Chứng minh
\(2x^2+2y^2-2xy-4x-4y+8\ge0\forall x;y\)
CMR
\(4x^2+4y^2-2xy-6x-6y+6\ge0\forall x;y\)
ta có : \(4x^2+4y^2-2xy-6x-6y+6\)
\(=x^2-2xy+y^2+3x^2-6x+3+3y^2-6y+3\)
\(=\left(x-y\right)^2+3\left(x-1\right)^2+3\left(y-1\right)^2\ge0\forall x;y\left(đpcm\right)\)
CHỨNG MINH :
a/ \(3x^2+y^2-2xy+4x+20\forall x,y\)
b/ \(5x^2+10y^2-6xy-4x-2y+3\forall x,y\)
AI GIÚP MK VS Ạ AI NHANH MK SẼ VOTE NHA
a) \(3x^2+y^2-2xy+4x+20=\left(x^2-2xy+y^2\right)+2\left(x^2+2x+1\right)+18=\left(x-y\right)^2+2\left(x+1\right)^2+18\ge18>0\forall x,y\)
\(ĐTXR\Leftrightarrow x=y=-1\)
Chứng minh đẳng thức sau:
a) (a + b + c) . (a - b + c) = a2 - b2 + c2 + 2ac
b) (3x + 2y) . (3x - 2y) - (4x - 2y) . (4x + 2y) = -7x2
c) (x - 1) . (x2 + x + 1) - (x +1) . (x2 - x +1) = -2
d) (2x + 1) . (4x2 - 2x + 1) - (2x - 1) . (4x2 +2x + 1) = 2
e) (x - 2y) . (x2 + 2xy + 4y2) - (2x - 1) . (x2 + 2xy + 4y2) = -16y3
\(a,VT=\left(a+b+c\right)\left(a-b+c\right)\)
\(=\left(a+c+b\right)\left(a+c-b\right)\)
\(=\left(a+c\right)^2-b^2\)
\(=a^2+2ac+c^2-b^2=VP\)
\(b,VT=\left(3x+2y\right)\left(3x-2y\right)-\left(4x-2y\right)\left(4x+2y\right)\)
\(=9x^2-4y^2-16x^2+4y^2=-7x^2=VP\)
\(c,VT=x^3-1-x^3-1=-2=VP\)
\(d,VT=8x^3+1-8x^3+1=2=VP\)
\(e,VT=\left(x^2+2xy+4y^2\right)\left(x-2y-2x+1\right)\)
\(=\left(x^2+2xy+4y^2\right)\left(-x-2y+1\right)\)
\(=-x^3-2x^2y+x^2-2x^2y-4xy^2+2xy-4xy^2-8y^3+4y^2\)
( bn kiểm tra lại đề nhé)
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: \(-x^2+6x-19\)
\(=-\left(x^2-6x+19\right)\)
\(=-\left(x^2-6x+9+10\right)\)
\(=-\left(x-3\right)^2-10< 0\forall x\)
phân tích thành nhân tử
`3x^2 -3xy-5x+5y`
`2x^3 y-2xy^3 -4xy^2 -2xy`
`x^2 -1+2x-y^2`
`x^2 +4x-2xy-4y+4y^2`
`x^3 -2x^2 +x`
`2x^2 +4x+2-2y^2`
a) \(3x^2-3xy-5x+5y\)
\(=\left(3x^2-3xy\right)-\left(5x-5y\right)\)
\(=3x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(3x-5\right)\)
b) \(2x^3y-2xy^3-4xy^2-2xy\)
\(=2xy\left(x^2-y^2-2y-1\right)\)
\(=2xy\left[x^2-\left(y^2+2y+1\right)\right]\)
\(=2xy\left[x^2-\left(y+1\right)^2\right]\)
\(=2xy\left(x-y-1\right)\left(x+y+1\right)\)
c) \(x^2+1+2x-y^2\)
\(=\left(x^2+2x+1\right)-y^2\)
\(=\left(x+1\right)^2-y^2\)
\(=\left(x+1+y\right)\left(x+1-y\right)\)
d) \(x^2+4x-2xy-4y+y^2\)
\(=\left(x^2-2xy+y^2\right)+\left(4x-4y\right)\)
\(=\left(x-y\right)^2+4\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y+4\right)\)
e) \(x^3-2x^2+x\)
\(=x\left(x^2-2x+1\right)\)
\(=x\left(x-1\right)^2\)
f) \(2x^2+4x+2-2y^2\)
\(=2\left(x^2+2x+1-y^2\right)\)
\(=2\left[\left(x^2+2x+1\right)+y^2\right]\)
\(=2\left[\left(x+1\right)^2-y^2\right]\)
\(=2\left(x-y+1\right)\left(x+y+1\right)\)
a: =3x(x-y)-5(x-y)
=(x-y)(3x-5)
b: \(=2xy\left(x^2-y^2-2y-1\right)\)
\(=2xy\left[x^2-\left(y^2+2y+1\right)\right]\)
\(=2xy\left(x-y-1\right)\left(x+y+1\right)\)
d:
Sửa đề: x^2+4x-2xy-4y+y^2
=x^2-2xy+y^2+4x-4y
=(x-y)^2+4(x-y)
=(x-y)(x-y+4)
e: =x(x^2-2x+1)
=x(x-1)^2
f: =2(x^2+2x+1-y^2)
=2[(x+1)^2-y^2]
=2(x+1+y)(x+1-y)
Bài 1: Tính giá trị:
A= x^2+4y^2-2x+10+4xy-4y tại x+2y=5
B= (x^2+4xy+4y^2)-2(x+2y)(y-1)+y^2-2y+1 tại x+y=5
C= x^2-y^2-4x tại x+y=2
D= x^2+y^2+2xy-4x-4y-3 tại x+y=4
E= 2x^6+3x^3y^3+y^6+y^3 tại x^3+y^3=1
Bài 2: Chứng minh rằng
a) -9x^2+12x-5<0
b) 4/9x^2-4x+9/2>0
Bài 3: Tìm giá trị lớn nhất:
A= 4-2x^2
B=(1-x)(2+x)(3+x)(6+x)
C=-2x^2-y^2-2xy+4x+2y+5
D=-9x^2+24x-18
E=-x^4+2x^3-3x^2+4x-1
Tính GTLN A=5-x^2+4x
B=-4x^2+12x-1
C=-x+2xy+4y^2+2x+10y+5
D=-x^2-2y^2-2xy+2x-2y-15
a: \(A=-x^2+4x+5\)
\(=-\left(x^2-4x-5\right)\)
\(=-\left(x^2-4x+4-9\right)\)
\(=-\left(x-2\right)^2+9\le9\)
Dấu '=' xảy ra khi x=2
b: \(B=-4x^2+12x-1\)
\(=-\left(4x^2-12x+1\right)\)
\(=-\left(4x^2-12x+9-8\right)\)
\(=-\left(2x-3\right)^2+8\le8\)
Dấu '=' xảy ra khi x=3/2