Question

1,Tìm giá trị nhỏ nhất:

a, \(2x^2+4y^2-4xy-4x-4y+2017\)

b, \(x^4+2x^3+3x^2+2x+1\)

2, Tìm giá trị lớn nhất

a, \(-x^2-y^2+xy+2x+2y\)

b,\(-x^2+2xy-4y^2+2x+10y-8\)

3, Tìm giá trị lớn nhất hoặc nhỏ nhất của biểu thức:

a,\(\dfrac{1}{x^2+x+1}\)

b, \(\dfrac{5}{x^2-6x+10}\)

c,\(\dfrac{-8}{x^2-2x+5}\)

Answer 1

1) a) Đặt biểu thức là A

\(A=2x^2+4y^2-4xy-4x-4y+2017\)

\(A=\left(x-2y\right)^2+x^2-4x-4y+2017\)

\(A=\left(x-2y\right)^2+2\left(x-2y\right)+x^2-6x+2017\)

\(A=\left(x-2y-1\right)^2+\left(x+3\right)^2+2008\)

Vậy: Min_A=2008 khi x=-3; y=-2

Answer 2

3) a) \(A=\dfrac{1}{x^2+x+1}\)

\(B=x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

\(\Rightarrow B\ge\dfrac{3}{4}\Rightarrow A\ge\dfrac{4}{3}\)

Vậy Min_A là \(\dfrac{4}{3}\) khi x=-0,5

Answer 3

b) \(A=\dfrac{5}{x^2-6x+10}\)

\(B=x^2-6x+10=\left(x-3\right)^2+1\)

\(B\ge1\Rightarrow A\ge5\)

Vậy Min_A = 5 khi x=3

Answer 4

3) c) \(A=\dfrac{-8}{x^2-2x+5}\)

\(B=x^2-2x+5=\left(x-1\right)^2+4\)

\(B\ge4\Rightarrow A\ge-2\)

Vậy: Min_A =-2 khi x=1

Answer 5

1) b) \(A=x^4+2x^3+3x^2+2x+1\)

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

Min_A=0 khi x=-1

Answer 6

3)

a\(\dfrac{1}{x^2+x+1}=\dfrac{1}{x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+1}=\dfrac{1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\)

=\(\dfrac{1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\ge\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}\)

Vậy \(Min=\dfrac{4}{3}=>x=-\dfrac{1}{2}\)

b)\(\dfrac{5}{x^2-6x+10}=\dfrac{5}{x^2-2.x.3+9+1}=\dfrac{5}{\left(x-3\right)^2+1}\ge5\)

Vậy Min=5 khi x=3

c)\(\dfrac{-8}{x^2-2x+5}=\dfrac{-8}{x^2-2x+1+4}=\dfrac{-8}{\left(x-1\right)^2+4}\ge-2\)

Vậy Min=-2 khi x=1

Answer 7

2) b) Đặt biểu thức là A

\(A=-x^2+2xy-4y^2+2x+10y-8\)

\(A=-\left(x^2-2xy+4y^2-2x-10y+8\right)\)

\(A=-\left[\left(x-y\right)^2+3y^2-2\left(x-y\right)-12y+8\right]\)

\(A=-\left[\left(x-y-1\right)^2+\left(\sqrt{3y}-2\sqrt{3}\right)^2\right]-12+8\)

Vậy Min_A là 4