Question

Bằng cách phân tích vế trái thành nhân tử, giải các PT sau:

d) \(x\left(2x-7\right)-4x+14=0\)

e) \(\left(2x-5\right)^2-\left(x+2\right)^2=0\)

f) \(x^2-x-\left(3x-3\right)=0\)

Answer 1

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

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-7=0\\x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=4\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{7}{2};4\right\}\)

e) \(PT\Leftrightarrow\left(2x-5-x-2\right)\left(2x-5+x+2\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\3x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=1\end{matrix}\right.\)

Vậy: \(S=\left\{7;1\right\}\)

f) \(PT\Leftrightarrow x\left(x-1\right)-3\left(x-1\right)=0\)

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

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

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

Vậy: \(S=\left\{1;3\right\}\)

Answer 2

\(d,x\left(2x-7\right)-4x+14=0\)

\(x\left(2x-7\right)-2\left(2x-7\right)=0\)

\(\left(x-2\right)\left(2x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{7}{2}\end{matrix}\right.\)

 

Answer 3

d: =>(2x-7)(x-2)=0

=>x=7/2 hoặc x=2

e: =>(2x-5-x-2)(2x-5+x+2)=0

=>(x-7)(3x-3)=0

=>x=7 hoặc x=1

f: =>x(x-1)-3(x-1)=0

=>(x-1)(x-3)=0

=>x=1 hoặc x=3

Answer 4

d, \(\Leftrightarrow x\left(2x-7\right)-2\left(2x-7\right)=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}2x-7=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{7}{2}\\x=2\end{matrix}\right.\)

Answer 5

e,\(\Leftrightarrow\left(2x-5-x-2\right)\left(2x-5+x+2\right)=0\)

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

\(\Leftrightarrow3\left(x-7\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-7=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=7\\x=1\end{matrix}\right.\)

Answer 6

\(d,x\left(2x-7\right)-4x+17=0\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}2x-7=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{7}{2}\\x=2\end{matrix}\right.\)

\(e,\left(2x-5\right)^2-\left(x+2\right)^2=0\)

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

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

\(\left\{{}\begin{matrix}x=7\\x=1\end{matrix}\right.\)

\(f,x^2-x-\left(3x-3\right)=0\)

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

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

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

Answer 7

f, \(\Leftrightarrow x\left(x-1\right)-3\left(x-1\right)=0\)

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

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