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\(\dfrac{\dfrac{2}{3}.\dfrac{7}{15}-\dfrac{2}{3}.\dfrac{5}{27}}{\dfrac{-9}{9}.\dfrac{7}{15}+\dfrac{4}{9}.\dfrac{5}{27}}\)
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`(6/11 +5/11) xx 3/7`
`= 11/11xx 3/7`
`=1xx3/7`
`=3/7`
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`3/5 xx 7/9 - 3/5 xx 2/9`
`= 3/5 xx (7/9-2/9)`
`= 3/5 xx 5/9`
`= 15/45`
`= 1/3`
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`(6/7 -4/7):2/5`
`= 2/7 : 2/5`
`= 2/7 xx 5/2`
`= 10/14`
`= 5/7`
\(\left(\dfrac{6}{11}+\dfrac{5}{11}\right)\times\dfrac{3}{7}\)
\(=\dfrac{11}{11}\times\dfrac{3}{7}\\ =1\times\dfrac{3}{7}=\dfrac{3}{7}\)
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\(\dfrac{3}{5}\times\dfrac{7}{9}-\dfrac{3}{5}\times\dfrac{2}{9}\\ =\dfrac{3}{5}\times\left(\dfrac{7}{9}-\dfrac{2}{9}\right)\\ =\dfrac{3}{5}\times\dfrac{5}{9}\\ =\dfrac{3}{9}\\ =\dfrac{1}{3}\)
_________
\(\left(\dfrac{6}{7}-\dfrac{4}{7}\right):\dfrac{2}{5}\\ =\dfrac{2}{7}\times\dfrac{5}{2}\\ =\dfrac{10}{14}\\ =\dfrac{5}{7}\)
\(\dfrac{3}{12}+\dfrac{1}{4}=\dfrac{3:3}{12:3}+\dfrac{1}{4}=\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{2}{4}=\dfrac{1}{2}\)
\(\dfrac{4}{10}+\dfrac{3}{5}=\dfrac{4:2}{10:2}+\dfrac{3}{5}=\dfrac{2}{5}+\dfrac{3}{5}=\dfrac{5}{5}=1\)
\(\dfrac{12}{27}+\dfrac{2}{9}=\dfrac{12:3}{27:3}+\dfrac{2}{9}=\dfrac{4}{9}+\dfrac{2}{9}=\dfrac{6}{9}=\dfrac{2}{3}\)
\(\dfrac{7}{3}+\dfrac{20}{15}=\dfrac{7}{3}+\dfrac{20:5}{15:5}=\dfrac{7}{3}+\dfrac{4}{3}=\dfrac{11}{3}\)
a: 4/9+3/7=28/63+27/63=55/63
3/4+7/24=18/24+7/24=25/24
1/3+2/9+4/27=9/27+6/27+4/27=19/27
b: 5/6-3/8=20/24-9/24=11/24
7/15-11/30=14/30-11/30=3/30=1/10
2/3+1/6-7/12
=8/12+2/12-7/12
=3/12=1/4
c: 18/25*15/6=15/25*18/6=3*3/5=9/5
30/49:6/7=30/49*7/6=210/294=5/7
1/2*3/4:6/5=3/8*5/6=15/48=5/16
d: 8*3/5:12/5=24/5*5/12=2
4:9/5:10/3=4*5/9*3/10=2/3
Giải:
a) A=7 4/15-(3 2/9+2 4/15)
A=109/15-29/9-34/15
A=(109/15-34/15)-29/9
A=5-29/9
A=16/9
b) B=4 1/4+3 1/5
B=17/4+16/5
B=149/20
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Gợi ý: Sử dụng tính chất phân phối của phép nhân đối với phép cộng để nhóm thừa số chung ra ngoài.
\(a,=\dfrac{3^6\cdot5^4\cdot9^4-5^{13}\cdot3^{13}\cdot5^{-9}}{3^{12}\cdot5^6+9^6\cdot5^6}=\dfrac{3^{14}\cdot5^4-5^4\cdot3^{13}}{3^{12}\cdot5^6+3^{12}\cdot5^6}\\ =\dfrac{3^{13}\cdot5^4\cdot2}{2\cdot3^{12}\cdot5^6}=\dfrac{3}{5^2}=\dfrac{3}{25}\)
\(b,=\dfrac{\left(\dfrac{2}{5}\cdot5\right)^7+\left(\dfrac{9}{4}\cdot\dfrac{16}{3}\right)^3}{2^7\cdot5^2+2^9}=\dfrac{2^7+12^3}{2^7\left(5^2+2^2\right)}=\dfrac{2^7+4^3\cdot3^3}{2^7\cdot29}=\dfrac{2^6\left(2+3^3\right)}{2^7\cdot29}=\dfrac{1}{2}\)
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