Question

b) (2x−1)2−(x+3)2=0;

c)  x2(x−3)+12−4x=0
 .

Answer 1

x^2(x-3) +4(3-x)=0

x^2(x-3)-4(x-3)=0

(x^2-4)(x-3)=0

xảy ra 2: Th1:x^2-4=0 =>x=-2 hoặc x=2

               Th2:x-3=0 =>x=3

Answer 2

b)(2x-1-x-3)2=0         

=>x-4=0

=>x=4

Answer 3

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{3}\end{matrix}\right.\)

b) \(x^2\left(x-3\right)+12-4x=0\)

\(\Leftrightarrow x^2\left(x-3\right)-4\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2-4\right)=0\)

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

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

Answer 4

b: Ta có: \(\left(2x-1\right)^2-\left(x+3\right)^2=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{3}\end{matrix}\right.\)

c: Ta có: \(x^2\left(x-3\right)-4x+12=0\)

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

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