Question

tìm GTLN

a) A= \(x^2+2x+3\)

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

C=\(5-8x-x^2\)

D=\(-x^2-x+1\)

E=\(5-x^2+2x-4y^2-4y\)

F=\(3-10^2-4xy-4y^2\)

Answer 1

\(A=x^2+2x+3=\left(x+1\right)^2+2>=2\)

Dấu '=' xảy ra khi x=-1

\(B=-\left(x^2+4x-1\right)\)

\(=-\left(x^2+4x+4-5\right)\)

\(=-\left(x+2\right)^2+5< =5\)

Dấu '=' xảy ra khi x=-2

\(C=-x^2-8x+5\)

\(=-\left(x^2+8x-5\right)\)

\(=-\left(x^2+8x+16-21\right)\)

\(=-\left(x+4\right)^2+21< =21\)

Dấu '=' xảy ra khi x=-4

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

\(=-\left(x^2+x+\dfrac{1}{4}-\dfrac{5}{4}\right)\)

\(=-\left(x+\dfrac{1}{2}\right)^2+\dfrac{5}{4}< =\dfrac{5}{4}\)

Dấu '=' xảy ra khi x=-1/2