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23 tháng 9 2017

\(\dfrac{x^3+y^3+z^3-3xyz}{xy^2+xz\left(2y+z\right)}.\dfrac{x\left(x+y\right)+y\left(x-xy\right)}{\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2}\\ =\dfrac{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)}{xy^2+2xyz+x^2z}.\dfrac{x^2+xy-xy-xy^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\\ =\dfrac{\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]}{2xy^2+4xyz+2x^2z}.\dfrac{x^2-xy^2}{\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2}\\ =\dfrac{\left(x+y+z\right)\left(x^2-xy\right)}{2xy^2+4xy+2x^2z}\)

@@ ko ra nữa

21 tháng 11 2017

d)

\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+.....+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)=\(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+.....-\dfrac{1}{x+99}+\dfrac{1}{x+100}\)=\(\dfrac{1}{x}-\dfrac{1}{x+100}\)

=\(\dfrac{x+100}{x\left(x+100\right)}-\dfrac{x}{x\left(x+100\right)}\)

=\(\dfrac{x+100-x}{x\left(x+100\right)}=\dfrac{100}{x\left(x+100\right)}\)

22 tháng 11 2017

Cảm ơn, mình làm được rồi :>

9 tháng 12 2018

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

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

\(\left\{{}\begin{matrix}\left(x^2-yz\right)\left(y+z\right)=x^2y+x^2z-y^2z-yz^2\\\left(y^2-xz\right)\left(x+z\right)=y^2x+y^2z-x^2z-xz^2\\\left(z^2-xy\right)\left(x+y\right)=z^2x+z^2y-x^2y-xy^2\end{matrix}\right.\)

Đa thức trên bằng 0

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

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

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

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

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

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

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

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

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

Thêm dấu - đằng trc nữa suy ra bt có giá trị bằng 1 :P

AH
Akai Haruma
Giáo viên
24 tháng 11 2018

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

AH
Akai Haruma
Giáo viên
24 tháng 11 2018

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:

\(\frac{(x^2-y)(y+1)+x^2y^2-1}{(x^2+y)(y+1)+x^2y^2+1}=\frac{(x^2-1)(y^2+y+1)}{(x^2+1)(y^2+y+1)}=\frac{x^2-1}{x^2+1}\)

2 tháng 7 2021

a) xy(x + y) + yz(y + z) + xz(z + x) + 3xyz

= xy(X + y + z)  + yz(x + y + z) + xz(X + y + z)

= (x + y +z)(xy + yz+ xz)

b) xy(x + y) - yz(y + z) - xz(z - x)

= x2y + xy2 - y2z - yz2 - xz2 + x2z

= x2(y + z) - yz(y + z) + x(y2 - z2)

= x2(y + z) - yz(y + z) + x(y + z)(y - z)

= (y + z)(x2 - yz + xy - xz)

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

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

c) x(y2 - z2) + y(z2 - x2) + z(x2 - y2)

 = x(y - z)(y + z) + yz2 - yx2 + x2z - y2z

= x(y - z)(y + z) - yz(y - z) - x2(y - z)

= (y - z)((xy + xz - yz - x2)

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

= (y - z)(x - z)(y -x)