Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(M=x^2+y^2-xy-x+y+1\)
\(4M=4x^2+4y^2-4xy-4x+4y+4\)
\(=\left(4x^2+y^2+1-4xy-4x+2y\right)+\left(3y^2+2y+3\right)\)
\(=\left(2x-y-1\right)^2+3\left(y^2+\dfrac{2}{3}y+\dfrac{1}{9}\right)+\dfrac{8}{3}\)
\(=\left(2x-y-1\right)^2+3\left(y+\dfrac{1}{3}\right)^2+\dfrac{8}{3}\ge\dfrac{8}{3}\)
\(\Rightarrow M\ge\dfrac{2}{3}\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}2x-y-1=0\\y+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=-\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(MinM=\dfrac{2}{3}\)
\(Q=\dfrac{x^2+xy+y^2+300}{x+y}=\dfrac{\dfrac{1}{2}\left(x+y\right)^2+\dfrac{1}{2}\left(x^2+y^2\right)+300}{x+y}\)
\(Q\ge\dfrac{\dfrac{1}{2}\left(x+y\right)^2+\dfrac{1}{4}\left(x+y\right)^2+300}{x+y}=\dfrac{\dfrac{3}{4}\left(x+y\right)^2+300}{x+y}\)
\(Q\ge\dfrac{2\sqrt{\dfrac{3}{4}\left(x+y\right)^2.300}}{x+y}=30\)
\(Q_{min}=30\) khi \(x=y=10\)
cho em hỏi là
chỗ này \(\dfrac{1}{2}\left(x+y^{ }\right)^{2
}+\dfrac{1}{2}\left(x^2+y^2\right)+300\)
tại sao lại ra như vậy ạ
Bài 1:
a) Ta có: \(A=-x^2-4x-2\)
\(=-\left(x^2+4x+2\right)\)
\(=-\left(x^2+4x+4-2\right)\)
\(=-\left(x+2\right)^2+2\le2\forall x\)
Dấu '=' xảy ra khi x=-2
b) Ta có: \(B=-2x^2-3x+5\)
\(=-2\left(x^2+\dfrac{3}{2}x-\dfrac{5}{2}\right)\)
\(=-2\left(x^2+2\cdot x\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{49}{16}\right)\)
\(=-2\left(x+\dfrac{3}{4}\right)^2+\dfrac{49}{8}\le\dfrac{49}{8}\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{3}{4}\)
c) Ta có: \(C=\left(2-x\right)\left(x+4\right)\)
\(=2x+8-x^2-4x\)
\(=-x^2-2x+8\)
\(=-\left(x^2+2x-8\right)\)
\(=-\left(x^2+2x+1-9\right)\)
\(=-\left(x+1\right)^2+9\le9\forall x\)
Dấu '=' xảy ra khi x=-1
Bài 2:
a) Ta có: \(=25x^2-20x+7\)
\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)
\(=\left(5x-2\right)^2+3>0\forall x\)
b) Ta có: \(B=9x^2-6xy+2y^2+1\)
\(=9x^2-6xy+y^2+y^2+1\)
\(=\left(3x-y\right)^2+y^2+1>0\forall x,y\)
c) Ta có: \(E=x^2-2x+y^2-4y+6\)
\(=x^2-2x+1+y^2-4y+4+1\)
\(=\left(x-1\right)^2+\left(y-2\right)^2+1>0\forall x,y\)
5a^2+2b^2=11ab
<=>5a^2+2b^2-11ab=0
<=>5a^2-10ab-ab+2b^2=0
<=>5a(a-2b)-b(a-2b)=0
<=>(5a-b)(a-2b)=0
<=>5a-b=0 hoặc a-2b=0 <=> 5a=b hoặc a=2b
Nhưng 0 < b/5 < a => b < 5a nên 5a=b là vô lí
Thay a=2b vào ,ta có M = 4.(2b)^2-5b^2/(2b)^2+3.2b.b=11b^2/10b^2=11/10
x^2 -6x +10 = x^2 -2.x.3 +3^2 +1 = (x-3)^2 +1
Ma (x-3)^2 >=0 <=> (x-3)^2 +1 >=1>0 (voi moi x)
b) 4x - x^2 -5 = -(x^2 -4x +5) =-[(x^2 -4x +4)+1] = -[(x-2)^2 +1]
Ma (x+2)^2 >=0 <=> (x-2)^2 +1 >=1 <=> -[(x-2)^2 +1] <=-1 => -[(x-2)^2 +1] <0
2) a) P= x^2 -2x +5 = x^2 -2x +1 +4 = (x-1)^2 +4
Ta co: (x-1)^2 >=0 <=> (x-1)^2 +4 >=4
Vay gia tri nho nhat P=4 khi x=1
b) Q= 2x^2 -6x = 2(x^2 -3x) = 2(x^2 - 2.x.3/2 + 9/4 -9/4)= 2[(x-3/2)^2 -9/4]
Ta co: (x-3/2)^2 >=0 <=>(x-3/2)^2 -9/4 >= -9/4 <=> 2[(x-3/2)^2 -9/4] >= -9/2
Vay gia tri nho nhat Q= -9/2 khi x= 3/2
c) M= x^2 +y^2 -x +6y +10 = (x^2 -2.x.1/2 + 1/4) +(y^2 +2.y.3+9)+3/4
= ( x-1/2)^2 + (y+3)^2 +3/4
M>= 3/4
Vay GTNN cua M = 3/4 khi x=1/2 va y=-3
3)a) A= 4x - x^2 +3 = -(x^2 -4x -3) = -( x^2 -4x+4 -7) =-[(x-2)^2 -7]
Ta co: (x-2)^2>=0 <=> (x-2)^2 -7 >=-7 <=> -[(x-2)^2 -7] <=7
Vay GTLN A=7 khi x=2
b) B= x-x^2 = -(x^2 -2.x.1/2+1/4-1/4) = -[(x-1/2)^2 -1/4]
GTLN B= 1/4 khi x=1/2
c) N= 2x - 2x^2 -5 =-2( x^2 -x+5/2) = -2(x^2 - 2.x.1/2 +1/4 +9/4)
= -2[(x-1/2)^2 +9/4]
GTLN N= -9/2 khi x=1/2
3, A=(x-3)^2+(x-11)^2
\(\Rightarrow\)(X^2-3^2)+(x^2-11^2)
\(\Rightarrow\)(X^2-9)+(X^2-121)
Ta có :X^2 \(\ge\)0 và X^2 \(\ge\)0
\(\Rightarrow\)X^2 - 9 \(\le\)-9 và X^2- 121 \(\le\)-121
\(\Rightarrow\)(X^2-9)+(X^2-121)\(\le\)-130
Dấu = xảy ra khi : X=0
Vậy : Min A = -130 khi x=0
Mình mới lớp 7 sai thì thôi nhé