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\(\left(19\frac{5}{8}:\frac{7}{12}-13\frac{1}{4}:\frac{7}{12}\right).\frac{4}{5}\)
\(=\left(\frac{157}{8}:\frac{7}{12}-\frac{53}{4}:\frac{7}{12}\right).\frac{4}{5}\)
\(=\left[\left(\frac{157}{8}-\frac{53}{4}\right):\frac{7}{12}\right].\frac{4}{5}\)
\(=\left[\frac{51}{8}:\frac{7}{12}\right].\frac{4}{5}\)
\(=\frac{153}{14}.\frac{4}{5}\)
\(=\frac{306}{35}\)
\(\left(\frac{-2}{5}+\frac{3}{7}\right)-\left(\frac{4}{9}+\frac{12}{20}-\frac{13}{35}\right)+\frac{7}{35}\)
\(=\frac{1}{35}-\frac{212}{315}+\frac{7}{35}\)
\(=\frac{1}{35}+\frac{-212}{315}+\frac{7}{35}\)
\(=\frac{9}{315}+\frac{-212}{315}+\frac{63}{315}\)
\(=\frac{-140}{315}=\frac{-4}{9}\)
\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\)
\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\)
\(=\dfrac{1}{1}\cdot\dfrac{1}{2}+\dfrac{1}{2}\cdot\dfrac{1}{3}+\dfrac{1}{3}\cdot\dfrac{1}{4}+\dfrac{1}{4}\cdot\dfrac{1}{5}+\dfrac{1}{5}\cdot\dfrac{1}{6}+\dfrac{1}{6}\cdot\dfrac{1}{7}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\)
\(=\dfrac{1}{1}-\dfrac{1}{7}=\dfrac{7}{7}-\dfrac{1}{7}=\dfrac{6}{7}\)
Ta có: \(A=\frac{4}{7.31}+\frac{6}{7.41}+\frac{9}{10.41}+\frac{7}{10.57}\)
\(=\frac{20}{35.31}+\frac{30}{35.41}+\frac{45}{50.41}+\frac{35}{50.57}\)
\(=5\left(\frac{4}{31.35}+\frac{6}{35.41}+\frac{9}{41.50}+\frac{7}{50.57}\right)\)
\(=5\left(\frac{1}{31}-\frac{1}{35}+\frac{1}{35}-\frac{1}{41}+\frac{1}{41}-\frac{1}{50}+\frac{1}{50}-\frac{1}{57}\right)\)
\(=5\left(\frac{1}{31}-\frac{1}{57}\right)\)
\(B=\frac{7}{19.31}+\frac{5}{19.43}+\frac{3}{23.43}+\frac{11}{23.57}\)
\(=\frac{14}{38.31}+\frac{10}{38.43}+\frac{6}{46.43}+\frac{22}{46.57}\)
\(=2\left(\frac{7}{31.38}+\frac{5}{38.43}+\frac{3}{43.46}+\frac{11}{46.57}\right)\)
\(=2\left(\frac{1}{31}-\frac{1}{38}+\frac{1}{38}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}+\frac{1}{46}-\frac{1}{57}\right)\)
\(=2\left(\frac{1}{31}-\frac{1}{57}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{5\left(\frac{1}{31}-\frac{1}{57}\right)}{2\left(\frac{1}{31}-\frac{1}{57}\right)}=\frac{5}{2}\)
số số hạng là: (20-1):1+1=20
số cặp là: 20:2=10(cặp)
tổng 1 cặp là: 1+(-2)=-1
tổng là:10*(-1)=-10
cho mình nha
Tách hai dấu ngoặc thành hai vế A và B
Ta có :
A=5.415.99-4.320.89=5.230.318-22.227.320
=229.318(2.5-32)=229.318
B=5.29.619-7.229.276=5.29.219.319-7.229.318
=228.318(5.3-7.2)=228.318
=>A:B=229.318:228:318=2
A = 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72
= 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9
= 1- 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9
= 1 - 1/9 = 8/9
Câu B, C dấu * là nhân hay công vậy?
`@` `\text {Ans}`
`\downarrow`
`c)`
`( 34 - 2x ) . ( 2x - 6 ) = 0`
`=>`\(\left[{}\begin{matrix}34-2x=0\\2x-6=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x=34\\2x=6\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=34\div2\\x=6\div2\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=17\\x=3\end{matrix}\right.\)
Vậy, `x \in {17; 3}`
`d)`
`( 2019 - x ) . ( 3x - 12 ) =0` `?`
`=>`\(\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=2019-0\\3x=12\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=2019\\x=12\div3\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=2019\\x=4\end{matrix}\right.\)
Vậy, `x \in {2019; 4}`
`e) `
`57 . ( 9x - 27 ) = 0`
`=>`\(9x-27=0\div57\)
`=> 9x - 27 = 0`
`=> 9x = 27`
`=> x = 27 \div 9`
`=> x = 3`
Vậy, `x = 3`
`f)`
`25 + ( 15 - x ) = 30`
`=> 15 - x = 30 - 25`
`=> 15 - x = 5`
`=> x = 15 -5 `
`=> x = 10`
Vậy, `x = 10`
`g) `
`43 - ( 24 - x ) = 20`
`=> 24 - x = 43 - 20`
`=> 24 - x = 23`
`=> x = 24 - 23`
`=> x = 1`
Vậy, `x = 1`
`h) `
`2 . ( x - 5 ) - 17 = 25`
`=> 2 ( x - 5) = 25+17`
`=> 2 ( x - 5) = 42`
`=> x - 5 = 42 \div 2`
`=> x - 5 = 21`
`=> x = 21 + 5`
`=> x = 26`
Vậy, `x = 26`
`i)`
`3 . ( x + 7 ) - 15 = 27`
`=> 3(x + 7) = 27 + 15`
`=> 3(x + 7) = 42`
`=> x +7 = 42 \div 3`
`=> x + 7 = 14`
`=> x = 14 - 7`
`=> x = 7`
Vậy, `x = 7`
`j)`
`15 + 4 . ( x - 2 ) = 95`
`=> 4(x - 2) = 95 - 15`
`=> 4(x - 2) = 80`
`=> x - 2 = 80 \div 4`
`=> x - 2 = 20`
`=> x = 20 + 2`
`=> x = 22`
Vậy, `x = 22`
`k)`
`20 - ( x + 14 ) = 5`
`=> x + 14 = 20 - 5`
`=> x + 14 = 15`
`=> x = 15 - 14`
`=> x = 1`
Vậy, `x = 1`
`l) `
`14 + 3 . ( 5 - x ) = 27`
`=> 3(5 - x) = 27 - 14`
`=> 3(5 - x) = 13`
`=> 5 - x = 13 \div 3`
`=> 5 - x = 13/3`
`=> x = 5- 13/3`
`=> x = 2/3`
Vậy, `x = 2/3.`
`@` `\text {Kaizuu lv uuu}`
GD se chet may
đề sai bn ơi
TINH nhanh
1+(-2)+3+(-4)+57+(-6)+....+19+(-20)
tại sao lại có số 57 ở đấy hả