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\(1.x-\dfrac{2}{3}\times\left(x+9\right)=1\)
\(x-\dfrac{2}{3}\times x-6=1\)
\(x\times\left(1-\dfrac{2}{3}\right)=7\)
\(x\times\dfrac{1}{3}=7\)
\(x=21\)
\(2.x-\dfrac{11}{15}=\dfrac{3+x}{5}\)
\(\dfrac{15x}{15}-\dfrac{11}{15}=\dfrac{9+3x}{15}\)
\(15x-11=9+3x\)
\(12x=20\)
\(x=\dfrac{5}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1.
=2/5 x 12/3 + 2/5 x 15/3 + 2/5 x 1
= 2/5 x (12/3 + 15/3 + 1)
=2/5 x 1
=2/5
2.a=1;2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{8.3.4}{8.2.3}=\dfrac{4}{2}=2\)
\(\dfrac{5.6.5.5.7.8}{5.5.3.8.2.6.2.7}=\dfrac{5}{12}\)
Bài 1:
\(\dfrac{7.3.7.11}{11.3.4}=\dfrac{49}{4}\)
\(b.\dfrac{3}{x}=\dfrac{3}{7}\Rightarrow x=7\)
Bài 2: ở phía dưới mình đã làm rồi
![](https://rs.olm.vn/images/avt/0.png?1311)
`a, x-2=7/15`
`=>x=7/15 +2`
`=>x= 7/15+ 30/15`
`=>x= 37/15`
`b, 9/20 +x=2/5 +3/20`
`=> 9/20 +x=8/20 +3/20`
`=> 9/20 +x=11/20`
`=>x=11/20-9/20`
`=>x= 2/20=1/10`
x = \(\dfrac{7}{15}\) + 2 = \(\dfrac{37}{15}\)
x = \(\dfrac{2}{5}\) + \(\dfrac{3}{20}\) - \(\dfrac{9}{20}\) = \(\dfrac{1}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,x-\dfrac{5}{7}=\dfrac{3}{5}\times\dfrac{1}{2}\)
\(x-\dfrac{5}{7}=\dfrac{3}{10}\)
\(x=\dfrac{3}{10}+\dfrac{5}{7}\)
\(x=\dfrac{21}{70}+\dfrac{50}{70}\)
\(x=\dfrac{71}{70}\)
\(Vayx=\dfrac{71}{70}\)
\(b,x\times\dfrac{2}{3}=\dfrac{5}{8}+\dfrac{3}{4}\)
\(x\times\dfrac{2}{3}=\dfrac{5}{8}+\dfrac{6}{8}\)
\(x\times\dfrac{2}{3}=\dfrac{11}{8}\)
\(x=\dfrac{11}{8}:\dfrac{2}{3}\)
\(x=\dfrac{11}{8}\times\dfrac{3}{2}\)
\(x=\dfrac{33}{16}\)
\(Vayx=\dfrac{33}{16}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>x-5/14=6/14-1/14=5/14
=>x=10/14=5/7
b; =>2/9:x=3/6+4/6=7/6
=>x=2/9:7/6=2/9*6/7=4/21
c: =>32:x=8
=>x=4
`@` `\text {Answer}`
`\downarrow`
`a,`
`x - 5/14 = 3/7 - 1/14`
`x - 5/14 = 5/14`
`=> x = 5/14 + 5/14`
`=> x = 5/7`
Vậy, `x = 5/7`
`b,`
`2/9 \div x = 1/2 + 2/3`
`2/9 \div x = 7/6`
`x = 2/9 \div 7/6`
`x = 4/21`
Vậy, `x = 4/21`
`c,`
\(\dfrac{6}{32\div x}=\dfrac{12}{16}\)
`6/(32 \div x) = 3/4`
`32 \div x = 6 \div 3/4`
`32 \div x = 8`
` x = 32 \div 8`
`x = 4`
Vậy, `x = 4`
\(\Leftrightarrow\dfrac{2}{3}x=\dfrac{1}{3}x\)
=>x=0
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