Question

[Nguyễn Văn Sang]

1, Tìm giá trị nhỏ nhất

A= x^2 - 10x +26

B= x^2 + 7x + 10

C= 4x^2 + 8x +15

D= 3x^2 - 7x + 20

E= x^2 - 4xy +5y^2 -22y+8

Answer 1

[T.Thùy Ninh]

\(A=x^2-10x+26\)

\(=\left(x^2-10x+25\right)+1\)

\(=\left(x-5\right)^2+1\ge1\)

Vậy \(Min_A=1\) khi \(x-5=0\Rightarrow x=5\)

\(B=x^2+7x+10=\left(x^2+7x+\dfrac{49}{4}\right)-\dfrac{9}{4}=\left(x+\dfrac{7}{2}\right)^2-\dfrac{9}{4}\ge\dfrac{-9}{4}\)Vậy \(Min_B=\dfrac{-9}{4}\) khi \(x+\dfrac{7}{2}=0\Rightarrow x=\dfrac{-7}{2}\)

\(C=4x^2+8x+15=4\left(x^2+2x+1\right)+11=4\left(x+1\right)^2+11\ge11\)Vậy \(Min_C=11\) khi \(x+1=0\Rightarrow x=-1\)

\(D=3x^2-7x+20=3\left(x^2-\dfrac{7}{3}x+\dfrac{49}{36}\right)+\dfrac{191}{12}=3\left(x-\dfrac{7}{6}\right)^2+\dfrac{191}{12}\ge\dfrac{191}{12}\)Vậy \(Min_D=\dfrac{191}{12}\) khi \(x-\dfrac{7}{6}=0\Rightarrow x=\dfrac{7}{6}\)

\(E=x^2-4xy+5y^2-22y+8\)

\(=\left(x^2-4xy+4y^2\right)+\left(y^2-22y+121\right)-113\)\(=\left(x-2y\right)^2+\left(y-11\right)^2-113\ge-113\)

Vậy \(Min_E=-113\) khi \(\left[{}\begin{matrix}x-2y=0\\x-11=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}11-2y=0\\x=11\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2y=11\\x=11\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=11\end{matrix}\right.\)