Question

[Hiếu Văn Huỳnh]

help meeeeeee!!!!!!

giải phương trình :

a) 3x(x-1)=(x-1)(x+2)

b) \(\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3\)

c) \(\frac{1}{2}\left(x+1\right)+\frac{1}{4}\left(x+3\right)=3-\frac{1}{2}\left(x+2\right)\)

d) ^** \(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\) ưu tiên câu d nha

Answer 1

[Nguyễn Lê Phước Thịnh]

a) Ta có: 3x(x-1)=(x-1)(x+2)

⇔3x(x-1)-(x-1)(x+2)=0

⇔(x-1)(3x-x-2)=0

⇔(x-1)(2x-1)=0

⇔2(x-1)²=0

mà 2≠0

nên (x-1)²=0

⇔x-1=0

hay x=1

Vậy: x=1

b) Ta có: \(\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3\)

\(\Leftrightarrow\frac{21\left(4x+3\right)}{105}-\frac{15\left(6x-2\right)}{105}-\frac{35\left(5x+4\right)}{105}-\frac{315}{105}=0\)

\(\Leftrightarrow84x+63-90x+30-175x-140-315=0\)

\(\Leftrightarrow-181x-362=0\)

\(\Leftrightarrow-181x=362\)

hay x=-2

Vậy: x=-2

c) Ta có: \(\frac{1}{2}\left(x+1\right)+\frac{1}{4}\left(x+3\right)=3-\frac{1}{2}\left(x+2\right)\)

\(\Leftrightarrow\frac{x}{2}+\frac{1}{2}+\frac{x}{4}+\frac{3}{4}=3-\frac{x}{2}-1\)

\(\Leftrightarrow\frac{x}{2}+\frac{x}{4}+\frac{5}{4}-\frac{-x}{2}-2=0\)

\(\Leftrightarrow\frac{x}{2}+\frac{x}{4}+\frac{x}{2}-\frac{3}{4}=0\)

\(\Leftrightarrow\frac{x}{4}+x-\frac{3}{4}=0\)

\(\Leftrightarrow\frac{x}{4}+\frac{4x}{4}-\frac{3}{4}=0\)

\(\Leftrightarrow5x-3=0\)

\(\Leftrightarrow5x=3\)

hay \(x=\frac{3}{5}\)

Vậy: \(x=\frac{3}{5}\)

d) Ta có: \(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)

\(\Leftrightarrow\frac{x+2}{98}+1+\frac{x+4}{96}+1=\frac{x+6}{94}+1+\frac{x+8}{92}+1\)

\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)

mà \(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0\)

nên x+100=0

hay x=-100

Vậy: x=-100