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A=\(\frac{3}{5}\): \(\frac{4}{5}\) + \(\frac{3}{7}\):\(\frac{4}{7}\)+ \(\frac{3}{13}\): \(\frac{4}{13}\)+\(\frac{3}{295}\): \(\frac{4}{295}\)
A= \(\frac{3}{4}\)+ 3/4 + 3/4 + 3/4
A=3/4 . 4 = 3
Ta có: \(\frac{3}{5}+\frac{3}{7}+\frac{3}{13}+\frac{3}{295}=3.\left(1+\frac{1}{5}+\frac{1}{7}+\frac{1}{13}+\frac{1}{295}\right)\)
\(\frac{4}{5}+\frac{4}{7}+\frac{4}{13}+\frac{4}{295}=4.\left(1+\frac{1}{5}+\frac{1}{7}+\frac{1}{13}+\frac{1}{295}\right)\)
\(\Rightarrow\frac{\frac{3}{5}+\frac{3}{7}+\frac{3}{13}+\frac{3}{295}}{\frac{4}{5}+\frac{4}{7}+\frac{4}{13}+\frac{4}{295}}=\frac{3.\left(1+\frac{1}{5}+\frac{1}{7}+\frac{1}{13}+\frac{1}{295}\right)}{4.\left(1+\frac{1}{5}+\frac{1}{7}+\frac{1}{13}+\frac{1}{295}\right)}=\frac{3}{4}\)
Để chứng minh tổng \( A = 1 - 4 - 7 + 10 + 13 - 16 - 19 + 22 + \dots - 295 + 298 + 301 - 304 \) chia hết cho 3, chúng ta có thể nhóm các số có cùng dấu và tính tổng của từng nhóm.
Nhóm các số cùng dấu:
\( (1 - 4 - 7) + (10 + 13 - 16) + (19 + 22 + \dots + 298 + 301) - 304 \)
Từ mẫu số 19 đến 301, có \( \frac{301 - 19}{3} + 1 = 95 \) số chia hết cho 3. Vì vậy, tổng của chúng là \( 95 \times 3 = 285 \).
Suy ra, tổng \( A \) sẽ là tổng các số đó trừ đi 304:
\( 285 - 304 = -19 \)
Vì -19 không chia hết cho 3, nên ta không thể chứng minh rằng tổng \( A \) chia hết cho 3.
A = 1 - 4 - 7 +10 +13 - 16 - 19 +22 + .... - 295 + 298 +301 - 304
A = (1-4) + (-7+10) + (13-16) + (-19+22) + ... + (-295+298) + (301-304)
A = (-3) + 3 + (-3) + 3 + ... + 3 + (-3) \(⋮\) 3
Vậy A\(⋮\) 3
TICK NHA! CÁCH NÀY DỄ HIỂU NÈ BẠN
\(1,-\dfrac{4}{7}+\dfrac{2}{3}\times\dfrac{-9}{14}\)
\(=\dfrac{-4}{7}+\dfrac{-18}{42}\)
\(=\dfrac{-4\times6}{7\times6}+\dfrac{-18}{42}\)
\(=\dfrac{-20}{42}+\dfrac{-18}{42}\)
\(=-\dfrac{38}{42}\)
\(=-\dfrac{19}{21}\)
\(2,\dfrac{17}{13}-\left(\dfrac{4}{13}-11\right)\)
\(=\dfrac{17}{13}-\dfrac{4}{13}+11\)
\(=\dfrac{13}{13}+11\)
\(=1+11\)
\(=12\)
\(3,8\dfrac{2}{7}-\left(3\dfrac{4}{9}+4\dfrac{2}{7}\right)\)
\(=\dfrac{58}{7}-\left(\dfrac{31}{9}+\dfrac{30}{7}\right)\)
\(=\dfrac{58}{7}-\dfrac{31}{9}-\dfrac{30}{7}\)
\(=\dfrac{58}{7}-\dfrac{30}{7}-\dfrac{31}{9}\)
\(=\dfrac{28}{7}-\dfrac{31}{9}\)
\(=\dfrac{28\times9}{7\times9}-\dfrac{31\times7}{9\times7}\)
\(=\dfrac{252}{63}-\dfrac{217}{63}\)
\(=\dfrac{35}{63}\)
\(=\dfrac{5}{9}\)
\(5,\left(\dfrac{2}{3}-1\dfrac{1}{2}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\left(\dfrac{2}{3}-\dfrac{3}{2}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\left(\dfrac{2\times2}{3\times2}-\dfrac{3\times3}{2\times3}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\left(\dfrac{4}{6}-\dfrac{9}{6}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\dfrac{-5}{6}:\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\dfrac{-5}{6}\times\dfrac{3}{4}+\dfrac{1}{2}\)
\(=\dfrac{-15}{24}+\dfrac{1}{2}\)
\(=\dfrac{-15}{24}+\dfrac{1\times12}{2\times12}\)
\(=\dfrac{-15}{24}+\dfrac{12}{24}\)
\(=\dfrac{-3}{24}\)
\(=-\dfrac{1}{8}\)
\(6,\dfrac{-5}{13}+\dfrac{2}{5}+\dfrac{-8}{13}+\dfrac{3}{5}-\dfrac{3}{7}\)
\(=\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)+\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{3}{7}\)
\(=\dfrac{-13}{13}+\dfrac{5}{5}-\dfrac{3}{7}\)
\(=-1+1-\dfrac{3}{7}\)
\(=-\dfrac{3}{7}\)
\(7,\dfrac{6}{5}\times\dfrac{3}{7}+\dfrac{6}{5}:\dfrac{7}{10}+\dfrac{6}{5}\)
\(=\dfrac{6}{5}\times\dfrac{3}{7}+\dfrac{6}{5}\times\dfrac{10}{7}+\dfrac{6}{5}\)
\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+1\right)\)
\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+\dfrac{1\times7}{1\times7}\right)\)
\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+\dfrac{7}{7}\right)\)
\(=\dfrac{6}{5}\times\dfrac{20}{7}\)
\(=\dfrac{120}{35}\)
\(=\dfrac{24}{7}\)
a)-84:4+3^9:3^7+5^0
=-21+(3^9:3^7)+1
=-21+3^2+1
=-21+9+1
=-12+1
=-11
b)295-(31-2^5.5)^2
=295-(31-32.5)^2
=295-(-129)^2
=295+129^2
=295-16641
=-16346
tick cho mk nha bạn
a) -84 : 4 + 3^ 9 : 3 ^ 7 + 5 ^0
=-84:4+3^2+1
=-84:4+9+1
=-21+9+1
=-12+1
=-11
b) 295 - ( 31 - 2^ 5. 5) ^2
= 295 - ( 31 -32. 5)^2
= 295 - (-129)^2
=295-16641
=-16346
Lời giải:
$-1+4-7+10-13+16-...-295+298-301$
$=(-1+4)+(-7+10)+(-13+16)+.....+(-295+298)-301$
$=3+3+3+....+3-301$
Số lần xuất hiện của 3 là: $[(298-1):3+1]:2=50$
$S=3.50-301=-151$
`@` `\text {Ans}`
`\downarrow`
`a.`
`A=(1/2-7/13-1/3)+(-6/13+1/2+1 1/3)`
`= 1/2 - 7/13 - 1/3 - 6/13 + 1/2 + 1 1/3`
`= (1/2 + 1/2) + (-7/13 - 6/13) + (-1/3 + 1 1/3) `
`= 1 - 1 + 1`
`= 1`
`b.`
`B=0,75+2/5+(1/9-1 1/2+5/4)`
`= 3/4 + 2/5 + 1/9 - 3/2 + 5/4`
`= (3/4+5/4)+ 1/9 + 2/5 - 3/2`
`= 2 + 1/9 - 11/10`
`= 19/9 - 11/10`
`= 91/90`
`c.`
`(-5/9).3/11+(-13/18).3/11`
`= 3/11*[(-5/9) + (-13/18)]`
`= 3/11*(-23/18)`
`= -23/66`
`d.`
`(-2/3).3/11+(-16/9).3/11`
`= 3/11* [(-2/3) + (-16/9)]`
`= 3/11*(-22/9)`
`= -2/3`
`e.`
`(-1/4).(-2/13)-7/24.(-2/13)`
`= (-2/13)*(-1/4-7/24)`
`= (-2/13)*(-13/24)`
`= 1/12`
`f.`
`(-1/27).3/7+(5/9).(-3/7)`
`= 3/7*(-1/27 - 5/9)`
`= 3/7*(-16/27)`
`= -16/63`
`g.`
`(-1/5+3/7):2/11+(-4/5+4/7):2/11`
`=[(-1/5+3/7)+(-4/5+4/7)] \div 2/11`
`= (-1/5+3/7 - 4/5 + 4/7) \div 2/11`
`= [(-1/5-4/5)+(3/7+4/7)] \div 2/11`
`= (-1+1) \div 2/11`
`= 0 \div 2/11 = 0`
B=-1+(4-7)+(10-13)+...+(292-295)+298
B=-1+(-3)+(-3)+...+(-3)+298
Có 49 số (-3)
B=-1+49*(-3)+298
B=-1+(-147)+298
B=(-148)+298
B=150
Vậy B=150