tim x biet a,25x^2 -9=0 b,(x+4)^2 -(x+1)*(x-6)=16 c,(2x-1)^2 +(x+3)^2 -5*(x+7)*(x-7)=0 21/07/2021 Bởi Melody tim x biet a,25x^2 -9=0 b,(x+4)^2 -(x+1)*(x-6)=16 c,(2x-1)^2 +(x+3)^2 -5*(x+7)*(x-7)=0
$a)25x² -9=0$ $(=)(5x-3)(5x+3)=0$ $(=)$\(\left[ \begin{array}{l}5x-3=0\\5x+3=0\end{array} \right.\) $(=)$\(\left[ \begin{array}{l}5x=3\\5x=-3\end{array} \right.\)$(=)$ \(\left[ \begin{array}{l}x=\frac{3}{5}\\x=-\frac{3}{5}\end{array} \right.\) $S={{\frac{3}{5};\frac{-3}{5}}}$ $b)(x+4)²-(x+1)(x-6)=16$ $(=)x²+8x+16-(x²-6x+x-6)=16$ $(=)x²+8x+16-x²+6x-x+6=16$ $(=)13x+22=16$ $(=)13x=16-22$ $(=)13x=-6$ $(=)x=\frac{-6}{13}$ $S$={$\frac{-6}{13}$} $c)(2x-1)² +(x+3)² -5(x+7)(x-7)=0$ $(=)4x²-4x+1+x²+6x+9-5(x²-49)=0$ $(=)4x²-4x+1+x²+6x+9-5x²+245=0$ $(=)2x+255=0$ $(=)2x=-255$ $(=)x=-\frac{255}{2}$ $S={-\frac{255}{2}}$ Bình luận
$a)25x² -9=0$
$(=)(5x-3)(5x+3)=0$
$(=)$\(\left[ \begin{array}{l}5x-3=0\\5x+3=0\end{array} \right.\) $(=)$\(\left[ \begin{array}{l}5x=3\\5x=-3\end{array} \right.\)$(=)$ \(\left[ \begin{array}{l}x=\frac{3}{5}\\x=-\frac{3}{5}\end{array} \right.\)
$S={{\frac{3}{5};\frac{-3}{5}}}$
$b)(x+4)²-(x+1)(x-6)=16$
$(=)x²+8x+16-(x²-6x+x-6)=16$
$(=)x²+8x+16-x²+6x-x+6=16$
$(=)13x+22=16$
$(=)13x=16-22$
$(=)13x=-6$
$(=)x=\frac{-6}{13}$
$S$={$\frac{-6}{13}$}
$c)(2x-1)² +(x+3)² -5(x+7)(x-7)=0$
$(=)4x²-4x+1+x²+6x+9-5(x²-49)=0$
$(=)4x²-4x+1+x²+6x+9-5x²+245=0$
$(=)2x+255=0$
$(=)2x=-255$
$(=)x=-\frac{255}{2}$
$S={-\frac{255}{2}}$