Question

tìm gtln

2x-x²-4

Answer 1

\(A=2x-x^2-4\)

\(=-\left(x^2-2x+4\right)\)

\(=-\left[\left(x-1\right)^2+2\right]\)

\(=-\left(x-1\right)^2-2\le-2,\forall x\)

Vậy \(MaxA=-2\) khi \(x=1\)

Nhớ tick

Answer 2

N =4x2+4x+1

=(2x+1)2

do (2x+1)2$\ge$ 0 $\forall$ x

Min N =0 khi 2x+1=0

=>2x=-1

=>x=-$\frac{1}{2}$

Answer 3

Đặt A = \(2x-x^2-4\)

\(-A=-2x+x^2+4\)

\(-A=x^2-2x+4\)

\(-A=x^2-2.x.1+1^2-1^2+4\)

\(-A=(x-1)^2-1+4\)

\(-A=(x-1)^2+3\) => \(A=-(x-1)^2-3 \) \(\le-3\)

=> max A = -3

A = 9 <=> \((x-1)^2=0\) => \(x-1=0\) => x = 1

Vậy max A = (-3) khi x = 1