`@` `\text {Ans}`
`\downarrow`
`a)`
\(\dfrac{2}{3}-4x=\dfrac{1}{2}-\dfrac{2}{5}\)
`=> 2/3 - 4x = 1/10`
`=> 4x = 2/3 - 1/10`
`=> 4x = 17/30`
`=> x=17/30 \div 4`
`=> x= 17/120`
Vậy, `x=17/120`
`b)`
\(\dfrac{1}{2}+\dfrac{2}{3}\div x=-\dfrac{5}{4}\)
`=> 2/3 \div x = -5/4 - 1/2`
`=> 2/3 \div x = -7/4`
`=> x = 2/3 \div (-7/4)`
`=> x=-8/21`
`c)`
\(0,01 \div 2,5 = 0,75x \div 0,75\)
`=> 1/250 = x`
Vậy, `x=1/250`
`d)`
\(3,8\div2x=\dfrac{1}{4}\div2\dfrac{2}{3}\)
`=> 3,8 \div 2x = 3/32`
`=> 2x = 3,8 \div 3/32`
`=> 2x = 608/15`
`=> x=608/15 \div 2`
`=> x=304/15`
Vậy, `x=304/15`
`e)`
\(1\dfrac{7}{9}\div\left[\left(1-x\right)\div\dfrac{2}{3}+\left(\dfrac{1}{2}\right)^2\cdot\dfrac{4}{5}\right]=\dfrac{5}{9}\)
`=> 1 7/9 \div [(1-x) \div 2/3 + 1/4*4/5] = 5/9`
`=> 1 7/9 \div [(1-x) \div 2/3 + 1/5] = 5/9`
`=> (1-x) \div 2/3 + 1/5 = 1 7/9 \div 5/9`
`=> (1-x) \div 2/3+ 1/5 = 16/5`
`=> (1-x) \div 2/3 = 16/5 - 1/5`
`=> (1-x) \div 2/3 = 15/5`
`=> 1-x = 3*2/3`
`=> 1 - x = 2`
`=> x= 1- 2`
`=> x = -1`
Vậy, `x=-1.`
`@` `\text {Kaizuulvu.}`
a: =>4x=2/3-1/2+2/5=40/60-30/60+24/60=34/60=17/30
=>x=17/120
b: =>2/3:x=-5/4-1/2=-7/4
=>x=-2/3:7/4=-2/3*4/7=-8/21
c: =>x:1=0,01:5/2=0,01*2/5=1/100*2/5=2/500=1/250
=>x=1/250
d: =>3,8:2x=1/4:8/3=1/4*3/8=3/32
=>2x=608/15
=>x=304/15
