Question

[Tuyết Ly]

Giải các phương trình sau:

Answer 1

[Vô danh]

\(a,3\left(x-1\right)-7=5\left(x+2\right)\\ \Leftrightarrow3x-3-7=5x+10\\ \Leftrightarrow3x-10=5x+10\\ \Leftrightarrow2x+20=0\\ \Leftrightarrow x=-10\\ b,\dfrac{3x+1}{2}-\dfrac{x+3}{5}=\dfrac{x}{10}+2\\ \Leftrightarrow\dfrac{5\left(3x+1\right)}{10}-\dfrac{2\left(x+3\right)}{10}-\dfrac{x}{10}-\dfrac{20}{10}\\ \Leftrightarrow15x+5-2x-6-x-20=0\\ \Leftrightarrow12x-21=0\\ \Leftrightarrow x=\dfrac{7}{4}\)

\(c,ĐKXĐ:x\ne\pm1\\ \dfrac{x-2}{x+1}-\dfrac{x}{x-1}=\dfrac{x-8}{x^2-1}\\ \Leftrightarrow\dfrac{\left(x-2\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x-8}{\left(x+1\right)\left(x-1\right)}=0\\ \Leftrightarrow\dfrac{x^2-3x+2-x^2-x-x+8}{\left(x+1\right)\left(x-1\right)}=0\\ \Rightarrow-5x+10=0\\ \Leftrightarrow x=2\left(tm\right)\)

Answer 2

[TranGiaHuy8A2PhuThai2022]

a,3x-3-7=5x+10

<=>3x-10-5x-10=0

<=>-2x-20=0

<=>x=-10

b,quy đồng mẫu là 10 , ta được

5(3x+1)-2(x+3)=x+20

<=>15x+5-2x-6=x+20

<=>13x-x-1-20=0

<=>12x=21

<=>x=\(\dfrac{7}{4}\)

c,đk:x khác 1,-1

quy đồng khử mẫu ta được

(x-2)(x-1)-x(x+1)=x-8

<=>x\(^2\) -3x+2-x\(^2\) -x=x-8

<=>-4x-x=-8-2

<=>-5x=-10

<=>x=2