Question

Giải phương trình

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

b) (5,5-11x)( \(\frac{7x+2}{5}\) + \(\frac{2\left(1-3x\right)}{3}\) ) = 0

Answer 1

\(A=\frac{1-x}{x+1}+3=\frac{2x+3}{x+1}\Leftrightarrow\frac{1-x+3x+3-2x-3}{x+1}=\frac{1}{x+1}=0\)

Vô nghiệm.

\(B=\left(5,5-11x\right)\left(\frac{7x+2}{5}+\frac{2\left(1-3x\right)}{3}\right)=0\)

\(\left[\begin{matrix}5,5-11x=0\\3\left(7x+2\right)+10-30x=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=\frac{5,5}{11}\\-9x+16=0\end{matrix}\right.\)\(\left[\begin{matrix}x=\frac{1}{2}\\x=\frac{16}{9}\end{matrix}\right.\)

Answer 2

b) (5,5-11x)(\(\frac{7x+2}{5}\)+ \(\frac{2\left(1-3x\right)}{3}\)) = 0

<=> (5,5 - 11x )(\(\frac{-9x+16}{15}\))=0

<=>\(\left[\begin{matrix}5,5-11x=0\\\frac{-9x+16}{15}=0\end{matrix}\right.\)

<=> \(\left[\begin{matrix}x=\frac{1}{2}\\x=\frac{16}{9}\end{matrix}\right.\)

Vậy pt có nghiệm là x=\(\frac{1}{2}\) và x= \(\frac{16}{9}\)

Answer 3

a) \(\frac{1-x}{x+1}\)+ 3 = \(\frac{2x+3}{x+1}\)ĐKXĐ x\(\ne\)-1

<=> \(\frac{1-x+3x+3-2x-3}{x+1}\)

<=> \(\frac{1}{x+1}\)=0 ( vô lí )

Vậy pt vô nghiệm