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Biểu thức thứ nhất
\(\dfrac{8}{5}:\dfrac{6}{5}=\dfrac{8}{6}=\dfrac{4}{3}\)
Biểu thức thứ hai:
\(\dfrac{5}{3}-\dfrac{5}{3}=0\)
Vậy biểu thức thứ nhất lớn hơn biểu thức thứ 2
`a, 3/4 + 1/2 xx 7/2`
`= 3/4 + 7/4`
`=10/4`
`=5/2`
`b, 6/15 - 1/3 : 5/3`
`= 6/15 - 1/3 xx 3/5`
`= 6/15 - 3/15`
`= 3/15`
`=1/5`
`c, x-4/9 = 3/7 : 9/4`
`=> x-4/9= 3/7 xx 4/9`
`=> x-4/9= 12/63`
`=> x-4/9=4/21`
`=> x= 4/21 +4/9`
`=>x= 40/63`
`d, 7/9 xx 3/5 -1/2=1/5`
`->` sao lại bằng có `x` ko vậy ạ?
`a,`
`3/4+1/2 \times 7/2=3/4+7/4=10/4=5/2`
`b,`
`6/15 - 1/3 \div 5/3=6/15-1/5=1/5`
`c,` Tìm x?
`x-4/9=3/7 \div 9/4`
`x-4/9=4/21`
`x=4/21+4/9`
`x=40/63`
`d, 7/9x \times 3/5-1/2=1/5`
`7/9x \times 3/5=1/5+1/2`
`7/9x \times 3/5=7/10`
`7/9x=7/10 \div 3/5`
`7/9x=7/6`
`x=7/6 \div 7/9=3/2`
A)\(\dfrac{7}{20}-\left(\dfrac{5}{8}-\dfrac{2}{5}\right)\)
\(=\dfrac{7}{20}-\left(\dfrac{25}{40}-\dfrac{16}{40}\right)\)
\(=\dfrac{7}{20}-\dfrac{9}{40}\)
\(=\dfrac{14}{40}-\dfrac{9}{40}=\dfrac{5}{40}=\dfrac{1}{8}\)
B) \(\dfrac{5}{6}+\left(\dfrac{5}{9}-\dfrac{1}{4}\right)\)
\(=\dfrac{5}{6}+\left(\dfrac{20}{36}-\dfrac{9}{36}\right)\)
\(=\dfrac{5}{6}+\dfrac{11}{36}\).
\(=\dfrac{30}{36}+\dfrac{11}{36}=\dfrac{41}{36}\)
C) \(\dfrac{9}{10}-\left(\dfrac{2}{5}-\dfrac{3}{10}\right)+\dfrac{7}{20}\)
\(=\dfrac{9}{10}-\left(\dfrac{4}{10}-\dfrac{3}{10}\right)+\dfrac{7}{20}\)
\(=\dfrac{9}{10}-\dfrac{1}{10}+\dfrac{7}{20}\)
\(=\dfrac{18}{20}-\dfrac{2}{20}+\dfrac{7}{20}=\dfrac{23}{20}\)
a: =7/20-5/8+2/5
=14/40-25/40+16/40
=5/40=1/8
b: =5/6+5/9-1/4
=30/36+20/36-9/36
=41/36
c: =9/10-2/5+3/10+7/20
=12/10-2/5+7/20
=7/20+6/5-2/5
=7/20+4/5
=7/20+16/20
=23/20
\(\left(\frac{9}{2}-\frac{5}{2}\right)-\frac{3}{4}=\frac{4}{2}-\frac{3}{4}=\frac{8}{4}-\frac{3}{4}=\frac{5}{4}\)\(\frac{5}{4}\)
\(\frac{9}{2}-\left(\frac{5}{2}+\frac{3}{4}\right)=\frac{9}{2}-\left(\frac{10}{4}+\frac{3}{4}\right)=\frac{9}{2}-\frac{13}{4}=\frac{18}{4}-\frac{13}{4}=\frac{5}{4}\)
Vậy : \(\left(\frac{9}{2}-\frac{5}{2}\right)-\frac{3}{4}=\frac{9}{2}-\left(\frac{5}{2}+\frac{3}{4}\right)\)
tk mk nha
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