K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Dưới đây là một vài câu hỏi có thể liên quan tới câu hỏi mà bạn gửi lên. Có thể trong đó có câu trả lời mà bạn cần!
S
16 tháng 8 2016
\(B=\frac{x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)}{x^2y-x^2z+y^2z-y^3}\)
\(=\frac{x^2y-x^2z+zy^2-xy^2+z^2x-z^2y}{x^2\left(y-z\right)-y^2\left(y-z\right)}\)
\(=\frac{\left(x^2y-z^2y\right)-\left(xy^2-zy^2\right)-\left(x^2z-z^2x\right)}{\left(x^2-y^2\right)\left(y-z\right)}\)
\(=\frac{\left[y\left(x+z\right)-y^2-xz\right]\left(x-z\right)}{\left(x-y\right)\left(x+y\right)\left(y-z\right)}\)
\(=\frac{\left(xy+zy-y^2-xz\right)\left(x-z\right)}{\left(x-y\right)\left(x+y\right)\left(y-z\right)}\)
\(=\frac{\left[\left(xy-y^2\right)-\left(xz-zy\right)\right]\left(x-z\right)}{\left(x-y\right)\left(x+y\right)\left(y-z\right)}\)
\(=\frac{\left[y\left(x-y\right)-z\left(x-y\right)\right]\left(x-z\right)}{\left(x-y\right)\left(x+y\right)\left(y-z\right)}\)
\(=\frac{\left(y-z\right)\left(x-y\right)\left(x-z\right)}{\left(x-y\right)\left(x+y\right)\left(y-z\right)}\)
\(=\frac{x-z}{x+y}\)
S
16 tháng 8 2016
\(A=\frac{\left(x^2-y\right)\left(y+1\right)+x^2y^2-1}{\left(x^2+y\right)\left(y+1\right)+x^2y^2+1}\)
\(=\frac{x^2y-y^2+x^2-y+x^2y^2-1}{x^2y+y^2+x^2+y+x^2y^2+1}\)
\(=\frac{\left(x^2y+x^2\right)+\left(x^2y^2-y^2\right)-\left(y+1\right)}{\left(x^2y+x^2\right)+\left(x^2y^2+y^2\right)+\left(y+1\right)}\)
\(=\frac{x^2\left(y+1\right)+y^2\left(x^2-1\right)-\left(y+1\right)}{x^2\left(y+1\right)+y^2\left(x^2+1\right)+\left(y+1\right)}\)
\(=\frac{\left(x^2-1\right)\left(y+1\right)+y^2\left(x^2-1\right)}{\left(x^2+1\right)\left(y+1\right)+y^2\left(x^2+1\right)}\)
\(=\frac{\left(x^2-1\right)\left(y^2+y+1\right)}{\left(x^2+1\right)\left(y^2+y+1\right)}\)
\(=\frac{x^2-1}{x^2+1}\)
S
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, \(\frac{x^5+x+1}{x^3+x^2+x}=\frac{x^5-x^2+x^2+x+1}{x\left(x^2+x+1\right)}=\frac{x^2\left(x-1\right)\left(x^2+x+1\right)+x^2+x+1}{x\left(x^2+x+1\right)}=\frac{\left(x^2+x+1\right)\left(x^3-x^2+1\right)}{x\left(x^2+x+1\right)}=\frac{x^3-x^2+1}{x}\)
HV
24 tháng 11 2017
\(B=\dfrac{x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)}{x^2y-x^2z+y^2z-y^3}\)
\(\Rightarrow B=\dfrac{x^2\left(y-z\right)-y^2\left[\left(y-z\right)+\left(x-y\right)\right]+z^2\left(x-y\right)}{x^2\left(y-z\right)-y^2\left(y-z\right)}\)
\(\Rightarrow B=\dfrac{\left(y-z\right)\left(x^2-y^2\right)-\left(x-y\right)\left(y^2-z^2\right)}{\left(y-z\right)\left(x^2-y^2\right)}\)
\(\Rightarrow B=\dfrac{\left(y-z\right)\left(x-y\right)\left(x+y\right)-\left(x-y\right)\left(y+z\right)\left(y-z\right)}{\left(y-z\right)\left(x-y\right)\left(x+y\right)}\)
\(\Rightarrow B=\dfrac{\left(y-z\right)\left(x-y\right)\left(x+y-y-z\right)}{\left(y-z\right)\left(x-y\right)\left(x+y\right)}\)
\(\Rightarrow B=\dfrac{\left(y-z\right)\left(x-y\right)\left(x-z\right)}{\left(y-z\right)\left(x-y\right)\left(x+y\right)}\)
\(\Rightarrow B=\dfrac{x-z}{x+y}\)
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: \(\frac{x^5+x+1}{x^3+x^2+1}=\frac{(x^2+x+1)(x^3-x^2+1)}{x(x^2+x+1)}=\frac{x^3-x^2+1}{x}\)
NL
0
CD
1
S
26 tháng 11 2018
Đặt B = \(bc\left(y-z\right)^2+ca\left(z-x\right)^2+ab\left(x-y\right)^2\)
\(=bcy^2+bcz^2+caz^2+cax^2+abx^2+aby^2-2\left(bcyz+acxz+abxy\right)\) (1)
Từ \(ax+by+cz=0\Rightarrow\left(ax+by+cz\right)^2=0\)
=>\(a^2x^2+b^2y^2+c^2z^2+2\left(bcyz+acxz+abxy\right)=0\)
=>\(a^2x^2+b^2y^2+c^2z^2=-2\left(bcyz+acxz+abxy\right)\) (2)
Thay (2) vào (1) ta được:
\(B=ax^2\left(b+c\right)+by^2\left(a+c\right)+cz^2\left(a+b\right)+a^2x^2+b^2y^2+c^2z^2\)
\(=ax^2\left(a+b+c\right)+by^2\left(a+b+c\right)+cz^2\left(a+b+c\right)\)
\(=\left(ax^2+by^2+cz^2\right)\left(a+b+c\right)\)
Vậy \(A=\frac{\left(ax^2+by^2+cz^2\right)\left(a+b+c\right)}{ax^2+by^2+cz^2}=a+b+c\)
CC
6
(x - y + z)^2 + (z - y)^2 + (x - y + z)(2y -2z)
\(<=>(x-y+z)^2+2(x-y+z)(y-z)+(z-y)^2\)
\(<=> (x-y+z+z-y)^2<=> ( x-2y-2z)^2\)
\(\left(x+y+z\right)^2+\left(z-y\right)^2+\left(x-y+z\right)\left(2y-2z\right)\)
\(=\left(x+y+z+y-z\right)^2\)
\(=\left(x+2y\right)^2\)