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24 tháng 11 2017

\(B=\dfrac{x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)}{x^2y-x^2z+y^2z-y^3}\)

\(\Rightarrow B=\dfrac{x^2\left(y-z\right)-y^2\left[\left(y-z\right)+\left(x-y\right)\right]+z^2\left(x-y\right)}{x^2\left(y-z\right)-y^2\left(y-z\right)}\)

\(\Rightarrow B=\dfrac{\left(y-z\right)\left(x^2-y^2\right)-\left(x-y\right)\left(y^2-z^2\right)}{\left(y-z\right)\left(x^2-y^2\right)}\)

\(\Rightarrow B=\dfrac{\left(y-z\right)\left(x-y\right)\left(x+y\right)-\left(x-y\right)\left(y+z\right)\left(y-z\right)}{\left(y-z\right)\left(x-y\right)\left(x+y\right)}\)

\(\Rightarrow B=\dfrac{\left(y-z\right)\left(x-y\right)\left(x+y-y-z\right)}{\left(y-z\right)\left(x-y\right)\left(x+y\right)}\)

\(\Rightarrow B=\dfrac{\left(y-z\right)\left(x-y\right)\left(x-z\right)}{\left(y-z\right)\left(x-y\right)\left(x+y\right)}\)

\(\Rightarrow B=\dfrac{x-z}{x+y}\)

24 tháng 11 2018

a, Xét tử thức \(x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\)

\(=x^2\left(y-z\right)-y^2\left(x-z\right)+z^2\left[\left(x-z\right)-\left(y-z\right)\right]\)

\(=x^2\left(y-z\right)-y^2\left(x-z\right)+z^2\left(x-z\right)-z^2\left(y-z\right)\)

\(=\left(x^2-z^2\right)\left(y-z\right)-\left(y^2-z^2\right)\left(x-z\right)\)

\(=\left(x-z\right)\left(x+z\right)\left(y-z\right)-\left(y-z\right)\left(y+z\right)\left(x-z\right)\)

\(=\left(x-z\right)\left(xy-xz+yz-z^2-y^2-yz+yz+z^2\right)\)

\(=\left(x-z\right)\left(xy-xz+yz-y^2\right)=\left(x-z\right)\left[x\left(y-z\right)-y\left(y-z\right)\right]\)

\(=\left(x-z\right)\left(x-y\right)\left(y-z\right)\)

Mẫu thức \(x^2y-x^2z+y^2z-y^3=x^2\left(y-z\right)-y^2\left(y-z\right)=\left(x-y\right)\left(x+y\right)\left(y-z\right)\)

Vậy \(\frac{x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)}{x^2y-x^2z+y^2z-y^3}=\frac{\left(x-y\right)\left(y-z\right)\left(x-z\right)}{\left(x-y\right)\left(x+y\right)\left(y-z\right)}=\frac{x-z}{x+y}\)

b, \(\frac{x^5+x+1}{x^3+x^2+x}=\frac{x^5-x^2+x^2+x+1}{x\left(x^2+x+1\right)}=\frac{x^2\left(x-1\right)\left(x^2+x+1\right)+x^2+x+1}{x\left(x^2+x+1\right)}=\frac{\left(x^2+x+1\right)\left(x^3-x^2+1\right)}{x\left(x^2+x+1\right)}=\frac{x^3-x^2+1}{x}\)

AH
Akai Haruma
Giáo viên
24 tháng 11 2018

Câu a:

Xét tử số:

\(x^2(y-z)+y^2(z-x)+z^2(x-y)\)

\(=x^2(y-z)-y^2[(y-z)+(x-y)]+z^2(x-y)\)

\(=x^2(y-z)-y^2(y-z)-y^2(x-y)+z^2(x-y)\)

\(=(x^2-y^2)(y-z)-(y^2-z^2)(x-y)\)

\(=(x-y)(y-z)[(x+y)-(y+z)]=(x-y)(y-z)(x-z)\)

Xét mẫu số:

\(x^2y-x^2z+y^2z-y^3=x^2(y-z)-y^2(y-z)=(x^2-y^2)(y-z)\)

\(=(x-y)(x+y)(y-z)\)

Do đó:
\(\frac{x^2(y-z)+y^2(z-x)+z^2(x-y)}{x^2y-x^2z+y^2z-y^3}=\frac{(x-y)(y-z)(x-z)}{(x-y)(x+y)(y-z)}=\frac{x-z}{x+y}\)

AH
Akai Haruma
Giáo viên
24 tháng 11 2018

Câu b:

Xét tử số:

\(x^5+x+1=x^5-x^2+x^2+x+1=x^2(x^3-1)+x^2+x+1\)

\(=x^2(x-1)(x^2+x+1)+(x^2+x+1)\)

\(=(x^2+x+1)(x^3-x^2+1)\)

Xét mẫu số:

\(x^3+x^2+x=x(x^2+x+1)\)

Do đó: \(\frac{x^5+x+1}{x^3+x^2+1}=\frac{(x^2+x+1)(x^3-x^2+1)}{x(x^2+x+1)}=\frac{x^3-x^2+1}{x}\)

21 tháng 11 2017

Hỏi đáp ToánHỏi đáp ToánHỏi đáp ToánHỏi đáp Toán

23 tháng 11 2017

Bn ko hiểu chỗ nào... Để mk giải thik cho...

\(\left(\dfrac{2x+2y-z}{3}\right)^2+\left(\dfrac{2y+2z-x}{3}\right)^2+\left(\dfrac{2z+2x-y}{3}\right)^2\)

\(=\dfrac{4y^2+4x^2+z^2+8xy-4xz-4yz+4y^2+4z^2+x^2+8yz-4xy-4xz}{9}+\dfrac{\left(2z+2x-y\right)^2}{9}\)

\(=\dfrac{8y^2+5x^2+5z^2+4xy-8xz+4yz+4z^2+4x^2+y^2+8xz-4yz-4xy}{9}\)

\(=\dfrac{9y^2+9z^2+9x^2}{9}=x^2+y^2+z^2\)

27 tháng 7 2017

b, \(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3\)

\(=\left(x-y\right)^2\left(x-y\right)-\left(y-z\right)^2\left[\left(x-y\right)+\left(z-x\right)\right]+\left(z-x\right)^2\left(z-x\right)\)

\(=\left(x-y\right)^2\left(x-y\right)-\left(y-z\right)^2\left(x-y\right)-\left(y-z\right)^2\left(z-x\right)+\left(z-x\right)^2\left(z-x\right)\)

\(=\left(x-y\right)\left[\left(x-y\right)^2-\left(y-z\right)^2\right]-\left(z-x\right)\left[\left(y-z\right)^2-\left(z-x\right)^2\right]\)

\(=\left(x-y\right)\left(x-y-y+z\right)\left(x-y+y-z\right)-\left(z-x\right)\left(y-z-z+x\right)\left(y-z+z-x\right)\)

\(=\left(x-y\right)\left(x-2y+z\right)\left(x-z\right)-\left(z-x\right)\left(y-2z+x\right)\left(y-x\right)\)

\(=\left(x-y\right)\left(x-2y+z\right)\left(x-z\right)-\left(x-z\right)\left(y-2z+x\right)\left(x-y\right)\)

\(=\left(x-y\right)\left(x-z\right)\left(x-2y+z-y+2z-x\right)\)

\(=\left(x-y\right)\left(x-z\right)\left(3z-3y\right)\)

\(=3\left(x-y\right)\left(x-z\right)\left(z-y\right)\)

c, \(x^2y^2\left(y-x\right)+y^2z^2\left(z-y\right)-z^2x^2\left(z-x\right)\)

\(=x^2y^2\left(y-x\right)-y^2z^2\left[\left(y-x\right)-\left(z-x\right)\right]-z^2x^2\left(z-x\right)\)

\(=x^2y^2\left(y-x\right)-y^2z^2\left(y-x\right)+y^2z^2\left(z-x\right)-z^2x^2\left(z-x\right)\)

\(=\left(x^2y^2-y^2z^2\right)\left(y-x\right)+\left(y^2z^2-z^2x^2\right)\left(z-x\right)\)

\(=y^2\left(x-z\right)\left(x+z\right)\left(y-x\right)+z^2\left(y-x\right)\left(x+y\right)\left(z-x\right)\)

\(=y^2\left(x-z\right)\left(x+z\right)\left(y-x\right)-z^2\left(y-x\right)\left(x+y\right)\left(x-z\right)\)

\(=\left(x-z\right)\left(y-x\right)\left[y^2\left(x+z\right)-z^2\left(x+y\right)\right]\)

\(=\left(x-z\right)\left(y-x\right)\left(y^2x+y^2z-z^2x-z^2y\right)\)

\(=\left(x-z\right)\left(y-x\right)\left[x\left(y^2-z^2\right)+yz\left(y-z\right)\right]\)

\(=\left(x-z\right)\left(y-x\right)\left[x\left(y-z\right)\left(y+z\right)+yz\left(y-z\right)\right]\)

\(=\left(x-z\right)\left(y-x\right)\left(y-z\right)\left(xy+xz+yz\right)\)

d, \(x^3+y^3+z^3-3xyz\)

\(=\left(x+y\right)^3+z^3-3xyz-3xy\left(x+y\right)\)

\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)

2 tháng 9 2018

\(yz\left(y+z\right)+zx\left(z-x\right)-xy\left(x+y\right)\)

\(=yz\left(y+z\right)+zx\left(z-x\right)-xy\left[\left(y+z\right)-\left(z-x\right)\right]\)

\(=yz\left(y+z\right)+zx\left(z-x\right)-xy\left(y+z\right)+xy\left(z-x\right)\)

\(=y\left(y+z\right)\left(z-x\right)+x\left(z-x\right)\left(z-y\right)\)

\(=\left(z-x\right)\left(yz-xy+xz-xy\right)\)

7 tháng 7 2023

Phân tích vế trái ta được: 2(x2 + y2 + z2 − (xy + yz + zx)

Phân tích vế phải ta được6(x2 + y2 + z2 − (xy + yz + zx)

VT = VP nên VP - VT=0

 4(x2 + y2 + z2 − (xy + yz + zx)) = 0

2(2 (x2 + y2 + z2 − (xy + yz + zx))) = 0

→2((x − y)2 + (y − z)2 + (z − x)2) = 0

→(x − y)2 + (y − z)2 + (z − x)2 = 0

→(x − y)2 = 0; (y − z)2 = 0; (z − x)2 = 0

→x = y = z

13 tháng 12 2018

\(2xyz+x^2y+xy^2+x^2z+xz^2+y^2z+yz^2\)

\(=x^2\left(y+z\right)+yz\left(y+z\right)+x\left(y^2+z^3\right)+2xyz\)

\(=\left(y+z\right)\left(x^2+yz\right)+x\left(y^2+z^2+2yz\right)\)

\(=\left(y+z\right)\left(x^2+yz\right)+x\left(y+z\right)^2\)

\(=\left(y+z\right)\left(x^2+yz\right)+xy+xz\)

\(=\left(y+z\right)\left[x\left(x+2\right)+y\left(x+2\right)\right]\)

\(=\left(y+z\right)\left(x+y\right)\left(x+2\right)\)

13 tháng 12 2018

\(b,x^2\left(y-z\right)+y^2\left(z-y\right)+z^2\left(x-y\right)\)

\(=x^2\left(y-z\right)+y^2z-y^2x+z^2x-z^2y\)

\(=x^2\left(y-z\right)+yz\left(y-z\right)-x\left(y^2-z^2\right)\)

\(=\left(y-z\right)\left[x^2+yz-x\left(y+z\right)\right]\)

\(=\left(y-z\right)\left[x\left(x-y\right)-z\left(x-y\right)\right]\)

\(=\left(y-z\right)\left[\left(x-z\right)\left(x-y\right)\right]\)