Question

giúp mik vs ạ. Cảm ơn nhìu nhé

Answer 1

\(PT\Leftrightarrow\left(3x+5\right)\dfrac{x+1}{2}+\left(3x+5\right)\dfrac{2x-2}{3}-\left(3x+5\right)=0\)

\(\Leftrightarrow\left(3x+5\right)\left(\dfrac{x+1}{2}+\dfrac{2x-2}{3}-1\right)=0\)

\(\Leftrightarrow\left(3x+5\right)\dfrac{3\left(x+1\right)+2\left(2x-2\right)-6}{6}=0\)

\(\Leftrightarrow\left(3x+5\right)\dfrac{7x-7}{6}=0\)

\(\Leftrightarrow\left(3x+5\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{3}\\x=1\end{matrix}\right.\)

KL: Phương trình có tập nghiệm \(S=\left\{\dfrac{-5}{3};1\right\}\)

Answer 2

g, cộng 1 vào cả 2 vế của pt , ta được :

<=>\(\dfrac{x+1}{7}+1+\dfrac{x+2}{6}=\dfrac{x+3}{5}+1-\dfrac{x+4}{4}+1=0\)

<=>\(\dfrac{x+8}{7}+\dfrac{x+8}{6}-\dfrac{x+8}{5}+\dfrac{x+8}{4}=0\)

<=>\(\left(x+8\right)\left(\dfrac{1}{7}+\dfrac{1}{6}-\dfrac{1}{5}+\dfrac{1}{4}\right)=0\)

<=> x+8 =0 

<=> x=-8

 

h, 

\(\dfrac{3x-2}{3}\left(x-1\right)+\dfrac{3x-2}{4}\left(2x-2\right)=0\)

<=>\(\dfrac{3x-2}{3}\left(x-1\right)+2.\dfrac{3x-2}{4}\left(x-1\right)=0\)

<=>\(\dfrac{3x-2}{3}\left(x-1\right)+\dfrac{6x-4}{4}\left(x-1\right)=0\)

<=>\(\left(x-1\right).\left(\dfrac{3x-2}{3}+\dfrac{6x-4}{4}\right)=0\)

<=>x-1 =0 

<=> x=1