LK
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
HK
3
20 tháng 10 2017
\(\text{a) }A=x^2-10x+25\\ A=x^2-2\cdot x\cdot5+5^2\\ A=\left(x-5\right)^2\\ Do\text{ }\left(x-5\right)^2\ge0\forall x\\ \Leftrightarrow A\ge0\forall x\\ \text{Dấu "=" xảy ra khi : }\\ \left(x-5\right)^2=0\\ \Leftrightarrow x-5=0\\ \Leftrightarrow x=5\\ \text{Vậy }A_{\left(Min\right)}=0\text{ }khi\text{ }x=5\)
\(\text{b) }B=x^2+y^2-x+6y+10\\ B=\left(x^2-x+\dfrac{1}{4}\right)+\left(y^2+6y+9\right)+\dfrac{3}{4}\\ B=\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\\ Do\text{ }\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\\ \left(y+3\right)^2\ge0\forall y\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2\ge0\forall x;y\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x;y\\ \text{Dấu "=" xảy ra khi: }\left\{{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2\\\left(y+3\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\y+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-3\end{matrix}\right.\\ \text{ Vậy }B_{\left(Min\right)}=\dfrac{3}{4}\text{ }khi\text{ }x=\dfrac{1}{2};y=-3\)
\(\text{c) }C=2x^2-6x+10\\ C=\left(2x^2-6x+\dfrac{9}{2}\right)+\dfrac{11}{2}\\ C=2\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{11}{2}\\ C=2\left[x^2-2\cdot x\cdot\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2\right]+\dfrac{11}{2}\\ C=2\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{2}\\ Do\text{ }\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\\ \Rightarrow2\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\\ \Rightarrow2\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{2}\ge\dfrac{11}{2}\\ \text{Dấu "=" xảy ra khi: }\\ \left(x-\dfrac{3}{2}\right)^2=0\\ \Leftrightarrow x-\dfrac{3}{2}=0\\ \Leftrightarrow x=\dfrac{3}{2}\\ \text{Vậy }C_{\left(Min\right)}=\dfrac{11}{2}khi\text{ }x=\dfrac{3}{2}\)
\(\)
19 tháng 10 2017
b)
\(B=x^2+y^2-x+6y+10=\left(x^2-x+\dfrac{1}{4}\right)+\left(y^2+6y+9\right)+\left(10-9-\dfrac{1}{4}\right)\)\(B=\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
26 tháng 7 2016
a. \(P=x^2-2x+5=x^2-2x+1+4=\left(x-1\right)^2+4\)
vì \(\left(x-1\right)^2\ge0\) với mọi x
=> (x-1)^2 +4 \(\ge\) vợi mọi x
Pmin=4 <=> x-1=0 <=>x=1
KL
26 tháng 7 2016
1.
b)\(M=\left(x^2-x+\frac{1}{4}\right)+\left(y^2+6y+9\right)+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Dấu = xảy ra \(\Leftrightarrow x-\frac{1}{2}=0\) và \(y+3=0\)
\(\Leftrightarrow x=\frac{1}{2}\) và \(y=-3\)
Vậy GTNN của M là \(\frac{3}{4}\Leftrightarrow x=\frac{1}{2}\)và \(y=-3\)
16 tháng 9 2017
M= x2+y2-x+6y+10=(y2+6y+9)+(x2-x+1/4)+3/4 = (y+3)2+(x-1/2)2+3/4>= 3/4 khi y=-3;x=1/2
16 tháng 9 2017
Ta có\(M=x^2+y^2-x+6y+10\)
\(=\left(x^2-x+\frac{1}{4}\right)+\left(y^2+6y+9\right)+\frac{3}{4}\)
\(=\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+\frac{3}{4}\)
\(\Rightarrow M\ge\frac{3}{4}\)\(\forall x;y\)
Dấu = xảy ra khi\(\hept{\begin{cases}x-\frac{1}{2}=0\\y+3=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=-3\end{cases}}}\)
Vậy MIN \(M=\frac{3}{4}\Leftrightarrow x=\frac{1}{2};y=-3\)
SG
1
TN
28 tháng 7 2017
Câu 1:
\(a,P=x^2-2x+5=\left(x^2-2x+1\right)+4\)
\(=\left(x-1\right)^2+4\ge4\forall x\)
Vậy Min \(P=4\) khi \(x-1=0\Rightarrow x=1\)
\(b,Q=2x^2-6x=2\left(x^2-\dfrac{3}{2}x+\dfrac{9}{4}\right)-\dfrac{9}{2}\)
\(=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\forall x\)
Vậy \(MinQ=-\dfrac{9}{2}\) khi \(x-\dfrac{3}{2}=0\Rightarrow x=\dfrac{3}{2}\)
\(c,M=x^2+y^2-x+6y+10\)
\(=\left(x^2-x+\dfrac{1}{4}\right)+\left(y^2+9y+9\right)+\dfrac{3}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)
Vậy Min \(M=\dfrac{3}{4}\) khi \(\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\y+3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\x=-3\end{matrix}\right.\)
11 tháng 7 2020
Bài làm:
+ \(C=10\left(x^2-2\right)+5=10x^2-20+5=10x^2-15\ge-15\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(10x^2=0\Rightarrow x=0\)
Vậy \(Min\left(C\right)=-15\Leftrightarrow x=0\)
+ \(D=\left(7-x\right)\left(2x+1\right)=-2x^2+13x+7=-2\left(x^2-\frac{13}{2}x+\frac{169}{16}\right)-\frac{225}{8}\)
\(=-2\left(x-\frac{13}{4}\right)^2-\frac{225}{8}\le-\frac{225}{8}\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(-2\left(x-\frac{13}{4}\right)^2=0\Rightarrow x=\frac{13}{4}\)
Vậy \(Max\left(D\right)=-\frac{225}{8}\Leftrightarrow x=\frac{13}{4}\)
+ \(H=x^2+y^2+2x-4y+10=\left(x^2+2x+1\right)+\left(y^2-4y+4\right)+5\)
\(=\left(x+1\right)^2+\left(y-2\right)^2+5\ge5\left(\forall x,y\right)\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x+1\right)^2=0\\\left(y-2\right)^2=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-1\\y=2\end{cases}}\)
Vậy \(Min\left(H\right)=5\Leftrightarrow\hept{\begin{cases}x=-1\\y=2\end{cases}}\)
+ \(E=-x^2-4x+6y-y^2-2021=-\left(x^2+4x+4\right)-\left(y^2-6y+9\right)-2008\)
\(=-\left(x+2\right)^2-\left(y-3\right)^2-2008\le-2008\left(\forall x,y\right)\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}-\left(x+2\right)^2=0\\-\left(y-3\right)^2=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-2\\y=3\end{cases}}\)
Vậy \(Max\left(E\right)=-2008\Leftrightarrow\hept{\begin{cases}x=-2\\y=3\end{cases}}\)
Học tốt!!!!
CP
3
15 tháng 6 2016
a) \(A=x^2+y^2-x+6y+10=x^2-x+\frac{1}{4}+y^2+6y+9+\frac{3}{4}\)
\(A=\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x;y\)
Vậy GTNN của A = 3/4 khi x=1/2 và y=-3.
b) \(B=2x-2x^2-5=-2\left(x^2-x+\frac{1}{4}\right)-\frac{9}{2}=-2\left(x-\frac{1}{2}\right)^2-\frac{9}{2}\ge-\frac{9}{2}\forall x\)
Vậy GTLN của B = -9/2 khi x=1/2.
KN
2
T
9 tháng 8 2017
\(P=x^2-2x+5=\left(x^2-2x+1\right)+4=\left(x-1\right)^2+4\ge4\)
Đẳng thức xảy ra \(\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)
\(Q=2x^2-6x=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)
Đẳng thức xảy ra \(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=0\Leftrightarrow x=\dfrac{3}{2}\)
\(M=x^2+y^2-x+6y+10\)
\(=\left(x^2-x+\dfrac{1}{4}\right)+\left(y^2+6y+9\right)+\dfrac{3}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Đẳng thức xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2=0\\\left(y+3\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-3\end{matrix}\right.\)
24 tháng 6 2018
P=\(\left(X-1\right)^2+4\) \(\ge4\)=> giá trị nhỏ nhất là 4
Dấu = xảy ra khi x=1
M=\(\left(X^2-X\right)+\left(y^2+6y+9\right)+1=X\left(X-1\right)+\left(Y+3\right)^2+1\ge1\)
Dấu = xảy ra khi X=1 và Y=-3
NA
1
6 tháng 12 2019
Sửa đề
\(P=\frac{x^2-6xy+6y^2}{x^2-2xy+y^2}\)
\(\Leftrightarrow P+3=\frac{x^2-6xy+6y^2}{x^2-2xy+y^2}+3=\frac{\left(3y-2x\right)^2}{\left(x-y\right)^2}\ge0\)
\(\Leftrightarrow P\ge-3\)
Ta có:
\(M=x^2+y^2-x+6y+10\)
\(M=\left(x^2-x+\frac{1}{4}\right)+\left(y^2+6y+9\right)+\frac{3}{4}\)
\(M=\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+\frac{3}{4}\ge\frac{3}{4}\left(\forall x,y\right)\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x-\frac{1}{2}\right)^2=0\\\left(y+3\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=-3\end{cases}}\)
M = x2 + y2 - x + 6y + 10
= ( x2 - x + 1/4 ) + ( y2 + 6y + 9 ) + 3/4
= ( x - 1/2 )2 + ( y + 3 )2 + 3/4 ≥ 3/4 ∀ x
Dấu "=" xảy ra <=> x = 1/2 ; y = -3
=> MinM = 3/4 <=> x = 1/2 ; y = -3