1, Tìm GTNN
A=x2+3x-1
B=2x2+5x+2
giải
A=(x^2+2.\(\dfrac{3}{2}\)x+\(\dfrac{9}{4}\))-\(\dfrac{13}{4}\)
A=(x+\(\dfrac{3}{2}\))^2-\(\dfrac{13}{4}\)
GTNN=-13/4
B=2(x^2+\(\dfrac{2}{5}\)x+1)
B=2(x^2+2.\(\dfrac{1}{5}\)x+\(\dfrac{1}{25}\))+\(\dfrac{24}{25}\)
2(x^2+\(\dfrac{1}{5}\))^2+\(\dfrac{24}{25}\)
GTNN=\(\dfrac{24}{25}\)
bạn tham khảo nhé
Ta có: \(A=x^2+3x-1\)
\(=x^2+2.1,5x+\left(1,5\right)^2-3,25\)
\(=\left(x+1,5\right)^2-3,25\)
Ta lại có: \(\left(x+1,5\right)^2\ge0\)
\(\Rightarrow\left(x+1,5\right)^2-3,25\ge-3,25\)
\(\Rightarrow A\ge-3,25\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x+1,5\right)^2=0\)
\(\Leftrightarrow x+1,5=0\)
\(\Leftrightarrow x=-1,5\)
Vậy MinA = -3,25 \(\Leftrightarrow x=-1,5\)
a,ta có:A=x^2+3x-1=x^2+2.x.3/2+(3/2)^2-9/4-1
=(x-3/2)^2-13/4
do (x-3/2)^2≥0 với mọi x( dấu "="xảy ra <=>x=3/2)
=>(x-3/2)^2-13/4 ≥-13/4
hay A≥-13/4( dấu "="xảy ra <=>x=3/2)
Vậy Min A = -13/4 tại x=3/2
b,ta có:B=2x^2+5x+2=2.(x^2+5/2.x+1)
=2.[x^2+2.x.5/4+(5/4)^2-25/16+1]
=2.[(x+5/4)^2-9/16]
=2.(x+5/4)^2-9/8
Do 2.(x+5/4)^2≥0 với mọi x (dấu"="xảy ra<=>x=-5/4)
=>2.(x+5/4)^2-9/8≥ -9/8
hayB≥-9/8 (dấu"="xảy ra<=>x=-5/4)
Vậy Min B=-9/8 tại x=-5/4
