Tim gtnn cua bieu thuc
C=5x^2-7x+4
D=x^2+y^2-2x-4y-6
Những câu hỏi liên quan
tim gia tri lon nhat cua bieu thuc :
b) D= 2 I 7x+5I +11/ I 7x+5I +4
tim GTNN cua bieu thuc :
c) C= 5+ -8/ 4x I5x+7I 24
tim gia tri lon nhat cua bieu thuc :
b) D= 2 I 7x+5I +11/ I 7x+5I +4
tim GTNN cua bieu thuc :
c) C= 5+ -8/ 4x I5x+7I 24
a,Tim GTNN cua bieu thuc \(C=\left(x+2\right)^2+\left(y-\frac{1}{5}\right)^2-10\)
b,Tim GTLN cua bieu thuc \(D=\frac{4}{\left(2x-3\right)^2+5}\)
\(\text{a)Để C đạt GTNN}\)
\(\Rightarrow\hept{\begin{cases}\left(x+2\right)^2\\\left(y-\frac{1}{5}\right)^2\end{cases}\ge0}\)
\(\Rightarrow\left(x+2\right)^2+\left(y-\frac{1}{5}\right)^2\ge0\)
\(\Rightarrow\left(x+2\right)^2+\left(y-\frac{1}{5}\right)^2-10\ge0-10\)
\(\Rightarrow C\ge-10\)
\(\text{Vậy minC=-10 khi x=-2;y= }\frac{1}{5}\)
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b)\(\text{Để D đạt GTLN}\)
=>(2x-3)2+5 đạt GTNN
Mà (2x-3)2\(\ge\)5
\(\Rightarrow GTLN\)của \(A=\frac{4}{5}\)khi \(x=\frac{3}{2}\)
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bai 1:tim GTNN cua bieu thucAx2+3x+7B(x-2)(x-5)(x2-7x-10)bai 2:tim GTLN cua bieu thucA11-10x-x2B[x-4](2-[x-4])bai 3:tim x,y sao choA2x2+9y2-6xy-6x-12y+2016 co GTNNB-x2+2xy-4y2+2x+10y-8 co GTLNbai 4 :a)cho x+y3;x2+y25.tinh x3+y3b)cho x-y5;x2+y215.tinh x3-y3
Đọc tiếp
bai 1:tim GTNN cua bieu thuc
A=x2+3x+7
B=(x-2)(x-5)(x2-7x-10)
bai 2:tim GTLN cua bieu thuc
A=11-10x-x2
B=[x-4](2-[x-4])
bai 3:tim x,y sao cho
A=2x2+9y2-6xy-6x-12y+2016 co GTNN
B=-x2+2xy-4y2+2x+10y-8 co GTLN
bai 4 :
a)cho x+y=3;x2+y2=5.tinh x3+y3
b)cho x-y=5;x2+y2=15.tinh x3-y3
tim gia tri lon nhat cua bieu thuc :
a) C= 5+ 15/ 4 I 3x+7 I +3
b) D= 2 I 7x+5I +11/ I 7x+5I +4
tim GTNN cua bieu thuc :
a) A= I x+1I + 1,7
b) B= I x-2/3I +3/7
c) C= 5+ -8/ 4x I5x+7I 24
tim gtnn cua bieu thuc
C=5x2+y2+10+4xy-14x-6y
tim gtnn cua bieu thuc
A= 5x^2+y^2 -4xy-2y+2023
\(=5\left(x^2-\dfrac{4}{5}xy+\dfrac{4}{25}y^2\right)+\dfrac{1}{5}y^2-2y+2023\)
\(=5\left(x-\dfrac{2}{5}y\right)^2+\dfrac{1}{5}\left(y^2-10y+25\right)+2018\)
\(=5\left(x-\dfrac{2}{5}y\right)^2+\dfrac{1}{5}\left(y-5\right)^2+2018>=2018\)
Dấu = xảy ra khi y=5 và x=2/5y=2
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1) Tim GTNN cua bieu thuc sau
a) M = x^2 + 4x + 9
b) N = x^2 - 20x +101
5) Tim GTLN cua bieu thuc sau
a) C = -y^2 + 6y -15
b) B = -x^2 + 9x - 12
c) D = 3x - x^2
Bài 1:
a: \(M=x^2+4x+4+5=\left(x+2\right)^2+5>=5\)
Dấu '=' xảy ra khi x=-2
b: \(N=x^2-20x+101=x^2-20x+100+1=\left(x-10\right)^2+1>=1\)
Dấu '=' xảy ra khi x=10
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1.cho phan thuc P=\(\frac{5x-4y}{5x+4y}\)voi 25x2+16y2=50xy.khi do gia tri cua bieu thuc P=\(\frac{1+P^2}{1-P^2}\)la
2.gia tri lon nhat cua bieu thuc B=\(\frac{14}{\frac{x^2}{4}-3x+16}la\)
3.gia tri cua a de da thuc A=2x3+7x2+ax+3 chia het cho B=(x+1)2la a=?
\(25x^2+16y^2=50xy\)
\(\Leftrightarrow\) \(\left(5x+4y\right)^2-40xy=50xy\)
\(\Leftrightarrow\) \(\left(5x+4y\right)^2=90xy\)
Mặt khác, ta cũng có: \(25x^2+16y^2=50xy\)
\(\Leftrightarrow\) \(\left(5x-4y\right)^2=10xy\)
Do đó:
\(P^2=\frac{\left(5x-4y\right)^2}{\left(5x+4y\right)^2}=\frac{10xy}{90xy}=\frac{1}{9}\)
Vậy, \(P'=\frac{1+\frac{1}{9}}{1-\frac{1}{9}}=1\frac{1}{4}\)
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1)
\(25x^2-40xy+16y^2=10xy\Leftrightarrow\left(5x-4y\right)^2=10xy\)
\(25x^2+40xy+16y^2=10xy\Leftrightarrow\left(5x+4y\right)^2=90xy\)
\(P^2=\frac{1}{9}\Leftrightarrow Q=\frac{1+P^2}{1-P^2}=\frac{1+\frac{1}{81}}{1-\frac{1}{81}}=\frac{82}{80}=\frac{41}{40}\)
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