giup minh bai toan nay
2x-2x^2-5(tim gia tri lon nhat)
giai chi tiet giup minh may bai nay nha
1 gia tri x>0 thoa man
(2x-3)2=(x+5)2
2 gia tri lon nhat cua -3x2-6x-4
3 tim gia tri cua x+y biet
x-y=2 ; x*y=99 va y <0
4 nghiem cua phuong trinh
(2x-3)2-4x2-297=0
giup giai cau nay voi
tim gia tri lon nhat
x2-2xy+y2+2x-10x+17
minh dang can gap lam.
giup to 3 bai nay
1. gia tri x lon nhat thoa man : x2(5x-20)=15x
2 gia tri x >0 de -3x4+12x2+1 dat gia tri lon nhat
3. tap hop cac gia tri cua x thoa man : x4+x3+2x-4=0
giup to 3 bai nay
1. gia tri cua x thoa man (2x+3)(x+1)2-(2x+3)(2x-3)
2. so nghiem cua da thuc x3+6x2+11x+6 la ?
3. bieu thuc C = 8-5x-2x2 dat gia tri lon nhat tai x = ?
giup to 3 bai nay
1. gia tri cua x thoa man (2x+3)(x+1)2-(2x+3)(2x-3)
2. so nghiem cua da thuc x3+6x2+11x+6 la ?
3. bieu thuc C = 8-5x-2x2 dat gia tri lon nhat tai x = ?
ai giup minh giai bai nay voi
1<=z, y, z<=2
tim gia tri lon nhat
A=(x+y+z)(1/x+1/y+1/z)
giai giup minh voi
cho 0 <= X<= 3/2.tim gia tri lon nhat cua y=x(3-2x)2
giup minh bai nay
tim gia tri nhỏ nhất của đa thuc 2x^2-6x
thanks
Đặt:
\(A=2x^2-6x\)
\(A=2x^2-6x+\dfrac{9}{2}-\dfrac{9}{2}\)
\(A=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}\)
\(A=2\left(x+\dfrac{3}{2}\right)^2-\dfrac{9}{2}\)
Vì \(2\left(x+\dfrac{3}{2}\right)^2\ge0\) nên \(A=2\left(x+\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)
Dấu "=" xảy ra khi:
\(x=-\dfrac{3}{2}\)
\(2x^2-6x\)
\(=2.\left(x^2-3x\right)\)
=\(2\left[x^2-2.x.\dfrac{3}{2}+\left(\dfrac{3^{ }}{2}\right)^2-\left(\dfrac{3}{2}\right)^2\right]\)
\(=2\left[\left(x-\dfrac{3}{2}\right)^2-\left(\dfrac{3}{2}\right)^2\right]\)
=\(2\left[\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{4}\right]\ge2\left(0-\dfrac{9}{4}\right)\ge0\)
Vậy GTNN của biểu thức là\(\dfrac{-9}{2}\) xẩy ra khi \(x=\dfrac{3}{2}\)
Nguồn: OLM
Bạn học tốt nhé!
\(2x^2-6x\\ =2x^2-6x+\dfrac{9}{2}-\dfrac{9}{2}\\ =2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}\\ =2\left[x^2-2\cdot x\cdot\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2\right]-\dfrac{9}{2}\\ =2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\)
\(\text{Ta có: }\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\\ \Rightarrow2\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\\ \Rightarrow2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\forall x\)
Dấu \("="\) xảy ra khi:
\(2\left(x-\dfrac{3}{2}\right)^2=0\\ \Leftrightarrow \left(x-\dfrac{3}{2}\right)^2=0\\\Leftrightarrow x-\dfrac{3}{2}=0\\ \Leftrightarrow x=\dfrac{3}{2}\)
Vậy \(GTNN\) của biểu thức là \(-\dfrac{9}{2}\) khi \(x=\dfrac{3}{2}\)
giup minh cau nay nhe
tim gia tri nho nhat cua bieu thuc A= 4x(2x+1) - 2(2x-3)2