Question

Tìm x biết:

a, 3x² (x - 5) + 12 (5 - x) = 0

b, (x - 2)² - (3x - 1)² = 0

c, \(\dfrac{\left(x+1\right)^2}{3}-\dfrac{\left(x-2\right)^2}{2}=\dfrac{2x+1}{2}-\dfrac{\left(x-3\right)^2}{6}\)

Làm được câu nào thì làm nha, ko cần làm hết đâu mà nếu làm hết được thì mk sẽ like nhìu hơn, giúp nha!

Answer 1

a)

\(3x^2(x-5)+12(5-x)=0\)

\(\Leftrightarrow 3x^2(x-5)-12(x-5)=0\)

\(\Leftrightarrow (x-5)(3x^2-12)=0\)

\(\Leftrightarrow 3(x-5)(x^2-4)=0\)

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

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

Answer 2

b)

\((x-2)^2-(3x-1)^2=0\)

\(\Leftrightarrow [(x-2)-(3x-1)][(x-2)+(3x-1)]=0\)

\(\Leftrightarrow (-2x-1)(4x-3)=0\Rightarrow \left[\begin{matrix} -2x-1=0\\ 4x-3=0\end{matrix}\right. \)

\(\Rightarrow \left[\begin{matrix} x=\frac{-1}{2}\\ x=\frac{3}{4}\end{matrix}\right.\)

Answer 3

c)

\(\frac{(x+1)^2}{3}-\frac{(x-2)^2}{2}=\frac{2x+1}{2}-\frac{(x-3)^2}{6}\)

\(\Leftrightarrow \frac{2(x+1)^2}{6}-\frac{3(x-2)^2}{6}=\frac{3(2x+1)}{6}-\frac{(x-3)^2}{6}\)

\(\Leftrightarrow 2(x+1)^2-3(x-2)^2=3(2x+1)-(x-3)^2\)

\(\Leftrightarrow 2(x^2+2x+1)-3(x^2-4x+4)=6x+3-(x^2-6x+9)\)

\(\Leftrightarrow -x^2+16x-10=-x^2+12x-6\)

\(\Leftrightarrow 4x=4\Rightarrow x=1\)