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a) Ta có: \(D=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{96}\)
\(=\dfrac{2}{3}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{24}+\dfrac{1}{24}-\dfrac{1}{48}+\dfrac{1}{48}-\dfrac{1}{96}\)
\(=\dfrac{2}{3}-\dfrac{1}{96}\)
\(=\dfrac{63}{96}=\dfrac{21}{32}\)
b)
Sửa đề: \(E=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{2048}\)
Ta có: \(E=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{2048}\)
\(\Leftrightarrow\dfrac{1}{2}\cdot E=\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+...+\dfrac{1}{4096}\)
\(\Leftrightarrow\dfrac{1}{2}\cdot E=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{16}+...+\dfrac{1}{2048}-\dfrac{1}{4096}\)
\(\Leftrightarrow\dfrac{E}{2}=\dfrac{1}{2}-\dfrac{1}{4096}=\dfrac{2047}{4096}\)
hay \(E=\dfrac{2047}{2048}\)
a. \(\dfrac{-2}{3}+\dfrac{-1}{5}+\dfrac{3}{4}-\dfrac{5}{6}-\dfrac{7}{10}\)
= \(\dfrac{-4}{6}+\dfrac{-2}{10}+\dfrac{3}{4}-\dfrac{5}{6}-\dfrac{7}{10}\)
= \(\dfrac{-3}{2}+\dfrac{1}{2}+\dfrac{3}{4}\)
= (-1) + \(\dfrac{3}{4}\)
= \(\dfrac{-4}{4}+\dfrac{3}{4}\)
= \(\dfrac{-1}{4}\)
gọi 1/547=a, 1/211=b ta đc:
\(2a.3b-546a.b-4ab< =>6ab-546ab-4ab=-544ab=\dfrac{-544}{547.211}\)
`@` `\text {Ans}`
`\downarrow`
`1)`
`5/7*37 13/23 - 51 13/23*5/7`
`= 5/7* (37 13/23 - 51 13/23)`
`= 5/7* (-14)`
`= -10`
`2)`
`-2/3 +1/3+0,5+2 1/2`
`= -2/3 + 1/3 + 1/2 + 5/2`
`= (-2/3+1/3) + (1/2+5/2)`
`= -1/3 + 3`
`=8/3`
`3)`
`-0,5+2/3+1/2`
`= -1/2 + 2/3 + 1/2`
`= (-1/2 + 1/2) + 2/3`
`= 2/3`
`4)`
`(8+2 1/3-3/5) -(5+0,4)-(3 1/2 -2)`
`= 8+ 7/3 - 3/5 - 5 - 0,4 - 7/2 + 2`
`= (8+2-5) + (-3/5 - 2/5) + (7/3 - 7/2)`
`= 5 - 1 - 7/6`
`= 4 - 7/6 = 17/6`
`5)`
`(2/9-7/12):3/4+(16/9-5/12):3/4`
`= (2/9 - 7/12) \times 4/3 + (16/9 - 5/12) \times 4/3`
`= 4/3 *(2/9 - 7/12 + 16/9 - 5/12)`
`= 4/3 * [(2/9 + 16/9) + (-7/12 - 5/12)]`
`= 4/3 * ( 2 - 1)`
`= 4/3 * 1 = 4/3`
`6)`
`-(2021.0,7+19,75) +0,7- (8-19,75)`
`= -2021*0,7 -19,75 + 0,7 - 8 + 19,75`
`= 0,7*(-2021 + 1) - 8`
`= -1414-8`
`= -1422`
`7)`
`15/34+7/21+19/34-20/15`
`= (15/34 + 19/34) + 7/21 - 20/15`
`= 1 + 7/21 - 20/15`
`= 4/3 - 20/15 =0`
`8)`
`2 5/6+1/6:(-5/8)`
`= 17/6 + (-4/15)`
`= 77/30`
`9)`
`(-2)^2 +2/9. (4/5-2/3)`
`= 4 + 2/9*2/15`
`= 4+4/135`
`= 544/135`
`10)`
`(-1/5+3/7):5/4+(-4/5+4/7):5/4`
`= (-1/5+3/7) * 4/5 + (-4/5+4/7) * 4/5`
`= 4/5*(-1/5 +3/7-4/5+4/7)`
`= 4/5*[(-1/5-4/5)+(3/7+4/7)]`
`= 4/5* (-1+1)`
`= 4/5*0=0`
`11)`
`2022,2021 . 1954,1945+ 2022,2021 . (-1954,1945)`
`= 2022,2021 * [1954,1945 + (-1954,1945)]`
`= 2022,2021*0 `
`= 0`
`12)`
`-5,2 .72 +69,1 +5,2 . (-28)+(-1,1)`
`= -5,2*72 + 69,1 - 5,2*28 - 1,1`
`= -5,2*(72+28) + (69,1 - 1,1)`
`= -5,2*100 + 68`
`= -520 + 68`
`= -452`
`13)`
`(7 -1/2-3/4) : (5-1/4-5/8)`
`= 23/4 \div 33/8`
`=46/33`
`14)`
`(8+ 2 1/3 -3/5) -(5+0,4) -( 3 1/3 - 2)`
`= 8+ 2 1/3 - 3/5 - 5 - 0,4 - 3 1/3 + 2`
`= (8+2-5) + (2 1/3 - 3 1/3) - (0,6 + 0,4) `
`= 5 - 1 - 1`
`= 3`
a: \(M=\dfrac{631}{315}\cdot\dfrac{1}{651}-\dfrac{1}{105}\cdot\dfrac{2603}{651}-\dfrac{4}{315\cdot651}+\dfrac{4}{105}\)
\(=\dfrac{1}{315\cdot651}\cdot\left(631-4\right)-\dfrac{1}{105}\left(\dfrac{2603}{651}-4\right)\)
\(=\dfrac{1}{105}\cdot\dfrac{1}{1953}\cdot627+\dfrac{1}{105\cdot651}\)
\(=\dfrac{1}{105\cdot651}\left(\dfrac{1}{3}\cdot627+1\right)=\dfrac{1}{105\cdot651}\cdot210=\dfrac{2}{651}\)
b: \(N=\dfrac{1095}{547}\cdot\dfrac{3}{211}-\dfrac{546}{547\cdot211}-\dfrac{4}{547\cdot211}\)
\(=\dfrac{1}{547\cdot211}\left(1095\cdot3-546-4\right)\)
\(=\dfrac{1}{547\cdot211}\cdot2735=\dfrac{5}{211}\)
=>\(B=\left(2\cdot\frac{1}{547}\cdot\frac{3}{211}\right)-\left(\frac{546}{547}\cdot\frac{1}{211}\right)-\left(\frac{4}{547\cdot211}\right)\)
=>\(B=\frac{6}{547\cdot211}-\frac{546}{547\cdot211}-\frac{4}{547\cdot211}\)
=>\(B=\frac{6}{115417}-\frac{546}{115417}-\frac{4}{115417}\)
=>\(B=\frac{-544}{115417}\)
\(2\dfrac{1}{547}.\dfrac{3}{211}-\dfrac{546}{547}.\dfrac{1}{211}-\dfrac{4}{547.211}\)
\(=\left(2+\dfrac{1}{547}\right).3.\dfrac{1}{211}-\left(1-\dfrac{1}{547}\right).\dfrac{1}{211}-4.\dfrac{1}{547}.\dfrac{1}{211}\)
Đặt \(a=\dfrac{1}{547};b=\dfrac{1}{211}\)
Thay \(a=\dfrac{1}{547};b=\dfrac{1}{211}\) vào biểu thức trên , ta được :
\(\left(2+a\right).3b-\left(1-a\right)b-4ab\)
\(=6b+3ab-b+ab-4ab\)
\(=5b\)
\(=5.\dfrac{1}{211}\)
\(=\dfrac{5}{211}\)
Vậy g/t biểu thức trên là : \(\dfrac{5}{211}\)
Giải:
A=1/10+1/40+1/88+1/154+1/238+1/340
A=1/2.5+1/5.8+1/8.11+1/11.14+1/14.17+1/17.20
A=1/2-1/5+1/5-1/8+1/8-1/11+1/11-1/14+1/14-1/17+1/17-1/20
A=1/2-1/20
A=9/20
D=1/3+1/6+1/12+1/24+1/48
D=1/3+1/2.3+1/3.4+1/4.6+1/6.8
D=1/3+1/2-1/3+1/3-1/4+1/2.(2/4.6+2/6.8)
D=1/3+1/2-1/4+1/2.(1/4-1/6+1/6-1/8)
D=1/3+1/4+1/2.(1/4-1/8)
D=1/3+1/4+1/2.1/8
D=1/3+1/4+1/16
D=31/48
F=0,5-1/3-0,4-4/7-1/6+4/35-1/41
F=1/2-1/3-2/5-4/7-1/6+4/35-1/41
F=1/6-(-6/35)-1/6+4/35-1/41
F=(1/6-1/6)+(6/35+4/35)-1/41
F=0+2/7-1/41
F=2/7+1/41
F=75/287
Chúc bạn học tốt!
Cảm ơn nhìu <3