Question

1.Tìm x:

96-3(x.1)=42

2.Cho a-b=8.Tính:

a.7a-7b

b.9a-3b+9b-3a

Answer 1

1. \(96-3\left(x.1\right)=42\)

\(96-3x=42\)

\(3x=96-42\)

\(3x=54\)

\(x=54:3\)

\(x=18\)

2.Theo bài ra ta có: \(a-b=8\)

a. Ta có: \(7a-7b=7\left(a-b\right)\)

\(=7.8=56\)

Answer 2

Bài 1:

96-3(x.1)=42

3.(x.1)= 96 - 42

3. (x.1)= 54

x.1= 54:3

x.1= 18

x=18:1=18

Vậy x = 18

Bài 2:

a) 7a-7b

Theo đề bài cho a-b=8

⇒ 7.a-7.b = 7. (a-b)=7.8=56

Answer 3

1) Tìm x:

\(96-3\left(x.1\right)=42\Leftrightarrow96-3x=42\)

\(\Leftrightarrow3x=96-42=54\Leftrightarrow x=\dfrac{54}{3}=18\)

2)Ta có: a-b=8

\(a)7a-7b=7\left(a-b\right)=7.8=56\)

\(b)9a-3b+9b-3a=6a+6b=6\left(a+b\right)\)