Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Câu a:
Xét tử số:
\(x^3-y^3+z^3+3xyz=(x-y)^3+3xy(x-y)+z^3+3xyz\)
\(=(x-y)^3+z^3+3xy(x-y+z)\)
\(=(x-y+z)[(x-y)^2-z(x-y)+z^2]+3xy(x-y+z)\)
\(=(x-y+z)(x^2+y^2+z^2-2xy-xz+yz)+3xy(x-y+z)\)
\(=(x-y+z)(x^2+y^2+z^2+xy+yz-xz)\)
Xét mẫu số:
\((x+y)^2+(y+z)^2+(z-x)^2\)
\(x^2+2xy+y^2+y^2+2yz+z^2+z^2-2zx+x^2\)
\(2(x^2+y^2+z^2+xy+yz-xz)\)
Do đó: \(\frac{x^3-y^3+z^3+3xyz}{(x+y)^2+(y+z)^2+(z-x)^2}=\frac{x-y+z}{2}\)
Câu b:
Xét tử số:
\((x^2-y)(y+1)+x^2y^2-1\)
\(=x^2y+x^2-y^2-y+x^2y^2-1\)
\(=(x^2y-y)+(x^2-1)+(x^2y^2-y^2)\)
\(=y(x^2-1)+(x^2-1)+y^2(x^2-1)=(x^2-1)(y^2+y+1)\)
Xét mẫu số:
\((x^2+y)(y+1)+x^2y^2+1\)
\(=x^2y+x^2+y^2+y+x^2y^2+1\)
\(=(x^2y+y)+(x^2+1)+(x^2y^2+y^2)\)
\(=y(x^2+1)+(x^2+1)+y^2(x^2+1)\)
\(=(x^2+1)(y+1+y^2)\)
Do đó:
\(\frac{(x^2-y)(y+1)+x^2y^2-1}{(x^2+y)(y+1)+x^2y^2+1}=\frac{(x^2-1)(y^2+y+1)}{(x^2+1)(y^2+y+1)}=\frac{x^2-1}{x^2+1}\)
\(M=4\left(x-1\right)\left(x+1\right)-5x\left(x-2\right)+x^2\)
\(=4x^2-4-5x^2+10x+x^2\)
\(=10x-4\)
\(M=\left(y^2+2\right)\left(y-4\right)-\left(2y^2+1\right)\left(\dfrac{1}{2}y-2\right)\)
\(=\left(y^2+2\right)\left(y-4\right)-\dfrac{1}{2}\left(2y^2+1\right)\left(y-4\right)\)
\(=\left(y-4\right)\left(y^2+2-y^2-\dfrac{1}{2}\right)\)
\(=\dfrac{3}{2}y-6\)
c)
\(C=\left(3-2x\right)\left(x-2\right)-4\left(x-1\right)\left(x-3\right)-\left(x-2\right)\left(x+2\right)\)
= 3x - 6 - 2x2 + 4x - 4x2 + 12x + 4x - 12 - x2 + 4
= - 7x2 + 23x - 14
a)\(x\left(2x^2-3\right)-x^2\left(5x+1\right)+x^2\)
=\(2x^3-3x-5x^3-x^2+x^2=-3x^3-3x\)
b) \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)
\(=3x^2-6x-5x+5x^2-8x^2+24=-11x+24\)
c) \(\dfrac{1}{2}x^2\left(6x-3\right)-x\left(x^2+\dfrac{1}{2}\right)+\dfrac{1}{2}\left(x+4\right)\)
\(=3x^3-\dfrac{3}{2}x^2-x^3-\dfrac{1}{2}x+\dfrac{1}{2}x+2=2x^3-\dfrac{3}{2}x^2+2\)
\(a,\frac{x}{xy-y^2}+\frac{2x-y}{xy-x^2}:\left(\frac{1}{x}+\frac{1}{y}\right)\)
\(=\left(\frac{x}{y\left(x-y\right)}+\frac{y-2x}{x\left(x-y\right)}\right):\left(\frac{y}{xy}+\frac{x}{xy}\right)\)
\(=\left(\frac{x-y}{x\left(x-y\right)}\right):\left(\frac{x+y}{xy}\right)\)
\(=\frac{1}{x}.\frac{xy}{x+y}=\frac{y}{x+y}\)
\(A=\dfrac{x^2}{\left(x+y\right)\left(1-y\right)}-\dfrac{y^2}{\left(x+y\right)\left(1+x\right)}-\dfrac{x^2y^2}{\left(1+x\right)\left(1-y\right)}\)đkxđ: \(y\ne1;x\ne-1;x\ne-y\)\(=\dfrac{x^2\left(1+x\right)}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}-\dfrac{y^2\left(1-y\right)}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}-\dfrac{x^2y^2\left(x+y\right)}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}\)\(=\dfrac{x^2+x^3-y^2+y^3-x^3y^2-x^2y^3}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}\)
\(=\dfrac{\left(x^3+y^3\right)+\left(x^2-y^2\right)-\left(x^3y^2+x^2y^3\right)}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}\)
\(=\dfrac{\left(x+y\right)\left(x^2-xy+y^2\right)+\left(x-y\right)\left(x+y\right)-x^2y^2\left(x+y\right)}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}\)\(=\dfrac{\left(x+y\right)\left(x^2-xy+y^2+x-y-x^2y^2\right)}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}\)\(=\dfrac{\left(x^2+x\right)-\left(xy+y\right)+\left(y^2-x^2y^2\right)}{\left(1-y\right)\left(x+1\right)}\)
\(=\dfrac{x\left(x+1\right)-y\left(x+1\right)-y^2\left(x-1\right)\left(x+1\right)}{\left(1-y\right)\left(x+1\right)}\) \(=\dfrac{\left(x+1\right)\left(x-y-y^2x+y^2\right)}{\left(1-y\right)\left(x+1\right)}\)
\(=\dfrac{-\left(y-y^2\right)+\left(x-y^2x\right)}{1-y}\)
\(=\dfrac{-y\left(1-y\right)+x\left(1-y\right)\left(1+y\right)}{1-y}\)
\(=\dfrac{\left(1-y\right)\left(x+xy-y\right)}{1-y}=x+xy-y\)
\(\dfrac{x^2}{\left(x+y\right)\left(1-y\right)}-\dfrac{y^2}{\left(x+y\right)\left(1+x\right)}-\dfrac{x^2y^2}{\left(1+x\right)\left(1-y\right)}\) MTC : (x+y)(1-y)(1+x)
A=
\(\dfrac{x^2\times\left(1+x\right)}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}-\dfrac{y^2\times\left(1-y\right)}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}-\dfrac{x^2y^2\times\left(x+y\right)}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}\)
A= \(\dfrac{x^2+x^3}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}-\dfrac{y^2}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}-\dfrac{x^3y^2+x^2y^3}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}\)
\(\dfrac{x^2+x^3-y^2-x^3y^2-x^2y^3}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}\)
a: \(\left(x-2y\right)^2+\left(x-\dfrac{1}{2}y\right)\left(x+\dfrac{1}{2}y\right)\)
\(=x^2-4xy+4y^2+x^2-\dfrac{1}{4}y^2\)
\(=2x^2-4xy+\dfrac{15}{4}y^2\)
b: \(\left(x-2\right)^2+\left(x+3\right)^2-2\left(x-1\right)\left(x+1\right)\)
\(=x^2-4x+4+x^2+6x+9-2\left(x^2-1\right)\)
\(=2x^2+2x+13-2x^2+2\)
=2x+15
a) \(=x^2-4xy+4y^2+x^2-\dfrac{1}{4}y^2=2x^2-4xy+\dfrac{15}{4}y^2\)
b) \(=x^2-4x+4+x^2+6x+9-2x^2+2\)
\(=2x+15\)