Question

tìm giá trị nhở nhất của biến

a,\(x^2+10x+27\)

b,\(x^2-12x+37\)

c, \(x^2+x+7\)

d,\(x^2+4xy-5y^2-8y-18\)

e,\(x^2+14x+y^2-2y+7\)

f,\(x^2+4xy+2y^2-22y+173\)

Answer 1

a.A= \(x^2+10x+27\)

\(=x^2+2.x.5+25+2\)

\(\left(x+5\right)^2+2\ge2\forall x\)

Dấu " = " xảy ra <=> x + 5 = 0

=> x = -5

Vậy Min A = 2 <=> x = -5

b.B = \(x^2-12x+37\)

\(=x^2-2.x.6+36+1\)

\(=\left(x-6\right)^2+1\ge1\forall x\)

Dấu " = " xảy ra <=> x - 6 = 0

=> x = 6

Vậy Min B = 1 <=> x = 6

c. \(x^2+x+7\)

\(=x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{27}{4}\)

\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{27}{4}\ge\dfrac{27}{4}\forall x\)

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

\(x=\dfrac{-1}{2}\)

Vậy Min C = \(\dfrac{27}{4}\Leftrightarrow x=\dfrac{-1}{2}\)

Answer 2

\(F=x^2+4xy+2y^2-22y+173\)

\(=\left(x^2+2xy+y^2\right)+\left(y^2-22y+121\right)+52\)

\(=\left(x+y\right)^2+\left(y-11\right)^2+52\)

\(\left(x+y\right)^2\ge0;\left(y-11\right)^2\ge0\forall x,y\)

\(\Rightarrow F\ge52\forall x,y\)

Dấu " =" xảy ra <=>

\(\left\{{}\begin{matrix}\left(x+y\right)^2=0\\\left(y-11\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\y-11=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-11\\y=11\end{matrix}\right.\)

Vậy Min F = 52 <=> x = -11; y = 11

Answer 3

Nguyễn Huy TúToshiro Kiyoshi giúp mk câu c và câu d

Answer 4

e, \(x^2+12x+y^2-2y+7\)

\(=x^2+12x+36+y^2-2y+1-30\)

\(=\left(x+6\right)^2+\left(y-1\right)^2-30\)

Với mọi giá trị của \(x;y\in R\) ta có: \(\left(x+6\right)^2+\left(y-1\right)^2\ge0\) \(\Rightarrow\left(x+6\right)^2+\left(y-1\right)^2-30\ge-30\) Để \(\left(x+6\right)^2+\left(y-1\right)^2-30=-30\) thì

\(\left\{{}\begin{matrix}\left(x+6\right)^2=0\\\left(y-1\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-6\\y=1\end{matrix}\right.\)

Vậy................