3+7=
3+4=
4+6=
<,>,=
3 7
5 3
hi hi giúp mình nhé
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`@` `\text {Ans}`
`\downarrow`
`1/3+6/9+4/5+5/25`
`= 1/3+2/3+4/5+1/5`
`= 1+1`
`= 2`
`4/7*2/5+4/7*3/5-1/7`
`= 4/7*(2/5+3/5)-1/7`
`= 4/7*1-1/7`
`= 4/7-1/7=3/7`
.....................................................
a.\(\dfrac{27}{8}\)
b.\(\dfrac{37}{40}\)
c.\(\dfrac{5}{2}\)
d.\(\dfrac{7}{3}\)
e.5
g.\(\dfrac{53}{16}\)
Bài 1 :
a) \(\dfrac{3}{2}+\dfrac{5}{4}+\dfrac{5}{8}=\dfrac{12}{8}+\dfrac{10}{8}+\dfrac{5}{8}=\dfrac{12+10+5}{8}=\dfrac{27}{8}\)
b) \(\dfrac{4}{5}-\dfrac{3}{8}+\dfrac{2}{4}=\dfrac{32}{40}-\dfrac{15}{40}+\dfrac{20}{40}=\dfrac{32-15+20}{40}=\dfrac{37}{40}\)
c) \(3+\dfrac{6}{8}-\dfrac{5}{4}=\dfrac{3}{1}+\dfrac{6}{8}-\dfrac{5}{4}=\dfrac{24}{8}+\dfrac{6}{8}-\dfrac{10}{8}=\dfrac{20}{8}=\dfrac{5}{2}\)
d) \(\dfrac{5}{6}-\dfrac{1}{2}+2=\dfrac{5}{6}-\dfrac{1}{2}+\dfrac{2}{1}=\dfrac{5}{6}-\dfrac{3}{6}+\dfrac{12}{6}=\dfrac{14}{6}=\dfrac{7}{3}\)
e) \(\dfrac{3}{5}+\dfrac{6}{11}+\dfrac{7}{13}+\dfrac{2}{5}+\dfrac{16}{11}+\dfrac{19}{13}=\left(\dfrac{3}{5}+\dfrac{2}{5}\right)+\left(\dfrac{6}{11}+\dfrac{16}{11}\right)+\left(\dfrac{7}{13}+\dfrac{19}{13}\right)=1+2+2=5\)
g) \(\dfrac{75}{100}+\dfrac{18}{21}+\dfrac{29}{32}+\dfrac{1}{4}+\dfrac{3}{21}+\dfrac{13}{32}=\dfrac{3}{4}+\dfrac{6}{7}+\dfrac{29}{32}+\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{13}{32}=\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+\left(\dfrac{6}{7}+\dfrac{1}{7}\right)+\left(\dfrac{29}{32}+\dfrac{13}{32}\right)=1+1+\dfrac{21}{16}=2+\dfrac{21}{16}=\dfrac{53}{16}\)
Số số hạng của dãy số trên là :
( 101 - 1 ) : 1 +1 = 101 ( số )
tổng trên là :
( 101 + 1 ) x 101 : 2 = 5151
Đáp số 5151
k tin thì bấm máy !! ( dại ) \
Tk nha !!
1. \(14-\left(\frac{17}{2}+\frac{7}{2}\right)=14-12=2\)
2. \(12+64:\left(3-\frac{5}{3}\right)=12+64:\frac{4}{3}=12+\frac{16}{3}=\frac{52}{3}\)
3. \(\left(\frac{2}{3}+\frac{3}{4}-\frac{4}{5}\right).\frac{5}{6}:\frac{7}{8}=\frac{47}{60}.\frac{5}{6}.\frac{8}{7}=\frac{47}{63}\)
Nguyễn Minh Thụ là có sai vài bước đó nhưng k sao , mk vẫn cảm ơn bạn
1: Rút gọn
a) \(-\dfrac{33}{51}=\dfrac{-33:3}{51:3}=\dfrac{-11}{17}\)
b) \(\dfrac{156}{-168}=\dfrac{-156}{168}=\dfrac{-156:12}{168:12}=\dfrac{-13}{14}\)
c) \(\dfrac{-75}{-100}=\dfrac{75}{100}=\dfrac{75:25}{100:25}=\dfrac{3}{4}\)
2: Quy đồng mẫu:
a) \(-\dfrac{3}{8}\) và \(\dfrac{5}{4}\)
MSC: 8
\(-\dfrac{3}{8}=\dfrac{-3}{8}\)
\(\dfrac{5}{4}=\dfrac{5\cdot2}{4\cdot2}=\dfrac{10}{8}\)
b) \(\dfrac{-7}{6}\) ; \(\dfrac{5}{12}\) và \(-\dfrac{5}{6}\)
MSC: 12
\(-\dfrac{7}{6}=\dfrac{-7\cdot2}{6\cdot2}=\dfrac{-14}{12}\)
\(\dfrac{5}{12}=\dfrac{5}{12}\)
\(\dfrac{-5}{6}=\dfrac{-5\cdot2}{6\cdot2}=\dfrac{-10}{12}\)
a) (38 - 60) + (20 - 38)
= 38 - 60 + 20 - 38
= (38 - 38) + (-60 + 20)
= 0 - 40
= -40
b) 75 - (20 + 75)
= 75 - 20 - 75
= (75 - 75) - 20
= 0 - 20
= -20
c) 32 + (60 - 32)
= 32 + 60 - 32
= (32 - 32) + 60
= 0 + 60
= 60
d) (81 - 36) - (81 - 36)
= 81 - 36 - 81 + 36
= (81 - 81) + (-36 + 36)
= 0 + 0
= 0
e) (2 + 4 + 6 + 8) - (1 + 3 + 5 + 7)
= 2 + 4 + 6 + 8 - 1 - 3 - 5 - 7
= (2 - 1) + (4 - 3) + (6 - 5) + (8 - 7)
= 1 + 1 + 1 + 1
= 4
f) (1 + 3 + 5 + 7 + ... + 99) - (2 + 4 + 6 + 8 + ... + 100)
= 1 + 3 + 5 + 7 + ... + 99 - 2 - 4 - 6 - 8 - ... - 100
= (1 - 2) + (3 - 4) + (5 - 6) + (7 - 8) + ... + (99 - 100)
= -1 - 1 - 1 - 1 - ... - 1 (50 chữ số 1)
= -50
a, 13/6+5/8 : -3/4 - 7/12.4
= 13/6 + -5/6-7/3
=8/6-7/3
= -6/6
= -1
b, ( 73/5 - 21/3) + ( 4/3-43/5 )
= 73/5-21/3+4/3-43/5
=( 73/5-43/5)-(21/3-4/3)
= 6-17/3
=1/3
c, 7/5.4/9 +7/5: 9/16- 14/10.2/9
= 7/5.4/9 +7/5.16/9 - 14/45
=7/5.(4/9+16/9)-14/45
=7/5.20/9-14/45
= 140/45 - 14/45
= 126/45
Xong rùi nè! Nhưng bạn kiểm tra lại giùm nhé vì làm vào ban đêm nên hơi bất tiện
a: \(=\dfrac{3}{5}+\dfrac{2}{7}=\dfrac{21+10}{35}=\dfrac{31}{35}\)
b: \(=\dfrac{6}{5}+\dfrac{3}{2}\cdot\dfrac{1}{3}=\dfrac{6}{5}+\dfrac{1}{2}=\dfrac{12+5}{10}=\dfrac{17}{10}\)
c: \(=\dfrac{5\cdot3}{9}=\dfrac{5}{3}\)
d: =5-2/3*6/5
=5-12/15
=5-4/5
=21/5
\(=\frac{\left(x^3\right)^2-\left(y^3\right)^2}{\left[\left(x^2\right)^2-\left(y^2\right)^2\right]-xy\left(x^2-y^2\right)}=\)
\(=\frac{\left(x^3-y^3\right)\left(x^3+y^3\right)}{\left(x^2-y^2\right)\left(x^2+y^2\right)-xy\left(x^2-y^2\right)}=\)
\(=\frac{\left(x+y\right)\left(x^2-xy+y^2\right)\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(x^2-y^2\right)\left(x^2+y^2-xy\right)}=\)
\(=\frac{\left(x^2-y^2\right)\left(x^2-xy+y^2\right)\left(x^2+xy+y^2\right)}{\left(x^2-y^2\right)\left(x^2-xy+y^2\right)}=x^2+xy+y^2\)
3+7=10
3+4=7
4+6=10
3<7
5>3
HT
3+7=10
3+4=7
4+6=10
3<7
5>3