Question

Answer 1

a,A=x²-4x+15=(x-2)²+11\(\ge11\)

Dấu "=" xảy ra <=> x-2=0 <=> x=2

Vậy Min A=11 <=> x=2

Answer 2

\(A=x^2-4x+15=\left(x-2\right)^2+11\ge11;\forall x\)

\(\Rightarrow A_{min}=11\Leftrightarrow x=2\)

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

\(\Rightarrow B_{min}=-\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{2}\)

\(C=x^2+5x+4=\left(x^2+2.\dfrac{5}{2}x+\dfrac{25}{4}\right)-\dfrac{9}{4}=\left(x+\dfrac{5}{2}\right)^2-\dfrac{9}{4}\ge-\dfrac{9}{4};\forall x\)

\(\Rightarrow C_{min}=-\dfrac{9}{4}\Leftrightarrow x=-\dfrac{5}{2}\)

\(D=4x^2+y^2-8x+6y+9=\left(4x^2-8x+4\right)+\left(y^2+6y+9\right)-4=4\left(x-1\right)^2+\left(y+3\right)^2-4\ge-4;\forall x,y\)

\(\Rightarrow D_{min}=-4\Leftrightarrow x=1;y=-3\)

Answer 3

B=x(x-1)= x²-x = \(\left(x^2-2.\dfrac{1}{2}x+\dfrac{1}{4}\right)-\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

Dấu "=" xảy ra <=> \(x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)

Vậy, Min B=-1/4 <=> x=1/2

Answer 4

C=x²+5x+4=\(\left(x^2+2.\dfrac{5}{2}x+\dfrac{25}{4}\right)-\dfrac{9}{4}=\left(x+\dfrac{5}{2}\right)^2-\dfrac{9}{4}\ge-\dfrac{9}{4}\)

Dấu "=" xảy ra <=> \(x+\dfrac{5}{2}=0\Leftrightarrow x=-\dfrac{5}{2}\)

Vậy Min \(C=-\dfrac{9}{4}\Leftrightarrow x=-\dfrac{5}{2}\)

Answer 5

a) Ta có: \(A=x^2-4x+15\)

\(=x^2-4x+4+11\)

\(=\left(x-2\right)^2+11\ge11\forall x\)

Dấu '=' xảy ra khi x=2

Vậy: \(A_{min}=11\) khi x=2

b) Ta có: \(B=x\left(x-1\right)\)

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

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

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

c) Ta có: \(C=x^2+5x+4\)

\(=x^2+2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{9}{4}\)

\(=\left(x+\dfrac{5}{2}\right)^2-\dfrac{9}{4}\ge-\dfrac{9}{4}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{5}{2}\)

d) Ta có: \(D=4x^2+y^2-8x+6y+9\)

\(=4x^2-8x+4+y^2+6y+9-4\)

\(=\left(2x-2\right)^2+\left(y+3\right)^2-4\ge-4\forall x,y\)

Dấu '=' xảy ra khi x=1 và y=-3

Answer 6

D=4x²+y²-8x+6y+9=4(x²-2x+1)+(y²+6y+9)-1=4(x-1)²+(y+3)²-1\(\ge\) -1

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

Vậy Min D= -1 <=>\(\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)