Question

 

Tìm x

1/a) \(\frac{10+x}{17+x}=\frac{3}{4}\)

 b) \(\frac{40+x}{77-x}=\frac{6}{7}\)

Answer 1

Ta có :\(\frac{40+x}{77-x}=\frac{6}{7}\)

\(\Rightarrow7\left(40+x\right)=\left(77-x\right).6\)

\(\Leftrightarrow280+7x=462-6x\)

\(\Rightarrow7x+6x=462-280\)

\(\Rightarrow13x=182\)

\(\Rightarrow x=14\)

Answer 2

Ta có : \(\frac{10+x}{17+x}=\frac{3}{4}\)

\(\Rightarrow4\left(10+x\right)=3\left(17+x\right)\)

\(\Leftrightarrow40+4x=51+3x\)

\(\Rightarrow4x-3x=51-40\)

\(\Rightarrow x=11\)

Answer 3

\(\frac{40+x}{77-x}=\frac{6}{7}\)

\(\Rightarrow7\left(40+x\right)=\left(77-x\right).6\)