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8)\(\frac{4}{9}:\left(-\frac{1}{7}\right)+6\frac{5}{9}:\left(-\frac{1}{7}\right)\)
=\(\frac{4}{9}:\left(-\frac{1}{7}\right)+\frac{59}{9}:\left(-\frac{1}{7}\right)\)
=\(\left(\frac{4}{9}+\frac{59}{9}\right).\left(-7\right)\)
=7.(-7)
=-49
A)0,25:(10,3-9,8)-3/4
=1/4:(103/10-49/5)-3/4
=1/4:1/2-3/4
=1/2-3/4
=2/4-3/4
=-1/4
B)-5/9.13/28-13/28.4/9
=-5/9-4/9.13/28
=-1.13/28
=-13/28
c)6/7+5/8:5-3/16
=6/7+1/8-3/16
=55/56-3/16
=89/112
d)-5/7.2/11+-5/7.9/11+1/5/7
=-5/7.(2/11+9/11)+12/7
=-5/7.1+12/7
=-5/7+12/7
=1
e)-7/12-8/15+11/20
=-67/60+11/20
=-17/30
f)-17/25.20/33+-17/25.13/33+-3/25
=-17/25.(20/33+13/33)-3/25
=-17/25.1-3/25
=-17/25-3/25
=-4/5
CHÚC BẠN HỌC TỐT...............
NẾU ĐÚNG THÌ TICK CHO MK VỚI NHA HELLO HELLO..........
1. Tính:
a. \(\dfrac{\text{−1 }}{\text{4 }}+\dfrac{\text{5 }}{\text{6 }}=\dfrac{-3}{12}+\dfrac{10}{12}=\dfrac{7}{12}\)
b. \(\dfrac{\text{5 }}{\text{12 }}+\dfrac{\text{-7 }}{8}=\dfrac{10}{24}+\dfrac{-21}{24}=\dfrac{-11}{24}\)
c. \(\dfrac{-7}{6}+\dfrac{-3}{10}=\dfrac{-35}{30}+\dfrac{-9}{30}=\dfrac{-44}{30}=\dfrac{-22}{15}\)
d.\(\dfrac{-3}{7}+\dfrac{5}{6}=\dfrac{-18}{42}+\dfrac{35}{42}=\dfrac{17}{42}\)
2. Tính :
a. \(\dfrac{2}{14}-\dfrac{5}{2}=\dfrac{2}{14}-\dfrac{35}{14}=\dfrac{-33}{14}\)
b.\(\dfrac{-13}{12}-\dfrac{5}{18}=\dfrac{-39}{36}-\dfrac{10}{36}=\dfrac{49}{36}\)
c.\(\dfrac{-2}{5}-\dfrac{-3}{11}=\dfrac{-2}{5}+\dfrac{3}{11}=\dfrac{-22}{55}+\dfrac{15}{55}=\dfrac{-7}{55}\)
d. \(0,6--1\dfrac{2}{3}=\dfrac{6}{10}--\dfrac{5}{3}=\dfrac{3}{5}+\dfrac{5}{3}=\dfrac{9}{15}+\dfrac{25}{15}=\dfrac{34}{15}\)
3. Tính :
a.\(\dfrac{-1}{39}+\dfrac{-1}{52}=\dfrac{-4}{156}+\dfrac{-3}{156}=\dfrac{-7}{156}\)
b.\(\dfrac{-6}{9}-\dfrac{12}{16}=\dfrac{2}{3}-\dfrac{3}{4}=\dfrac{8}{12}-\dfrac{9}{12}=\dfrac{-17}{12}\)
c. \(\dfrac{-3}{7}-\dfrac{-2}{11}=\dfrac{-3}{7}+\dfrac{2}{11}=\dfrac{-33}{77}+\dfrac{14}{77}=\dfrac{-19}{77}\)
d.\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\dfrac{1}{1}+\dfrac{1}{10}\)
\(=\dfrac{10}{10}-\dfrac{1}{10}\)
= \(\dfrac{9}{10}\)
Chế Kazuto Kirikaya thử tham khảo thử đi !!!
Mấy câu trên kia dễ rồi mình chữa mình câu \(c\) bài \(3\) thôi nhé Kazuto Kirikaya
d) \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)
a/Ta có: \(\dfrac{4}{3}-\left[\left(\dfrac{-11}{6}\right)-\left(\dfrac{2}{9}+\dfrac{5}{3}\right)\right]\)
\(=\) \(\dfrac{4}{3}-\left[\dfrac{-11}{6}-\dfrac{2}{9}-\dfrac{5}{3}\right]\)
\(=\) \(\dfrac{4}{3}+\dfrac{11}{6}+\dfrac{2}{9}+\dfrac{5}{3}\)
\(=\) \(\dfrac{24}{18}+\dfrac{33}{18}+\dfrac{4}{18}+\dfrac{30}{18}\)
\(=\) \(\dfrac{91}{18}\)
b/Ta có: \(\left(8-\dfrac{9}{4}+\dfrac{2}{7}\right)-\left(-6-\dfrac{3}{7}+\dfrac{5}{4}\right)-\left(3+\dfrac{2}{4}-\dfrac{9}{7}\right)\)
\(=\) \(8-\dfrac{9}{4}+\dfrac{2}{7}+6+\dfrac{3}{7}-\dfrac{5}{4}-3-\dfrac{2}{4}+\dfrac{9}{7}\)
\(=\) \(8+6-3-\dfrac{9}{4}-\dfrac{5}{4}-\dfrac{2}{4}+\dfrac{2}{7}+\dfrac{3}{7}+\dfrac{9}{7}\)
\(=\) \(11-\dfrac{2}{4}+\dfrac{14}{7}\)
\(=\) \(11-\dfrac{1}{2}+2\)
\(=\) \(9-\dfrac{1}{2}\)
\(=\) \(\dfrac{17}{2}\)
Chúc bn học tốt!!!
\(a)\dfrac{3}{4}+\dfrac{6}{12}-\dfrac{5}{24}\)
\(=\dfrac{18}{24}+\dfrac{12}{24}+\left(-\dfrac{5}{24}\right)\)
\(=\dfrac{18+12+\left(-5\right)}{24}\)
\(=\dfrac{25}{24}\)
\(b)\dfrac{-5}{7}.\dfrac{2}{13}-\dfrac{5}{7}.\dfrac{11}{13}+\dfrac{5}{7}\)
\(=\dfrac{5}{7}.\dfrac{-2}{13}-\dfrac{5}{7}.\dfrac{11}{13}+\dfrac{5}{7}\)
\(=\dfrac{5}{7}\left(\dfrac{-2}{13}+\dfrac{-11}{13}+\dfrac{13}{13}\right)\)
\(=\dfrac{5}{7}.0=0\)
\(c)\dfrac{27}{23}+\dfrac{5}{21}+\dfrac{1}{2}-\dfrac{4}{23}+\dfrac{16}{21}\)
\(=\left(\dfrac{27}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+\dfrac{1}{2}\)
\(=1+1+\dfrac{1}{2}\)
\(=2\dfrac{1}{2}\)
\(d)\dfrac{15}{34}+\dfrac{7}{21}+\dfrac{19}{34}.\dfrac{20}{15}+\dfrac{3}{7}\)
\(=\dfrac{315}{714}+\dfrac{238}{714}+\dfrac{38}{51}+\dfrac{306}{714}\)
\(=\dfrac{315}{714}+\dfrac{238}{714}+\dfrac{532}{714}+\dfrac{306}{714}\)
\(=\dfrac{1391}{714}\)
a)\(\dfrac{3}{4}+\dfrac{6}{12}-\dfrac{5}{24}=\dfrac{18}{24}+\dfrac{12}{24}-\dfrac{5}{24}=\dfrac{25}{24}\)
b)\(\dfrac{-5}{7}.\dfrac{2}{13}-\dfrac{5}{7}.\dfrac{11}{13}+\dfrac{5}{7}=\dfrac{5}{7}\left(\dfrac{-2}{13}-\dfrac{11}{13}+1\right)=\dfrac{5}{7}.0=0\)
c)\(\dfrac{27}{23}+\dfrac{5}{21}+\dfrac{1}{2}-\dfrac{4}{23}+\dfrac{16}{21}=\left(\dfrac{27}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+\dfrac{1}{2}=1+1+\dfrac{1}{2}=2,5\)
d)\(\dfrac{15}{34}+\dfrac{7}{21}+\dfrac{19}{34}.\dfrac{20}{15}+\dfrac{3}{7}=\dfrac{15}{34}+\left(\dfrac{1}{3}+\dfrac{38}{51}+\dfrac{3}{7}\right)=\dfrac{15}{34}+\dfrac{538}{357}=\dfrac{1391}{714}\)
Ta có : a, 25/7 + 13/21 - 11/7 + 17/21 + 1/3 .
= ( 25/7 - 11/7 ) + ( 13/21 + 17/21 + 1/3 ) .
= 2 + ( 20/21 + 7/21 ) .
= 2 + 9/7 .
= 23/7 .
b, ( 1/3 + 12/67 + 13/41 ) - ( 79/67 - 28/41 ) .
= 1/3 + 12/67 + 13/41 - 79/67 + 28/41 .
= 1/3 + ( 12/67 - 79/67 ) + ( 13/41 + 28/41 ) .
= 1/3 - 1 + 1 .
= 1/3 .
c, ( 11/4 . -5/9 - 4/9 . 11/4 ) . 8/33 .
= [ 11/4 . ( -5/9 - 4/9 ) ] . 8/33 .
= [ 11/4 . ( - 1 ) ] . 8/33 .
= -11/4 . 8/33 .
= -2/3 .
d, 38/45 - ( 8/45 - 17/51 - 3/11 ) .
= 38/45 - 8/45 + 17/51 + 3/11 .
= 2/3 + 17/51 + 3/11 .
= 374/561 + 187/561 + 153/561 .
= 14/11 .
a)\(\dfrac{8^8.3^{14}}{9^6.2^{20}}=\dfrac{^{24}.3^{14}}{3^{12}.2^{20}}=\dfrac{2^4.3^2}{1}=144\)
b),c) tự tính nha bn
a) \(\dfrac{8^8.3^{14}}{9^6.2^{20}}=\dfrac{\left(2^3\right)^8.3^{14}}{\left(3^2\right)^6.2^{20}}=\dfrac{2^{24}.3^{14}}{3^{12}.2^{20}}=\dfrac{2^4.3^2}{1}=144\)
b,c bấm máy tính
1: \(\Leftrightarrow3x+4=2\)
=>3x=-2
=>x=-2/3
2: \(\Leftrightarrow7x-7=6x-30\)
=>x=-23
3: =>\(5x-5=3x+9\)
=>2x=14
=>x=7
4: =>9x+15=14x+7
=>-5x=-8
=>x=8/5
Bài 1:
\(A=\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}:\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\)
\(A=\dfrac{2.\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7.\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}:\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{2}{7}.\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}\right)}\)
\(A=\dfrac{2}{7}:\dfrac{2}{7}=1\)
Bài 2: Here
Chúc bạn học tốt!!!
1. Giải:
Gọi A =M : N
Ta có:M=\(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}\)= \(\dfrac{2.\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7.\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}\)=\(\dfrac{2}{7}\)
N=\(\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\)=\(\dfrac{2.\left(\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{10}\right)}{7.\left(\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{10}\right)}\)=\(\dfrac{2}{7}\)
Vậy A=M: N \(\Rightarrow\)A=\(\dfrac{2}{7}\):\(\dfrac{2}{7}\)=\(\dfrac{2}{7}\).\(\dfrac{7}{2}\)=\(\dfrac{2.7}{7.2}\)=1
2. Giải:
Với mọi x \(\in\)Q, ta luôn có \(x\) \(\le\) \(|x|\)(dấu bằng xảy ra khi x\(\ge\)0)
a)Nếu \(x+y\)\(\ge\)0 thì\(|x+y|=x+y\).
Vì \(x\le|x|,y\le|y|\)với mọi x, y\(\in\)Q nên:\(|x+y|=x+y\le|x|+|y|\)
b)Nếu x+y < 0 thì\(|x+y|=-\left(x+y\right)\)=\(-x-y\)
Mà -x\(\le\)\(|x|\), -y\(\le\)\(|y|\) nên: \(|x+y|\)= -x-y\(\le\)\(|x|+|y|\)
Vậy với mọi x, y\(\in\)Q ta đều có:\(|x+y|\le|x|+|y|\). Dấu bằng xảy ra khi x, y cùng dấu hoặc ít nhất có một số bằng 0.
a) \(=\left(13\dfrac{2}{7}+2\dfrac{5}{7}\right):\left(-\dfrac{8}{9}\right)\)
\(=16:\dfrac{-8}{9}=\dfrac{-8\cdot\left(-2\right)\cdot9}{-8}=-18\)
b)
\(=\left(\dfrac{-6}{11}\cdot\dfrac{11}{-6}\right)\cdot\dfrac{7\cdot10\cdot\left(-2\right)}{10}\)
\(=-14\)
c) \(=\dfrac{-1}{2}\cdot\dfrac{4}{3}\cdot\dfrac{-7}{2}\)
\(=\dfrac{-1\cdot2\cdot2\cdot\left(-7\right)}{2\cdot3\cdot2}=\dfrac{7}{3}\)