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Question

Chứng minh đa thức vô nghiệm f(x) = -x2 -5x -10

Lightning Farron · Accepted Answer

$f\left(x\right)=-x^2-5x-10=-\left(x^2+5x+10\right)$

$=-\left[x^2+5x+\dfrac{40}{4}\right]=-\left[x^2+5x+\dfrac{25}{4}+\dfrac{15}{4}\right]$

$=-\left[\left(x+\dfrac{5}{2}\right)^2+\dfrac{15}{4}\right]=-\left(x+\dfrac{5}{2}\right)^2-\dfrac{15}{4}$

Ta thấy: $-\left(x+\dfrac{5}{2}\right)^2\le0\forall x$

$\Rightarrow-\left(x+\dfrac{5}{2}\right)^2-\dfrac{15}{4}< 0\forall x$

Đa thức vô nghiệmCâu trả lời: $f\left(x\right)=-x^2-5x-10=-\left(x^2+5x+10\right)$

$=-\left[x^2+5x+\dfrac{40}{4}\right]=-\left[x^2+5x+\dfrac{25}{4}+\dfrac{15}{4}\right]$

$=-\left[\left(x+\dfrac{5}{2}\right)^2+\dfrac{15}{4}\right]=-\left(x+\dfrac{5}{2}\right)^2-\dfrac{15}{4}$

Ta thấy: $-\left(x+\dfrac{5}{2}\right)^2\le0\forall x$

$\Rightarrow-\left(x+\dfrac{5}{2}\right)^2-\dfrac{15}{4}< 0\forall x$

Đa thức vô nghiệm

Đại số lớp 7