Question

1) ( 3x - 7)(x^2 + 4) =0
2) [( -2x + 6)^2 -1]( x+8) = 0

 

Answer 1

\(1.\left(3x-7\right)\left(x^2+4\right)=0\)

\(\Leftrightarrow3x-7=0\)

\(\Leftrightarrow x=\dfrac{7}{3}\)

\(2.\left[\left(-2x+6\right)^2-1\right]\left(x+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(-2x+6\right)^2-1=0\\x+8=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x+6=1\\x=-8\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-8\end{matrix}\right.\)

Answer 2

a, Ta có x^2 + 4 > 0 

=> 3x - 7 = 0 <=> x = 7/3 

2, TH1 : x + 8 = 0 <=> x = -8 

TH2 : (-2x+5)(-2x+7) = 0 <=> x = 5/2 ; x = 7/2