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\(\dfrac{7}{5}+\dfrac{4}{7}-\dfrac{9}{10}=\dfrac{49-20}{35}-\dfrac{9}{10}=\dfrac{19}{35}-\dfrac{9}{10}=\dfrac{190-315}{350}=\dfrac{-125}{350}\)
\(\dfrac{2}{1}+\dfrac{3}{4}\text{×}\dfrac{8}{5}=\dfrac{8+3}{4}\text{×}\dfrac{8}{5}=\dfrac{11\text{×}8}{4\text{×}5}=\dfrac{88}{20}\)
mấy câu kia áp dụng là dc!
a)
`127+246+273+354`
`=(127+273)+(246+354)`
`=400+600`
`=1000`
b)
`1,58+3,04+6,96+3,42`
`=(1,58+3,42)+(3,04+6,96)`
`=5+10`
`=15`
c)
`1/2+1/3+1/5+1/6`
`=(1/2+1/3+1/6)+1/5`
`=(3/6+2/6+1/6)+1/5`
`=6/6+1/5`
`=1+1/5`
`=5/5+1/5`
`=6/5`
a, 127 + 246 + 273 + 354
= ( 127 + 273) + ( 246 + 354)
= 400 + 600
= 1000
b, 1,58 + 3,04 + 6,96 + 3,42
= ( 1,58 + 3,42) + ( 3,04 + 6,96)
= 5 + 10
= 15
c, \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + \(\dfrac{1}{5}\) + \(\dfrac{1}{6}\)
= ( \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\)) + 0,2
= 1 + 0,2
= 1,2
Bài 1 :
a, 5/7 - 3/6 + 2/7 + 9/6 - 4/3 - 1/3
= (5/7 + 2/7) + (9/6 - 3/6) - (4/3 + 1/3)
= 1 + 1 - 5/3
= 2- 5/3
= 1/3
b. 0,75 + 5,64 + ( 7,25 - 3,14 )
= (0,75 + 7,25) + (5,64 - 3,14)
= 8 + 2,5
= 10,5
c, 15/26 x 39/25 x 54/18 x 36/27 ( Làm = cách triệt tiêu )
\(=\frac{15\cdot39\cdot54\cdot36}{26\cdot25\cdot18\cdot27}\)
\(=\frac{3\cdot5\cdot3\cdot13\cdot27\cdot2\cdot2\cdot18}{2\cdot13\cdot5\cdot5\cdot18\cdot27}\)
\(=\frac{3\cdot3\cdot2}{5}\)
\(=\frac{18}{5}\)
Bài 2 : Tìm x
a, ( 6,73 + 5,26 - 3,5 ) - x = 4,32 - 2,56
8,49 - x = 1,76
x= 8,49 - 1,76
x= 6,73
b, x : ( 7,25 - 5,42 ) = 4,63 +1,37
x : 1,83 = 6
x= 6 * 1,83
x= 10,98
a) Ta thấy \(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};...;\frac{99}{100}< \frac{100}{101}\)
\(\Rightarrow A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}< B=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)
b) \(A.B=\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\right).\left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\right)\)
\(A.B=\frac{1.\left(3.5...99\right).\left(2.4.6...100\right)}{\left(2.4.6...100\right).\left(3.5.7...99\right).101}=\frac{1}{101}\)
c) vì A < b nên A . A < A . B < \(\frac{1}{101}< \frac{1}{100}\)
do đó : A . A < \(\frac{1}{10}.\frac{1}{10}\)suy ra A < \(\frac{1}{10}\)
