|x^2 +1|-x^2+1/2.x=-3 tìm x |1/2+x^2|+|1/3+x^2|=2.(x^2-1/5+x) 18/07/2021 Bởi Josie |x^2 +1|-x^2+1/2.x=-3 tìm x |1/2+x^2|+|1/3+x^2|=2.(x^2-1/5+x)
Đáp án: a)$x=4$b)$x=-\dfrac{25}{12}$ Giải thích các bước giải: a) $|x^2+1|-x^2+\dfrac{1}{2}x=3$$⇔x^2+1-x^2+\dfrac{1}{2}x=3$$⇔\dfrac{1}{2}x=2$$⇔x=4$b)$|x^2+\dfrac{1}{2}|+|x^2+\dfrac{1}{3}|=2(x^2-\dfrac{1}{5}x)$⇔$x^2+\dfrac{1}{2}+x^2+\dfrac{1}{3}=2x^2-\dfrac{2}{5}x$$⇔-\dfrac{2}{5}x=\dfrac{5}{6}$$⇔x=-\dfrac{25}{12}$ Bình luận
Đáp án:
a)$x=4$
b)$x=-\dfrac{25}{12}$
Giải thích các bước giải:
a) $|x^2+1|-x^2+\dfrac{1}{2}x=3$
$⇔x^2+1-x^2+\dfrac{1}{2}x=3$
$⇔\dfrac{1}{2}x=2$
$⇔x=4$
b)$|x^2+\dfrac{1}{2}|+|x^2+\dfrac{1}{3}|=2(x^2-\dfrac{1}{5}x)$
⇔$x^2+\dfrac{1}{2}+x^2+\dfrac{1}{3}=2x^2-\dfrac{2}{5}x$
$⇔-\dfrac{2}{5}x=\dfrac{5}{6}$
$⇔x=-\dfrac{25}{12}$