\(\left(x+y\right)\left(x^2-y^2\right)+\left(y+z\right)\left(y^2-z^2\right)+\left(x+z\right)\left(z^2-x^2\right)\) pt đt thành nt
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19 tháng 8 2018
\(x\left(y^2+z^2\right)+y\left(z^2+x^2\right)+z\left(x^2+y^2\right)+2xyz=xy^2+xz^2+yz^2+x^2y++zx^2+zy^2+2xyz=xy\left(x+y+z\right)+yz\left(x+y+z\right)+xz\left(x+z\right)=\left(x+y+z\right)\left(xy+yz\right)+xz\left(x+z\right)=y\left(x+y+z\right)\left(z+x\right)+xz\left(x+z\right)=\left(x+z\right)\left(xy+y^2+yz+xz\right)=\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
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
7 tháng 7 2023
Phân tích vế trái ta được: 2(x2 + y2 + z2 − (xy + yz + zx)
Phân tích vế phải ta được: 6(x2 + y2 + z2 − (xy + yz + zx)
Vì 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
21 tháng 11 2017
d)
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+.....+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)=\(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+.....-\dfrac{1}{x+99}+\dfrac{1}{x+100}\)=\(\dfrac{1}{x}-\dfrac{1}{x+100}\)
=\(\dfrac{x+100}{x\left(x+100\right)}-\dfrac{x}{x\left(x+100\right)}\)
=\(\dfrac{x+100-x}{x\left(x+100\right)}=\dfrac{100}{x\left(x+100\right)}\)
\(\left(x+y\right)\left(x^2-y^2\right)+\left(y+z\right)\left(y^2-z^2\right)+\left(x+z\right)\left(z^2-x^2\right)\)
\(=\left(x+y\right)\left(x^2-y^2\right)-\left(y+z\right)\left[\left(x^2-y^2\right)+\left(z^2-x^2\right)\right]+\left(x+z\right)\left(z^2-x^2\right)\)
\(=\left(x+y\right)\left(x^2-y^2\right)-\left(y+z\right)\left(x^2-y^2\right)-\left(y+z\right)\left(z^2-x^2\right)+\left(x+z\right)\left(z^2-x^2\right)\)
\(=\left(x^2-y^2\right)\left(x+y-y-z\right)-\left(z^2-x^2\right)\left(y+z-x-z\right)\)
\(=\left(x^2-y^2\right)\left(x-z\right)-\left(z^2-x^2\right)\left(y-x\right)\)
\(=\left(x-y\right)\left(x+y\right)\left(x-z\right)-\left(z-x\right)\left(z+x\right)\left(y-x\right)\)
\(=-\left(y-x\right)\left(x+y\right)\left(x-z\right)+\left(x-z\right)\left(z+x\right)\left(y-x\right)\)
\(=\left(y-x\right)\left(x-z\right)\left[-\left(x+y\right)+\left(z+x\right)\right]\)
\(=\left(y-x\right)\left(x-z\right)\left(-x+y+z+x\right)\)
\(=\left(y-x\right)\left(x-z\right)\left(y+z\right)\)