Question

a) (3x+2)(x^2-1)=(9x^2-4)(x+1)

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

c) (2x-5)^2-(x+2)^2=0

d) (x^2-2x+1)-4=0

Answer 1

a) \(\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)

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

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

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

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

Vậy ................

Answer 2

b) \(\left(x+2\right)\left(3-4x\right)=x^2+4x+4\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{5}\end{matrix}\right.\)

Vậy ................

Answer 3

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\3x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=1\end{matrix}\right.\)

Vậy ...........

p/s: câu d dùng đẳng thức

Answer 4

d.

\(\left(x^2-2x+1\right)-4=0\)

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

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

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

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

Vậy.........

Answer 5

b)\(\left(x+2\right)\left(3-4x\right)=x^2+4x+4\)

\(< =>\left(x+2\right)\left(3-4x\right)-\left(x+2\right)^2=0\)

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

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

\(< =>\left[{}\begin{matrix}x+2=0\\-5x-5=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=-2\\x=-1\end{matrix}\right.\)

S=\(\left\{-2;-1\right\}\)

Answer 6

a)\(\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)

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

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

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

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

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

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

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

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

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

S=\(\left\{\dfrac{-2}{3};1;\dfrac{-1}{2}\right\}\)