phan tich da thuc thanh nhan tu :
a^2+b^2-a^2*b^2+ab-a-b
x^2*y+xy^2+x^2*z+xz^2+y^2*z+yz^2+2xyz
Những câu hỏi liên quan
phan tich da thuc sau thanh nhan tu
1)x2-2xy+y2+xz+yz
2)a3x-ab+b-x
phan tich da thuc sau thanh nhan tu
xy(x+y) + yz(y+z) + xz(x+z) + 2xyz
tinh gia tri bieu thuc
3(x-3)(x+7) + (x-4)2 + 48 tai x = 0,5
chung minh rang
x2 - 6x + 10 >0 voi moi x
4x - x2 - 5 <0 voi moi x
1, xy(x+y)+yz(y+z)+xz(x+z)+2xyz
= x2y+xy2+y2z+yz2+x2z+xz2+2xyz
=(x2y+x2z+xz2+xyz) + ( xy2+y2z+yz2+xyz)
=x(xy+xz+z2+yz)+y(xy+yz+z2+xz)
=(xy+xz+yz+z2).(x+y)
=(x(y+z)+z(y+z)).(x+y)
=((y+z).(x+z)).(x+y)= (x+y)(x+z)(y+z)
2. 3(x-3)(x-7)+(x-4)2+48
=3(x2+4x-21)+x2-8x+16+48
=4x2-4x+1 = (2x-1)2
Thay x=0,5 vào bt trên, ta có : (2.0,5 -1)2=0
3, x2-6x+10
= x2-2.3.x+9+1
=(x-3)2+1 \(\ge\)1 >0 ( do (x-3)2 >=0 với mọi x)
=> x26x+10 >0 với mọi x
4x-x2-5
=-(x2-4x+5)
=- (x2-2.2x+4+1)
= - ((x-2)2+1) = -(x-2)2-1\(\le\)-1 < 0 ( do (x-2)2\(\ge\)0 với mọi x => - (x-2)2\(\le\)0 với mọi x)
vậy, 4x-x2-5<0 với mọi x
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Ta có : x2 - 6x + 10
= x2 - 6x + 9 + 1
= (x - 3)2 + 1
Mà (x - 3)2 \(\ge0\forall x\)
Nên : (x - 3)2 + 1 \(\ge1\forall x\)
=> (x - 3)2 + 1 \(>0\)(đpcm)
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Phan tich da thuc thanh nhan tu
xy(x+y)+yz(y+z)+zx(x+z)+2xyz
Ta có : \(xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)+2xyz\)
\(=\left[xy\left(x+y\right)+xyz\right]+\left[yz\left(y+z\right)+xyz\right]+xz\left(x+z\right)\)
\(=xy\left(x+y+z\right)+yz\left(x+y+z\right)+xz\left(x+z\right)\)
\(=y\left(x+y+z\right)\left(x+z\right)+xz\left(x+z\right)\)
\(=\left(x+z\right)\left(xy+y^2+yz+xz\right)\)
\(=\left(x+z\right)\left(x+y\right)\left(y+z\right)\)
phan tich da thuc sau thanh nhan tu ab(x^2+y^2)-xy(a^2+b^2)
\(ab\left(x^2+y^2\right)-xy\left(a^2+b^2\right)\)
\(=abx^2+aby^2-a^2xy-b^2xy\)
\(=ax\left(bx-ay\right)+by\left(ay-bx\right)\)
\(=ax\left(bx-ay\right)-by\left(bx-ay\right)\)
\(\left(bx-ay\right)\left(ax-by\right)\)
hãy k nếu bạn thấy đây là câu tl đúng :)
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ai giup minh vs can gap
pt da thuc thanh nhan tu
a](x^2+y^2+z^2)*(x+y+z)^2+(xy+yz+xz)
b]2x^3-x^2+5x+3
c]x^3-7x^2-3
phan tich da thuc thanh nhan tu :xy(x-y)-xz(x+z)+yz(2x+z-y)
Phan tich da thuc thanh nhan tu :
a) x^2 - x - y^2 - y
b) x^2 - 2 xy + y^2 - z^2
c) 5x - 5y + ax - ay
d) a^3 - a^2 x - ay + xy
e) 4x^2 - y^2 +4x + 1
a) x2-x-y2-y=(x2-y2)-(x+y)=(x+y)(x-y)-(x+y)=(x+y)(x-y-1)
b)x2-2xy+y2-z2=(x-y)2-z2=(x-y-z)(x-y+z)
c)5x-5y+ax-ay=5(x-y)+a(x-y)=(x-y)(5+a)
d)a3-a2x-ay+xy=a2(a-x)-y(a-x)=(a-x)(a2-y)
e)4x2-y2+4x+1=[(2x)2+2.2x.1+12]-y2=(2x+1)2-y2=(2x+1-y)(2x+1+y)
Chúc bạn học tốt!
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phan tich da thuc sau thanh nhan tu:
a) (3x-2)(4x-3)-(2-3x)(x-1)-2(3x-2)(x+1)
b) x^2(y-z)+y^2(z-x)+z^2(x-y)
phan tich da thuc thanh nhan tu: (x+y+z)(xy+yz+xz)-xyz
Xem chi tiết
\(\left(x+y+z\right)\left(xy+yz+xz\right)-xyz=xy\left(x+y+z\right)-xyz+\left(yz+xz\right)\left(x+y+z\right)\)
\(=xy\left(x+y+z-z\right)+z\left(x+y\right)\left(x+y+z\right)\)
\(=xy\left(x+y\right)+z\left(x+y\right)\left(x+y+z\right)\)
\(=\left(x+y\right)\left[xy+z\left(x+y+z\right)\right]\)
\(=\left(x+y\right)\left(xy+xz+yz+z^2\right)\)
\(=\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\)
\(=\left(x+y\right)\left(y+z\right)\left(x+z\right)\)
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phan tich da thuc thanh nhan tu
x^2-x-y^2-y
x^2-2xy+y^2-z^2
bai 32 va 33 sbt
lop 8 bai phan tich da thuc thanh nhan tu bang cach nhom hang tu
Ta có
a, x2-x-y2-y
=x2-y2-(x+y)
=(x-y)(x+y) - (x+y)
=(x+y)(x-y-1)
b, x2-2xy+y2-z2
=(x-y)2-z2
=(x-y-z)(x-y+z)
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con bai 32, 33 neu ban tra loi duoc minh h them
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