e)\(\dfrac{2xy-x^2+z^2-y^2}{x^2+2-y^2+2xz}=\)
g)\(\dfrac{x^3+2x^2-x-z}{x^3-3x+2}=\)
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13 tháng 11 2017
a) \(\dfrac{x^2-y^2}{x^2-y^2+xz-yz}=\dfrac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)\left(x-y\right)+z\left(x-y\right)}\)
\(=\dfrac{\left(x-y\right)\left(x+y\right)}{\left(x-y\right)\left(x+y+z\right)}=\dfrac{x+y}{x+y+z}\)
b) \(\dfrac{x^2+y^2-z^2+2xy}{x^2+z^2-y^2-2xz}=\dfrac{\left(x+y\right)^2-z^2}{\left(x-z\right)^2-y^2}=\dfrac{\left(x+y-z\right)\left(x+y+z\right)}{\left(x-y-z\right)\left(x-z+y\right)}\)\(=\dfrac{x+y+z}{x-y-z}\)
c) \(\dfrac{x^2\left(x-3\right)-\left(x-3\right)}{x\left(x-3\right)}=\dfrac{\left(x-3\right)\left(x^2-1\right)}{x\left(x-3\right)}=\dfrac{x^2-1}{x}\)
d) \(\dfrac{4x^2\left(x-2\right)+3\left(x-2\right)}{4x^2\left(3x+1\right)+3\left(3x+1\right)}=\dfrac{\left(x-2\right)\left(4x^2+3\right)}{\left(3x+1\right)\left(4x^2+3\right)}=\dfrac{x-2}{3x+1}\)
30 tháng 12 2021
\(5x^2+2xy+2y^2-\left(4x^2+4xy+y^2\right)=\left(x-y\right)^2\ge0\\ \Leftrightarrow5x^2+2xy+2y^2\ge4x^2+4xy+y^2=\left(2x+y\right)^2\)
\(\Leftrightarrow P\le\dfrac{1}{2x+y}+\dfrac{1}{2y+z}+\dfrac{1}{2z+x}=\dfrac{1}{9}\left(\dfrac{9}{x+x+y}+\dfrac{9}{y+y+z}+\dfrac{9}{z+z+x}\right)\\ \Leftrightarrow P\le\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{z}+\dfrac{1}{z}+\dfrac{1}{x}\right)\\ \Leftrightarrow P\le\dfrac{1}{9}\left(\dfrac{3}{x}+\dfrac{3}{y}+\dfrac{3}{z}\right)=\dfrac{1}{3}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=1\)
Dấu \("="\Leftrightarrow x=y=z=1\)
19 tháng 1 2021
Đẳng thức đã cho tương đương với:
\(\dfrac{x^2z+y^2z-z^3+y^2x+z^2x-x^3+z^2y+x^2y-y^3}{2yxz}=1\)
\(\Leftrightarrow x^3+y^3+z^3+2xyz-x^2y-y^2z-z^2x-xy^2-yz^2-zx^2=0\)
\(\Leftrightarrow\left(x+y-z\right)\left(y+z-x\right)\left(z+x-y\right)=0\Leftrightarrow z+x=y\) (Do x + y khác z và y + z khác x).
Từ đó P = 2y (Biểu thức của P phụ thuộc vào biến y).
6 tháng 9 2021
a) x2+y2-4x+4y+8=0
⇔ (x-2)2+(y+2)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\)
b)5x2-4xy+y2=0
⇔ x2+(2x-y)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\2x-y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
c)x2+2y2+z2-2xy-2y-4z+5=0
⇔ (x-y)2+(y-1)2+(z-2)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-1=0\\z-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y=1\\z=2\end{matrix}\right.\)
6 tháng 9 2021
b: Ta có: \(5x^2-4xy+y^2=0\)
\(\Leftrightarrow x^2-\dfrac{4}{5}xy+y^2=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{2}{5}y+\dfrac{4}{25}y^2+\dfrac{21}{25}y^2=0\)
\(\Leftrightarrow\left(x-\dfrac{2}{5}y\right)^2+\dfrac{21}{25}y^2=0\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
5 tháng 11 2017
\(\text{1) }\dfrac{x^7+x^6+x^5+x^4+x^3+x^2+x+1}{x^2-1}\\ =\dfrac{\left(x^7+x^6\right)+\left(x^5+x^4\right)+\left(x^3+x^2\right)+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^6\left(x+1\right)+x^4\left(x+1\right)+x^2\left(x+1\right)+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{\left(x^6+x^4+x^2+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^6+x^4+x^2+1}{x-1}\)
\(\text{3) }\dfrac{x^2+y^2+z^2-2xy+2xz-2yz}{x^2-2xy+y^2-z^2}\\ =\dfrac{\left(x^2-2xy+y^2\right)+\left(2xz-2yz\right)+z^2}{\left(x^2-2xy+y^2\right)-z^2}\\ =\dfrac{\left(x-y\right)^2+2\left(x-y\right)z+z^2}{\left(x-y\right)^2-z^2}\\ =\dfrac{\left(x-y+z\right)^2}{\left(x-y+z\right)\left(x-y-z\right)}\\ =\dfrac{x-y+z}{x-y-z}\)
