Question

[Sứ Tử Thiên]

Tìm x biết:

a) \(x.0,\left(2\right)+0,\left(3\right)=0,\left(77\right)\)
b)\(0,\left(153\right):0,\left(123\right)=1\dfrac{10}{41}.x\)

Answer 1

[JakiNatsumi]

\(a,\) \(x.0,\left(2\right)+0,\left(3\right)=0,\left(77\right)\)

⇔ \(x.2.0,\left(1\right)+3.0,\left(1\right)=77.0,\left(01\right)\)

⇔ \(2x.\dfrac{1}{9}+3.\dfrac{1}{9}=77.\dfrac{1}{99}\)

⇔ \(2x.\dfrac{1}{9}+\dfrac{1}{3}=\dfrac{7}{9}\)

⇔ \(2x.\dfrac{1}{9}=\dfrac{7}{9}-\dfrac{1}{3}=\dfrac{4}{9}\)

⇔ \(2x=\dfrac{4}{9}:\dfrac{1}{9}=4\)

⇔ \(x=4:2=2\)

Vậy \(x=2\)

\(b,\) \(0,\left(153\right):0,\left(123\right)=1\dfrac{10}{41}.x\)

⇔ \(153.0,\left(001\right):\left[123.0,\left(001\right)\right]=\dfrac{51}{41}.x\)

⇔ \(153.\dfrac{1}{999}:\left(123.\dfrac{1}{999}\right)=\dfrac{51}{41}.x\)

⇔ \(\dfrac{17}{111}:\dfrac{41}{333}=\dfrac{51}{41}.x\)

⇔ \(\dfrac{51}{41}=\dfrac{51}{41}x\)

⇔ \(x=\dfrac{51}{41}:\dfrac{51}{41}=1\)

Vậy \(x=1\)

Answer 2

[Huyền Linh Channel]

a)x.0,(2)+0,(3)=0,(77)

x.0,(2)=0,(77)-0,(3)

x.0,(2)=0,47

x=0,47:0,(2)

x=0,77

b) 0,(153):0,(123)=1/10/41.x

1,24390=1/10/41.x

x=1/10/41:1,24390

x=1