Question

Giải các pt sau:

a) (x-3)-(x-3)(2x-5)/6=(x-3)(3-x)/4

b) (2x-7)^2-x^2+8x-16=0

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

Answer 1

\(\left(3x+1\right)\left(x-3\right)=\left(3x+1\right)\left(2x-5\right)\)

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

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

\(\Leftrightarrow\left(3x+1\right)\left(2-x\right)=0\)

\(\Leftrightarrow\left[\begin{matrix}3x+1=0\\2-x=0\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left[\begin{matrix}3x=-1\\x=2\end{matrix}\right.\)

\(\Leftrightarrow\left[\begin{matrix}x=-\frac{1}{3}\\x=2\end{matrix}\right.\)

Vậy tập nghiệm của pt là \(S=\left\{-\frac{1}{3};2\right\}\)

Answer 2

Có : \(\left(3x+1\right)\left(x-3\right)=\left(3x+1\right)\left(2x-5\right)\)

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

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

\(\Leftrightarrow\) \(\left(3x+1\right)\left(-x+2\right)=0\)

\(\Leftrightarrow\) \(\left[\begin{matrix}3x+1=0\\-x+2=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left[\begin{matrix}3x=-1\\-x=-2\end{matrix}\right.\) \(\Leftrightarrow\) \(\left[\begin{matrix}x=\frac{-1}{3}\\x=2\end{matrix}\right.\)

Vậy phương trình có tập nghiệm \(S=\left\{\frac{-1}{3};2\right\}\)

Answer 3

\(\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}=\frac{\left(x-3\right)\left(3-x\right)}{4}\)

\(\Leftrightarrow\frac{24\left(x-3\right)}{24}-\frac{4\left(x-3\right)\left(2x-5\right)}{24}=-\frac{6\left(x-3\right)\left(x-3\right)}{24}\)

\(\Leftrightarrow24\left(x-3\right)-4\left(x-3\right)\left(2x-5\right)+6\left(x-3\right)^2=0\)

\(\Leftrightarrow\left(x-3\right)\left[24-4\left(2x-5\right)+6\left(x-3\right)\right]=0\)

\(\Leftrightarrow\left(x-3\right)\left(24-8x+20+6x-18\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(26-2x\right)=0\)

\(\Leftrightarrow2\left(x-3\right)\left(13-x\right)=0\)

\(\Leftrightarrow\left[\begin{matrix}x-3=0\\13-x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[\begin{matrix}x=3\\x=13\end{matrix}\right.\)

Vậy tập nghiệm của pt là \(S=\left\{3;13\right\}\)

Answer 4

\(\left(2x-7\right)^2-x^2+8x-16=0\)

\(\Leftrightarrow\)\(\left(2x-7\right)^2-\left(x-4\right)^2=0\)

\(\Leftrightarrow\)\(\left(2x-7+x-4\right)\left(2x-7-x+4\right)=0\)

\(\Leftrightarrow\)\(\left(3x-11\right)\left(x-3\right)=0\)

\(\Leftrightarrow\)\(\left[\begin{matrix}3x-11=0\\x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left[\begin{matrix}3x=11\\x=3\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left[\begin{matrix}x=\frac{11}{3}\\x=3\end{matrix}\right.\)

Vậy tập nghiệm của pt là \(S=\left\{3;\frac{11}{3}\right\}\)

Answer 5

Có : \(\left(2x-7\right)^2-x^2+8x-16=0\)

\(\Leftrightarrow\) \(\left(2x-7\right)^2-\left(x^2-2.x.4+4^2\right)=0\)

\(\Leftrightarrow\) \(\left(2x-7\right)^2-\left(x-4\right)^2=0\)

\(\Leftrightarrow\) \(\left(2x-7+x-4\right)\left(2x-7-x+4\right)=0\)

\(\Leftrightarrow\) \(\left(3x-11\right)\left(x-3\right)=0\)

\(\Leftrightarrow\) \(\left[\begin{matrix}3x-11=0\\x-3=0\end{matrix}\right.\)\(\Leftrightarrow\) \(\left[\begin{matrix}3x=11\\x=3\end{matrix}\right.\) \(\Leftrightarrow\) \(\left[\begin{matrix}x=\frac{11}{3}\\x=3\end{matrix}\right.\)

Vậy tập nghiệm của phương trình \(S=\left\{\frac{11}{3};3\right\}\)