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a) \(\dfrac{2x}{x^2+2xy}+\dfrac{y}{xy-2y^2}+\dfrac{4}{x^2-4y^2}\)
\(=\dfrac{2x}{x\left(x+2y\right)}+\dfrac{y}{y\left(x-2y\right)}+\dfrac{4}{\left(x-2y\right)\left(x+2y\right)}\) MTC: \(xy\left(x-2y\right)\left(x+2y\right)\)
\(=\dfrac{2x.y\left(x-2y\right)}{xy\left(x+2y\right)\left(x-2y\right)}+\dfrac{y.x\left(x+2y\right)}{xy\left(x-2y\right)\left(x+2y\right)}+\dfrac{4.xy}{xy\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{2xy\left(x-2y\right)+xy\left(x+2y\right)+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{2x^2y-4xy^2+x^2y+2xy^2+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{3x^2y-2xy^2+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
b) \(\dfrac{1}{x-y}+\dfrac{3xy}{y^3-x^3}+\dfrac{x-y}{x^2+xy+y^2}\)
\(=\dfrac{1}{x-y}-\dfrac{3xy}{x^3-y^3}+\dfrac{x-y}{x^2+xy+y^2}\)
\(=\dfrac{1}{x-y}-\dfrac{3xy}{\left(x-y\right)\left(x^2+xy+y^2\right)}+\dfrac{x-y}{x^2+xy+y^2}\) MTC: \(\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=\dfrac{x^2+xy+y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}-\dfrac{3xy}{\left(x-y\right)\left(x^2+xy+y^2\right)}+\dfrac{\left(x-y\right)\left(x-y\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{\left(x^2+xy+y^2\right)-3xy+\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{x^2+xy+y^2-3xy+x^2-2xy+y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{2x^2-4xy+2y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{2\left(x^2-2xy+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{2\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{2\left(x-y\right)}{x^2+xy+y^2}\)
\(x^2+y^2+z^2=xy+yz+zx\)
\(2.\left(x^2+y^2+z^2\right)=2.\left(xy+yz+zx\right)\)
\(\Rightarrow2.\left(x^2+y^2+z^2\right)-2xy-2yz-2zx=0\)
\(\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)=0\)
\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
Ta có: \(VT\ge0\forall x;y;z\)( tự c/m. nếu b ko c/m được thì bảo mình )
Mà \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}\left(x-y\right)^2=0\\\left(y-z\right)^2=0\\\left(z-x\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}\Leftrightarrow}}\hept{\begin{cases}x=y\\y=z\\z=x\end{cases}\Leftrightarrow x=y=z}\)
Có \(x^{2014}+y^{2014}+z^{2014}=3\)
\(\Rightarrow3.x^{2014}=3\)
\(\Rightarrow x^{2014}=1\)
\(\Rightarrow x=1\)
\(\Rightarrow x=y=z=1\)
Có: \(P=x^{25}+y^4+z^{2015}\)
\(\Rightarrow P=1^{25}+1^4+1^{2015}\)
\(P=1+1+1\)
\(P=3\)
Vậy \(P=3\)
Tham khảo nhé~
Ta có: x2+y2+z2=xy+yz+zx
<=>2x2+2y2+2z2=2xy+2yz+2zx
<=>2x2+2y2+2z2-2xy-2yz-2zx=0
<=>(x2-2xy+y2)+(y2-2yz+z2)+(z2-2zx+x2)=0
<=>(x-y)2+(y-z)2+(z-x)2=0
Vì \(\hept{\begin{cases}\left(x-y\right)^2\ge0\\\left(y-z\right)^2\ge0\\\left(z-x\right)^2\ge0\end{cases}\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0}\)
=>\(\hept{\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}\Rightarrow x=y=z}\)
=>x2014=y2014=z2014
Lại có: x2014+y2014+z2014 = 3
=>3x2014 = 3 => x2014 = 1 => \(x=\pm1\)
=>\(x=y=z=\pm1\)
Thay x,y,z vào P rồi tính
1. \(x^3-x^2+x-1=(x^3-x^2)+(x-1)\)
\(=x^2(x-1)+(x-1)=(x^2+1)(x-1)\)
2. \(6x^2y-2xy^2+3x-y=2xy(3x-y)+(3x-y)\)
\(=(3x-y)(2xy+1)\)
3. \(4x^2+1\) thì còn cái gì để phân tích hả bạn? Hay ý bạn là \(4x^4+1\)?
\(4x^4+1=(2x^2)^2+1=(2x^2)^2+1+4x^2-4x^2\)
\(=(2x^2+1)^2-(2x)^2=(2x^2+1-2x)(2x^2+1+2x)\)
4. \(x^2-9x+8=(x^2-x)-(8x-8)\)
\(=x(x-1)-8(x-1)=(x-1)(x-8)\)
5. \(x^3-2x^2y+3xy^2=x(x^2-2xy+3y^2)\)
6. \(x^2-6x+y-y^2\) (sai đề)
7. \(x^2-xy-2x+2y=(x^2-xy)-(2x-2y)\)
\(=x(x-y)-2(x-y)=(x-y)(x-2)\)
\(a,x^2+y^2-x-y=8\)
\(\Rightarrow x^2-x+\frac{1}{4}+y^2-y+\frac{1}{4}-8,5=0\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2-8,5=0\)
Ta có : \(\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2-8,5\ge-8,5\forall x;y\)
Để VP=0 và là các số nguyên
=>\(\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2=8,5\)
a/ x^2 + y^2 - x - y = 8
<=> 4x^2 + 4y^2 - 4x - 4y = 32
<=> (2x - 1)^2 + (2y - 1)^2 = 34
<=> (2x - 1)^2 = 9 và (2y - 1)^2 = 25
Hoặc (2x - 1)^2 = 25 và (2y - 1)^2 = 9
Bài 2 .
a) \(\dfrac{2x}{x^2+2xy}+\dfrac{y}{xy-2y^2}+\dfrac{4}{x^2-4y^2}\)
\(=\dfrac{2x}{x\left(x+2y\right)}+\dfrac{y}{y\left(x-2y\right)}+\dfrac{4}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{2xy\left(x-2y\right)+xy\left(x+2y\right)+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{2x^2y-2xy^2+x^2y+2xy^2+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{3x^2y+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
b) Sai đề hay sao ý
c) \(\dfrac{2x+y}{2x^2-xy}+\dfrac{16x}{y^2-4x^2}+\dfrac{2x-y}{2x^2+xy}\)
\(=\dfrac{2x+y}{x\left(2x-y\right)}+\dfrac{-16x}{\left(2x-y\right)\left(2x+y\right)}+\dfrac{2x-y}{x\left(2x+y\right)}\)
\(=\dfrac{\left(2x+y\right)^2-16x^2+\left(2x-y\right)^2}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\dfrac{4x^2+4xy+y^2-16x^2+4x^2-4xy+y^2}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\dfrac{-8x^2}{x\left(2x-y\right)\left(2x+y\right)}\)
d) \(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
.....
\(=\dfrac{16}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{32}{1-x^{32}}\)
a,(x-y)^2-2(x+y)+1 b, x^2-y^2+4x+4 c, 4x^2-y^2+8(y-2)
=(x-y-1)^2 =(x^2+4x+4)-y^2 =4x^2-y^2+8y-16
=(x+2)^2-y^2 =4x^2-(y^2-8y+16)
=(x+2-y)(x+2+y) =4x^2-(y-4)^2
a) (x+y)2-2(x+y)+1=(x+y-1)2
b) x2-y2+4x+4 = (x2+4x+4)-y2=(x+2)2-y2=(x+y+2)(x-y+2)
c)4x2-y2+8(y-2) = 4x2-(y2-8y+16) = (2x)2-(y-4)2=(2x+y-4)(2x-y+4)
d)x3-2x2+2x-4 = x2(x-2)+2(x-2) = (x-2)(x2+2)
e)xy-4+2x-2y=x(y+2) - 2(y+2) = (x-2)(y+2)