfrac{11}{9}:\dfrac{2}{5}=\dfrac{11}{9}\times\dfrac{5}{2}=\dfrac{55}{18}\)
a) 2/5 + 1/2 + 7/10
= 4/10 + 5/10 + 7/10
= 9/10 + 7/10
= 16/10 = 8/5
b) 4/9 + 11/8 - 5/6
= 32/72 + 99/72 - 60/72
= 191/72
c) 9/20 - 8/15 x 5/12
= 9/20 - 2/9
= 41/180
d) 2/3 : 4/5 : 7/12
= 5/6 : 7/12
= 10/7
5/8 x 6 + 1/3 = 15/4 + 1/3 = 49/12
5/4- 1/3 = 15/12 - 4/12 = 11/12
a) = 15/4 + 1/3 = 49/12
b) = 5/4 - 1/3 = 11/12