Câu hỏi của Nguyễn Hà Giang

Question

Tìm GTNN của biểu thức : A=(x^2+3x+4)^2.

mk cảm ơn các bn nhiều nha.❤💚💙💛💜💓💕💖💗💘💝💞💟

Truong Viet Truong · Accepted Answer

$A=\left(x^2+3x+4\right)^2$

ta có:

$x^2+3x+4=x^2+2\cdot\dfrac{3}{2}x+\left(\dfrac{3}{2}\right)^2+\dfrac{7}{4}\
=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}$

vậy $minA=\left(\dfrac{7}{4}\right)^2=\dfrac{49}{16}\Leftrightarrow x=-\dfrac{3}{2}$Câu trả lời: $A=\left(x^2+3x+4\right)^2$

ta có:

$x^2+3x+4=x^2+2\cdot\dfrac{3}{2}x+\left(\dfrac{3}{2}\right)^2+\dfrac{7}{4}\
=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}$

vậy $minA=\left(\dfrac{7}{4}\right)^2=\dfrac{49}{16}\Leftrightarrow x=-\dfrac{3}{2}$

Violympic toán 8