Question

giải phương trình 

a, 1-\(\frac{2x+3}{7}=0\)

b, \(\frac{3x-5}{4x+1}-\frac{x-2}{3x-5}=0\)

c ,

Answer 1

a) \(1-\frac{2x+3}{7}=0\)

=> \(\frac{7-2x-3}{7}=0\)

=> \(\frac{4-2x}{7}=0\)

=> 4 - 2x = 0

=> 2x = 4

=> x = 4 : 2 = 2

Answer 2

a) \(1-\frac{2x+3}{7}=0\)

\(\Leftrightarrow\frac{7}{7}-\frac{2x-3}{7}=0\)

\(\Leftrightarrow\frac{7-2x+3}{7}=0\)

\(\Leftrightarrow7-2x+3=0\)

\(\Leftrightarrow-2x+10=0\)

\(\Leftrightarrow-2\left(x-10\right)=0\)

\(\Leftrightarrow x-10=0\)

\(\Leftrightarrow x=10\)

Answer 3

a ) \(1-\frac{2x+3}{7}=0\)

\(\Leftrightarrow\frac{7}{7}-\frac{2x+3}{7}=0\)

\(\Leftrightarrow7-2x+3=0\)

\(\Leftrightarrow2x+3=7\)

\(\Leftrightarrow2x=4\)

\(\Leftrightarrow x=2\)

Answer 4

mk nhầm chút , quên chưa đổi dấu.......sorry nhé

\(\frac{7-2x-3}{7}=0\)

\(\Leftrightarrow7-2x-3=0\)

\(\Leftrightarrow4-2x=0\)

\(\Leftrightarrow2\left(2-x\right)=0\)

\(\Leftrightarrow2-x=0\)

\(\Leftrightarrow x=2\)