Question

Tìm GTNN hoặc GTLN :
a,x^2+y^2-2x+6y-30
b,4-2x^2
c,-x^2+10x-5
d,-3x^2+2x-5
giup mk với mk cần gấp lắm..rất gấp T^T

Answer 1

a, \(x^2+y^2-2x+6y-30\)

\(=x^2-2x+1+y^2+6y+9-40\)

\(=\left(x-1\right)^2+\left(y+3\right)^2-40\ge-40\)

\(min=-40\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)

Answer 2

a)x^2+y^2-2x+6y-30=(x-1)^2+(y+3)^2-40\(\ge\) -40

dấu = xảy ra khi x=1,y=-3

Answer 3

b, \(4-2x^2\le4\)

\(max=4\Leftrightarrow x=0\)

c, \(-x^2+10x-5=-\left(x^2-10x+25\right)+20=-\left(x-5\right)^2+20\le2\text{​​}0\)

\(max=20\Leftrightarrow x=5\)

Answer 4

b,4-2x^2\(\le\)4-2.0=4

dấu = xảy ra khi x=0

 

Answer 5

-x^2+10x-5=-(x^2-10x+25)+20\(\le\)20

dấu = xảy ra khi x=5

Answer 6

d, \(-3x^2+2x-5\)

\(=-3\left(x^2-\dfrac{2}{3}x+\dfrac{1}{9}\right)-\dfrac{14}{3}\)

\(=-3\left(x-\dfrac{1}{3}\right)^2-\dfrac{14}{3}\le-\dfrac{14}{3}\)

\(max=-\dfrac{14}{3}\Leftrightarrow x=\dfrac{1}{3}\)

Answer 7

a) \(x^2+y^2-2x+6y-30=\left(x-1\right)^2+\left(y+3\right)^2-40\ge-40\)

\(ĐTXR\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)

b) \(-2x^2\le0\Rightarrow4-2x^2\le4\)

\(ĐTXR\Leftrightarrow x=0\)

c) \(-x^2+10x-5=-\left(x-5\right)^2+20\le20\)

\(ĐTXR\Leftrightarrow x=5\)

d) \(-3x^2+2x-5=-3\left(x-\dfrac{1}{3}\right)^2-\dfrac{14}{3}\le-\dfrac{14}{3}\)

\(ĐTXR\Leftrightarrow x=\dfrac{1}{3}\)

Answer 8

d,-3x^2+2x-5=-3(x^2-2/3x)-5=-3(x^2-2x.1/3)-5=-3(x^2-2x.1/3+1/9)-5+1/3

                    =-3(x-1/3)^2-14/3\(\le\)-14/3

dấu = xảy ra khi x=1/3

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Answer 9

a: Ta có: \(x^2+y^2-2x+6y-30\)

\(=\left(x^2-2x+1\right)+\left(y^2+6y+9\right)-40\)

\(=\left(x-1\right)^2+\left(y+3\right)^2-40\ge-40\forall x,y\)

Dấu '=' xảy ra khi x=1 và y=-3

c: Ta có: \(-x^2+10x-5\)

\(=-\left(x^2-10x+5\right)\)

\(=-\left(x^2-10x+25-20\right)\)

\(=-\left(x-5\right)^2+20\le20\forall x\)

Dấu '=' xảy ra khi x=5