CMR
\(4x^2+4y^2-2xy-6x-6y+6\ge0\forall x;y\)
Những câu hỏi liên quan
Chứng minh
\(2x^2+2y^2-2xy-4x-4y+8\ge0\forall x;y\)
\(2x^2+2y^2-2xy-4x-4y+8\)
\(=x^2-2xy+y^2+x^2-4x+y^2-4y+8\)
\(=\left(x-y\right)^2+x^2-4x+4+y^2-4x+4\)
\(=\left(x-y\right)^2+\left(x-2\right)^2+\left(y-2\right)^2\ge0\)
\(\RightarrowĐPCM\)
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CMR với mọi x,y thì
a)\(x^2+4y^2-2x+4y+2\ge0\)
b)\(3y^2+y^2+2xy+2x+6y+3\ge0\)
a)\(x^2+4y^2-2x+4y+2\)
\(=\left(x^2-2x+1\right)+\left(4y^2+4y+1\right)\)
\(=\left(x-1\right)^2+\left(2y+1\right)^2\ge0\)(đúng)
b) Sửa đề
\(3y^2+x^2+2xy+2x+6y+3\)
\(=\left(x^2+y^2+2xy\right)+2y^2+2x+6y+3\)
\(=\left(x+y\right)^2+2\left(x+y\right)+1+2y^2+4y+2\)
\(=\left(x+y+1\right)^2+2\left(y+1\right)^2\ge0\) (đúng)
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a, x2-2xy+y2-4x+6y+15
b, 2x2-4xy+4y2+6x-6y+40
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answer= foethe www
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Bộ bạn ko biết ghi đề bài hả xuống tiểu học đi nha
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Chứng minh rằng \(x^4+4x^3+6x^2+4x+1\ge0,\forall x\in R\)
\(x^4+4x^3+6x^2+4x+1\)
\(=\left(x^4+2x^3+x^2\right)+\left(2x^3+4x^2+2x\right)+\left(x^2+2x+1\right)\)
\(=x^2\left(x^2+2x+1\right)+2x\left(x^2+2x+1\right)+\left(x^2+2x+1\right)\)
\(=\left(x^2+2x+1\right)\left(x^2+2x+1\right)=\left(x+1\right)^4\ge0;\forall x\in R\)
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1/ PTDTTNT:
a) x² +6x–y² +9
b) 2x² –4x +2
c)2xy +3z +6y +xz
d)x² +4x –2xy –4y +y²
e)x² +2x +1 –16y²
d/ \(x^2+4x-2xy-4y+y^2=\left(x-y\right)^2+4\left(x-y\right)=\left(x-y\right)\left(x-y+4\right)\)
e/ \(x^2+2x+1-16y^2=\left(x+1\right)^2-\left(4y\right)^2=\left(x+1-4y\right)\left(x+1+4y\right)\)
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CHỨNG MINH :
a/ \(x^2-8x+20>0\forall x\)
b/ \(6x-x^2-19< 0\forall x\)
c/ \(3x^2+y^2-2xy+4x+20>0\forall x,y\)
d/ \(5x^2+10y^2-6xy-4x-2y+3>0\forall x,y\)
AI GIÚP MK VS Ạ AI NHANH MK SẼ VOTE NHA
a: Ta có: \(x^2-8x+20\)
\(=x^2-8x+16+4\)
\(=\left(x-4\right)^2+4>0\forall x\)
b: Ta có: \(-x^2+6x-19\)
\(=-\left(x^2-6x+19\right)\)
\(=-\left(x^2-6x+9+10\right)\)
\(=-\left(x-3\right)^2-10< 0\forall x\)
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Tim x,y biet:
1)x^2-2x+5+y^2-4y0
2)4x^2+y^2-20x+26-2y0
3)x^2+4y^2+13-6x-8y0
4)4x^2+4x-6y+9x^2+20
5)x^2+y^2+6x-10y+340
6)25x^2-10x+9y^2-12y+50
7)x^2+9y^2-10x-12y+290
89x^2+12x+4y62+8y+80
9)4x^2+9y^2+20x-6y+260
10)3x^2+3y^2+6x-12y+150
11)x^2+4y^2+4x-4y+50
12)4x^2-12x+y^2-4y+130
13)x^2+y^2+2x-6y+100
14)4x^2+9y^2-4x+6y+20
15)y^2+2y+5-12x+9x^20
16)x^2+26+6y+9y^2-10x0
17)10-6x+12y+9x^2+4y^20
18)16x^2+5+8x-4y+y^20
19)x^2+9y^2+4x+6y+50
20)5+9x^2+9y^2+6y-12x0
21)x^2+20+9y62+8x-12y0...
Đọc tiếp
Tim x,y biet:
1)x^2-2x+5+y^2-4y=0
2)4x^2+y^2-20x+26-2y=0
3)x^2+4y^2+13-6x-8y=0
4)4x^2+4x-6y+9x^2+2=0
5)x^2+y^2+6x-10y+34=0
6)25x^2-10x+9y^2-12y+5=0
7)x^2+9y^2-10x-12y+29=0
89x^2+12x+4y62+8y+8=0
9)4x^2+9y^2+20x-6y+26=0
10)3x^2+3y^2+6x-12y+15=0
11)x^2+4y^2+4x-4y+5=0
12)4x^2-12x+y^2-4y+13=0
13)x^2+y^2+2x-6y+10=0
14)4x^2+9y^2-4x+6y+2=0
15)y^2+2y+5-12x+9x^2=0
16)x^2+26+6y+9y^2-10x=0
17)10-6x+12y+9x^2+4y^2=0
18)16x^2+5+8x-4y+y^2=0
19)x^2+9y^2+4x+6y+5=0
20)5+9x^2+9y^2+6y-12x=0
21)x^2+20+9y62+8x-12y=0
22)x^2=4y+4y^2+26-10x=0
23)4y^2+34-10x+12y+x^2=0
24)-10x+y^2-8y+x^2+41=0
25)x^2+9y^2-12y+29-10x=0
26)9x^2+4y^2+4y+5-12x=0
27)4y^2-12x+12y+9x^2=13=0
28)4x^2+25-12x-8y+y^2=0
29)x62+17+4y^2+8x+4y=0
30)4y^2+12y+25+8x+x^2=0
31)x^2+20+9y^2+8x-12y=0
giup mk voi minh can gap ak, cam on cac ban
Tìm GTLN hoặc GTNN:
a) x2 + x + 1
b) 2 + x - x2
c) x2 - 4x + 1
d) 4x2 + 4x + 11
e) 3x2 - 6x + 1
f) x2 - 2x + y2 - 4y +6
g) -x4 + 4x - 5
h) A = x2 - 2xy + 6y2 - 12x + 2y + 45 (GTNN)
a) đặt \(A=x^2+x+1\)
\(=x^2+2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{1}{4}+1\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Dấu "=' xảy ra khi \(x=-\dfrac{1}{2}\)
Vậy \(MIN_A=\dfrac{3}{4}\) khi \(x=-\dfrac{1}{2}\)
b) đặt \(B=2+x-x^2\)
\(=-x^2+x+2\)
\(=-\left(x^2-x-2\right)\)
\(=-\left[x^2-2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{1}{4}-2\right]\)
\(=-\left[\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{4}\right]\)
\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}\le\dfrac{9}{4}\)
Dấu "=" xảy ra khi \(x=\dfrac{1}{2}\)
Vậy \(MAX_B=\dfrac{9}{4}\) khi \(x=\dfrac{1}{2}\)
c) đặt \(C=x^2-4x+1\)
\(=x^2-2\cdot x\cdot2+2^2-4+1\)
\(=\left(x-2\right)^2-3\ge-3\)
Dấu "=" xảy ra khi \(x=2\)
Vậy \(MIN_c=-3\) khi \(x=2\)
d) đặt \(D=4x^2+4x+11\)
\(=\left(2x\right)^2+2\cdot2x\cdot1+1^2-1+11\)
\(=\left(2x+1\right)^2+10\ge10\)
Dấu "=" xảy ra khi \(x=-\dfrac{1}{2}\)
Vậy \(MIN_D=10\) khi \(x=-\dfrac{1}{2}\)
mấy câu còn lại tương tự
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2) tính a)3x(3x+6). b)-1/2xy(4x^2+6x). c)-2x^2y^3(1/2xy+4y^2). d)-6x^2(1/3xy^2-1/2y)
\(a,3x\left(3x+6\right)=9x^2+18x\)
\(b,-\dfrac{1}{2}xy\left(4x^2+6x\right)\)
\(=-2x^3y-3x^2y\)
\(c,-2x^2y^3\left(\dfrac{1}{2}xy+4y^2\right)\)
\(=-x^3y^4-8x^2y^5\)
\(d,-6x^2\left(\dfrac{1}{3}xy^2-\dfrac{1}{2}y\right)\)
\(=-2x^3y^2+3x^2y\)
#\(Urushi\)
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a) 3x(3x+6) = 9x^2 + 18x b) -1/2xy(4x^2+6x) = -2x^3y - 3xy c) -2x^2y^3(1/2xy+4y^2 ) = -x^2y^2 - 8x^2y^5 d) -6x^2(1/3xy^2-1/2y) = -2xy + 3x^2y
Vậy kết quả của các biểu thức là: a) 9x^2 + 18x b) -2x^3y - 3xy c) -x^2y^2 - 8x^2y^5 d) -2xy + 3x^2y
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