Question

bài 1 : tìm x 

a, 2*x⁴ + 3*x² - 16*x - 24 = 0

bài 2 : tìm x,y biết 

4*x^2 + 9*y^2 - 12*x - 32*y - 2*x*y + 44 = 0

Answer 1

a) \(2x^4+3x^3-16x-24=0\)

\(\left(2x^4+3x^3\right)-\left(16x+24\right)=0\)

\(x^3.\left(2x+3\right)-8\left(2x+3\right)=0\)

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

\(\Rightarrow\orbr{\begin{cases}x^3-8=0\\2x+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x^3=8\\2x=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{-3}{2}\end{cases}}\)

vậy \(\orbr{\begin{cases}x=2\\x=-\frac{3}{2}\end{cases}}\)