Question

[Anatole NGô]

Bài 1. (3 điểm) Giải các phương trình sau:

a) (3x - 7)(x + 5) = (5 + x)(3 - 2x)     

giúp mình với mai thi rồi

Answer 1

[YangSu]

\(a,\left(3x-7\right)\left(x+5\right)=\left(5+x\right)\left(3-2x\right)\)

\(\Leftrightarrow\left(3x-7\right)\left(x+5\right)-\left(x+5\right)\left(3-2x\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(3x-7-3+2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\5x-10=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

\(b,\dfrac{-x+3}{2}=\dfrac{x-2}{3}\left(MSC=6\right)\)

Suy ra :

\(3\left(-x+3\right)=2\left(x-2\right)\)

\(\Leftrightarrow-3x+9-2x+4=0\)

\(\Leftrightarrow-5x+13=0\)

\(\Leftrightarrow x=\dfrac{13}{5}\)

\(c,\dfrac{x-1}{x-2}+\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)\(\left(dkxd:x\ne\pm2\right)\)

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x+2\right)+5\left(x-2\right)-12-x^2+4}{x^2-4}=0\)

\(\Leftrightarrow x^2+2x-x-2+5x-10-12-x^2+4=0\)

\(\Leftrightarrow6x-20=0\)

\(\Leftrightarrow x=\dfrac{10}{3}\)\(\left(n\right)\)

Vậy \(S=\left\{\dfrac{10}{3}\right\}\)

Answer 2

[Tuyến Ngô]

a
b

phải ko bn