Question

a) (x - 1)³ + 3(x + 1)² = (x² - 2x + 4)(x + 2)

b) x² - 4 = 8(x - 2)

c) x² - 4x + 4 = 9(x - 2)

d) 4x² - 12x + 9 = (5 - x)²

Answer 1

a) \(\left(x-1\right)^3+3\left(x+1\right)^2=\left(x^2-2x+4\right)\left(x+2\right)\)
\(\Leftrightarrow\left(x^3-3x^2+3x-1\right)+3\left(x^2+2x+1\right)=x^3+8\)
\(\Leftrightarrow x^3-3x^2+3x-1+3x^2+2x+1=x^3+8\)
\(\Leftrightarrow x^3-3x^2+3x+3x^2+2x-x^3=1-1+8\)
\(\Leftrightarrow5x=8\)
\(\Leftrightarrow x=\dfrac{8}{5}\)
Vậy \(S=\left\{\dfrac{8}{5}\right\}\)

b) \(x^2-4=8\left(x-2\right)\)
\(\Leftrightarrow\left(x+2\right)\left(x-2\right)-8\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2-8\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-6\right)=0\)
\(\Leftrightarrow x-2=0\) hoặc \(x-6=0\)
:) \(x-2=0\Leftrightarrow x=2\)
:) \(x-6=0\Leftrightarrow x=6\)
Vậy \(S=\left\{2;6\right\}\)

c) \(x^2-4x+4=9\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)^2=9\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x-2\right)=9\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x-2\right)-9\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-2-9\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-11\right)=0\)
\(\Leftrightarrow x-2=0\) hoặc \(x-11=0\)
:) \(x-2=0\Leftrightarrow x=2\)
:) \(x-11=0\Leftrightarrow x=11\)
Vậy \(S=\left\{2;11\right\}\)
(d ko bít lèm)
#IDOL

Answer 2

d, \(4x^2-12x+9=\left(5-x\right)^2\)

\(\Rightarrow4x^2-12x+9=25-10x+x^2\)

\(\Rightarrow4x^2-x^2-12x+10x+9-25=0\)

\(\Rightarrow3x^2-2x-16=0\)

\(\Rightarrow\left(-2-x\right)\left(8-3x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}-2-x=0\\8-3x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}-x=2\\-3x=-8\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{8}{3}\end{matrix}\right.\)