Tìm GTNN `A=x^2-2x+6` `B=m^2-m+1` `C=x^2+3x` 01/07/2021 Bởi Valentina Tìm GTNN `A=x^2-2x+6` `B=m^2-m+1` `C=x^2+3x`
Đáp án: $min_A=5 \Leftrightarrow x=1\\ min_B=\dfrac{3}{4} \Leftrightarrow x=\dfrac{1}{2}\\ min_C=-\dfrac{9}{4} \Leftrightarrow x=-\dfrac{3}{2}$ Giải thích các bước giải: $A=x^2-2x+6\\ =x^2-2x+1+5\\ =(x-1)^2+5 \ge 0 \ \forall \ x\\ min_A=5 \Leftrightarrow x=1\\ B=m^2-m+1\\ =m^2-m+\dfrac{1}{4}+\dfrac{3}{4}\\ =\left(m-\dfrac{1}{2}\right)^2+\dfrac{3}{4} \ge \dfrac{3}{4} \ \forall \ m\\ min_B=\dfrac{3}{4} \Leftrightarrow x=\dfrac{1}{2}\\ C=x^2+3x\\ =x^2+3x+\dfrac{9}{4}-\dfrac{9}{4}\\ =\left(x+\dfrac{3}{2}\right)-\dfrac{9}{4} \ge -\dfrac{9}{4} \ \forall \ x\\ min_C=-\dfrac{9}{4} \Leftrightarrow x=-\dfrac{3}{2}$ Bình luận
Đáp án:
$min_A=5 \Leftrightarrow x=1\\ min_B=\dfrac{3}{4} \Leftrightarrow x=\dfrac{1}{2}\\ min_C=-\dfrac{9}{4} \Leftrightarrow x=-\dfrac{3}{2}$
Giải thích các bước giải:
$A=x^2-2x+6\\ =x^2-2x+1+5\\ =(x-1)^2+5 \ge 0 \ \forall \ x\\ min_A=5 \Leftrightarrow x=1\\ B=m^2-m+1\\ =m^2-m+\dfrac{1}{4}+\dfrac{3}{4}\\ =\left(m-\dfrac{1}{2}\right)^2+\dfrac{3}{4} \ge \dfrac{3}{4} \ \forall \ m\\ min_B=\dfrac{3}{4} \Leftrightarrow x=\dfrac{1}{2}\\ C=x^2+3x\\ =x^2+3x+\dfrac{9}{4}-\dfrac{9}{4}\\ =\left(x+\dfrac{3}{2}\right)-\dfrac{9}{4} \ge -\dfrac{9}{4} \ \forall \ x\\ min_C=-\dfrac{9}{4} \Leftrightarrow x=-\dfrac{3}{2}$