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a.\(\dfrac{17}{15}\div\dfrac{4}{3}=\dfrac{17}{20}\)
b.\(\dfrac{-12}{21}\div\dfrac{34}{43}=\dfrac{-86}{119}\)
c.\(\dfrac{-5}{9}\times\dfrac{3}{11}+\dfrac{13}{18}\times\dfrac{3}{11}\)
=\(\dfrac{3}{11}\times(\dfrac{-5}{9}+\dfrac{13}{18})=\dfrac{3}{11}\times\dfrac{1}{6}=\dfrac{1}{22}\)
d.\(\dfrac{-2}{9}\times\dfrac{5}{11}+\dfrac{-16}{9}\times\dfrac{5}{11}=\dfrac{5}{11}\times(\dfrac{-2}{9}+\dfrac{-16}{9})\)
=\(\dfrac{5}{11}\times(-2)=\dfrac{-10}{11}\)
a) 2/3 + 3/4 . (-4/9)
= 2/3 - 1/3
= 1/3
b) -5/7 . 31/33 + (-5/7) : 33/2
= -5/7 . 31/33 - 5/7 . 2/33
= -5/7 . (31/33 + 2/33)
= -5/7 . 1
= -5/7
c) -3/5 . 13/11 - (-3/5) . 2/11
= -3/5 . (13/11 - 2/11)
= -3/5 . 1
= -3/5
a) \(A=\dfrac{2}{3}+\dfrac{3}{4}.\left(\dfrac{-4}{9}\right)=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\)
b) \(B=2\dfrac{3}{11}.1\dfrac{1}{12}.\left(-2,2\right)=\dfrac{25}{11}.\dfrac{13}{12}.\dfrac{-11}{5}=-\dfrac{65}{12}\)
c) \(C=\left(\dfrac{3}{4}-0,2\right)\left(0,4-\dfrac{4}{5}\right)=\left(\dfrac{3}{4}-\dfrac{1}{5}\right)\left(\dfrac{2}{5}-\dfrac{4}{5}\right)=\dfrac{11}{20}\left(\dfrac{-2}{5}\right)=\dfrac{-11}{50}\)
`@` `\text {Ans}`
`\downarrow`
`a)`
`-9/34*17/4`
`=`\(\dfrac{-9}{17\cdot2}\cdot\dfrac{17}{4}\)
`=`\(-\dfrac{9}{2}\cdot\dfrac{1}{4}\)
`=`\(-\dfrac{9}{8}\)
`b)`
\(\dfrac{17}{15}\div\dfrac{4}{3}\)
`=`\(\dfrac{17}{15}\cdot\dfrac{3}{4}\)
`=`\(\dfrac{17}{3\cdot5}\cdot\dfrac{3}{4}\)
`=`\(\dfrac{17}{5}\cdot\dfrac{1}{4}\)
`=`\(\dfrac{17}{20}\)
`c)`
\(4\dfrac{1}{5}\div\left(-2\dfrac{4}{5}\right)\)
`=`\(4\dfrac{1}{5}\cdot\left(-\dfrac{5}{14}\right)\)
`=`\(\dfrac{21}{5}\cdot\left(-\dfrac{5}{14}\right)\)
`=`\(-\dfrac{21}{14}=-\dfrac{3}{2}\)
a) \(\dfrac{-9}{34}\cdot\dfrac{17}{4}\)
\(=\dfrac{-9\cdot17}{34\cdot4}\)
\(=-\dfrac{153}{136}\)
\(=\dfrac{9}{8}\)
b) \(\dfrac{17}{15}:\dfrac{4}{3}\)
\(=\dfrac{17}{15}\cdot\dfrac{3}{4}\)
\(=\dfrac{17\cdot3}{15\cdot4}\)
\(=\dfrac{51}{60}=\dfrac{17}{20}\)
c) \(4\dfrac{1}{5}:\left(-2\dfrac{4}{5}\right)\)
\(=\dfrac{21}{5}:-\dfrac{14}{5}\)
\(=\dfrac{21}{5}\cdot-\dfrac{5}{14}\)
\(=\dfrac{21\cdot-5}{5\cdot14}\)
\(=-\dfrac{105}{70}=\dfrac{3}{2}\)
a) Ta có: \(\dfrac{a}{2}=\dfrac{b}{3}\)
\(\Leftrightarrow\dfrac{a}{8}=\dfrac{b}{12}\)(1)
Ta có: \(\dfrac{b}{4}=\dfrac{c}{5}\)
nên \(\dfrac{b}{12}=\dfrac{c}{15}\)(2)
Từ (1) và (2) suy ra \(\dfrac{a}{8}=\dfrac{b}{12}=\dfrac{c}{15}\)
mà a+b+c=2
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a}{8}=\dfrac{b}{12}=\dfrac{c}{15}=\dfrac{a+b+c}{8+12+15}=\dfrac{2}{35}\)
Do đó:
\(\left\{{}\begin{matrix}\dfrac{a}{8}=\dfrac{2}{35}\\\dfrac{b}{12}=\dfrac{2}{35}\\\dfrac{c}{15}=\dfrac{2}{35}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{16}{35}\\b=\dfrac{24}{35}\\c=\dfrac{30}{35}=\dfrac{6}{7}\end{matrix}\right.\)
Vậy: \(a=\dfrac{16}{35}\); \(b=\dfrac{24}{35}\); \(c=\dfrac{6}{7}\)
b) Ta có: 2a=3b=5c
nên \(\dfrac{a}{\dfrac{1}{2}}=\dfrac{b}{\dfrac{1}{3}}=\dfrac{c}{\dfrac{1}{5}}\)
mà a+b-c=3
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a}{\dfrac{1}{2}}=\dfrac{b}{\dfrac{1}{3}}=\dfrac{c}{\dfrac{1}{5}}=\dfrac{a+b-c}{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}=\dfrac{3}{\dfrac{19}{30}}=\dfrac{90}{19}\)
Do đó:
\(\left\{{}\begin{matrix}2a=\dfrac{90}{19}\\3b=\dfrac{90}{19}\\5c=\dfrac{90}{19}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{45}{19}\\b=\dfrac{30}{19}\\c=\dfrac{18}{19}\end{matrix}\right.\)
Vậy: \(a=\dfrac{45}{19}\); \(b=\dfrac{30}{19}\); \(c=\dfrac{18}{19}\)
b) Ta có : \(\dfrac{2a}{3}=\dfrac{3b}{4}=\dfrac{4c}{5}\)
\(\Leftrightarrow\dfrac{a}{\dfrac{3}{2}}=\dfrac{b}{\dfrac{4}{3}}=\dfrac{c}{\dfrac{5}{4}}=\dfrac{a+b+c}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{49}{\dfrac{49}{12}}=12\)
Khi đó \(a=12.\dfrac{3}{2}=18;b=12.\dfrac{4}{3}=16;c=12.\dfrac{5}{4}=15\)
Vậy (a,b,c) = (18,16,15)
`#3107`
a)
\(\dfrac{11}{12}-\left(\dfrac{2}{5}+\dfrac{3}{4}x\right)=\dfrac{2}{3}?\\ \Rightarrow\dfrac{2}{5}+\dfrac{3}{4}x=\dfrac{11}{12}-\dfrac{2}{3}\\ \Rightarrow\dfrac{2}{5}+\dfrac{3}{4}x=\dfrac{1}{4}\\ \Rightarrow\dfrac{3}{4}x=\dfrac{1}{4}-\dfrac{2}{5}\\ \Rightarrow\dfrac{3}{4}x=-\dfrac{3}{20}\\ \Rightarrow x=-\dfrac{3}{20}\div\dfrac{3}{4}\\ \Rightarrow x=-\dfrac{1}{5}\)
Vậy, \(x=-\dfrac{1}{5}\)
b)
\(\dfrac{-2}{5}+\dfrac{5}{3}\cdot\left(\dfrac{3}{2}-\dfrac{4}{15}x\right)=\dfrac{-7}{6}\\ \Rightarrow\dfrac{5}{3}\cdot\left(\dfrac{3}{2}-\dfrac{4}{15}x\right)=\dfrac{-7}{6}-\dfrac{-2}{5}\\ \Rightarrow\dfrac{5}{3}\cdot\left(\dfrac{3}{2}-\dfrac{4}{15}x\right)=-\dfrac{23}{30}\\ \Rightarrow\dfrac{3}{2}-\dfrac{4}{15}x=-\dfrac{23}{30}\div\dfrac{5}{3}\\ \Rightarrow\dfrac{3}{2}-\dfrac{4}{15}x=-\dfrac{23}{50}\\ \Rightarrow\dfrac{4}{15}x=\dfrac{3}{2}-\left(-\dfrac{23}{50}\right)\\ \Rightarrow\dfrac{4}{15}x=\dfrac{49}{25}\\ \Rightarrow x=\dfrac{147}{20}\)
Vậy, \(x=\dfrac{147}{20}\)
c)
\(\dfrac{1}{2}+\dfrac{3}{4}x=\dfrac{1}{4}\\ \Rightarrow\dfrac{3}{4}x=\dfrac{1}{4}-\dfrac{1}{2}\\ \Rightarrow\dfrac{3}{4}x=-\dfrac{1}{4}\\ \Rightarrow x=-\dfrac{1}{4}\div\dfrac{3}{4}\\ \Rightarrow x=-\dfrac{1}{3}\)
Vậy, \(x=-\dfrac{1}{3}.\)
\(#Emyeu1aithatroi...\)
(2/5 + 3/4 . x)= 11/12 -2/3
(2/5 +3/4 . x)= 1/4
3/4 . x = 1/4 - 2/5
3/4 . x = -3/20
x = -3/20 : 3/4
x = -1/5
Vậy .....
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a}{15}=\dfrac{b}{10}=\dfrac{c}{8}=\dfrac{a+b+c}{15+10+8}=\dfrac{11}{33}=\dfrac{1}{3}\)
Do đó: a=5; b=10/3; c=8/3
Có \(\dfrac{a}{3}\)=\(\dfrac{b-3}{4}\)=\(\dfrac{c+5}{15}\)
=>\(\dfrac{a}{3}\)=\(\dfrac{b-3+3}{4+3}\)=\(\dfrac{c+5-15}{15-5}\)
=>\(\dfrac{a}{3}\)=\(\dfrac{b}{7}\)=\(\dfrac{c}{10}\)
Áp dụng tính chất dãy tỉ số bằng nhau và a+b+c=34 có:
\(\dfrac{a}{3}\)=\(\dfrac{b}{7}\)=\(\dfrac{c}{10}\)=>\(\dfrac{a+b+c}{3+7+10}\)=\(\dfrac{34}{20}\)=1,7
Với \(\dfrac{a}{3}\)=1,7 => a=1,7 x 3 = 5,1
Với \(\dfrac{b}{7}\)=1.7 => b= 1,7 x 7 =11.9
Với\(\dfrac{c}{10}\)=1,7=> c= 1,7 x 10 = 17
KL:Vậy a=5,1 ; b=11,9 ; c=17