Question

Answer 1

Bài 1 :

a, \(A=x^2-20x+101=x^2-2.10x+100+1=\left(x+10\right)^2+1\ge1\)

Dấu ''='' xảy ra khi x = -10 

b, \(B=4a^2+4a+2=4a^2+2.2a+1+1=\left(2a+1\right)^2+1\ge1\)

Dấu ''='' xảy ra khi a = -1/2 

tương tự ... 

Bài 2 : 

a, \(A=-x^2+4x+3=-\left(x^2-4x+4-4\right)+3=-\left(x-2\right)^2+7\le7\)

Dấu ''='' xảy ra khi x = 2

b, \(B=-x^2+x=-\left(x^2-x+\dfrac{1}{4}-\dfrac{1}{4}\right)=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)

Dấu ''='' xảy ra khi x = 1/2 

 

Answer 2

Bài 3.

a)Ta có:A=x²-20x+101=(x-10)²+1≥1 (do (x-10)²≥0)

Dấu "=" xảy ra ⇔ x=10

b)Ta có:B=4a²+4a+2=(2a+1)²+1≥1 (do (2x+1)²≥0)

Dấu "=" xảy ra ⇔ x=-1/2

c)Ta có:C=x²-4xy+5y²-22y+10x+28

              = [x²-2x(2y-5)+(4y²-20y+25)]+(y²-2y+1)+2

              = (x-2y+5)²+(y-1)²+2≥2

Dấu "=" xảy ra ⇔ \(\left\{{}\begin{matrix}x-2y+5=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)

Answer 3

Bài 4: 

a) Ta có: \(A=-x^2+4x+3\)

\(=-\left(x^2-4x-3\right)\)

\(=-\left(x^2-4x+4-7\right)\)

\(=-\left(x-2\right)^2+7\le7\forall x\)

Dấu '=' xảy ra khi x=2

b) Ta có: \(B=-x^2+x\)

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

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

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

Answer 4

Bài 3:

a) Ta có: \(A=x^2-20x+101\)

\(=x^2-20x+100+1\)

\(=\left(x-10\right)^2+1\ge1\forall x\)

Dấu '=' xảy ra khi x=10

b) Ta có: \(B=4a^2+4a+2\)

\(=4a^2+4a+1+1\)

\(=\left(2a+1\right)^2+1\ge1\forall a\)

Dấu '=' xảy ra khi \(a=-\dfrac{1}{2}\)