Biết \(x+y=1\)và\(x.y\ne0\)
Chứng minh \(\frac{x}{y^3-1}+\frac{y}{x^3-1}+\frac{2\left(x-y\right)}{x^2y^2+3}\)
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13 tháng 5 2020
Ta có \(y^3-1=\left(y-1\right)\left(y^2+y+1\right)=-x\left(y^2+y+1\right)\)
(vì \(xy\ne0\Rightarrow x,y\ne0\))
\(\Rightarrow x-1\ne0;y-1\ne0\)
\(\Rightarrow\frac{x}{y^3-1}=\frac{-1}{y^2+y+1}\)
\(x^3-1=\left(x-1\right)\left(x^2-x+1\right)=-y\left(x^2-x+1\right)\Rightarrow\frac{y}{x^3-1}=\frac{-1}{x^2+x+1}\)
\(\Rightarrow\frac{x}{y^3-1}+\frac{y}{x^3-1}=\frac{-1}{y^2+y+1}+\frac{-1}{x^2+x+1}\)
\(=-\left(\frac{x^2+x+1+y^2+y+1}{\left(x^2+x+1\right)\left(y^2+y+1\right)}\right)=-\left(\frac{\left(x+y\right)^2-2xy+\left(x+y\right)+2}{x^2y^2+\left(x+y\right)^2-2xy+xy\left(x+y\right)+xy+\left(x+y\right)+1}\right)\)
\(=-\frac{4-2xy}{x^2y^2+3}\Rightarrow\frac{x}{y^3-1}+\frac{y}{x^3-1}-\frac{2\left(xy-2\right)}{x^2y^2+3}=0\)
26 tháng 3 2017
để cm thì ta cần cm nó đúng khi x+y=1
x+y=1
y=-(x-1) và x=-(y-1)
thế vào ta được
-(x-1)/(x^3-1)--(y-1)/(y^3-1)=2(x-y)/(x^2y^2+3)
ta có x^3-1=(x-1)(x^2+x+1),y^3-1=(y-1)(y^2+y+1)
từ đó rút gọn ta được -1/(x^2+x+1)+1/(y^2+y+1)=2(x-y)/(x^2y^2+3)
1/(y^2+y+1)-1/(x^2+x+1)=2(x-y)/(x^2y^2+3)
(x^2+x+1-y^2-y-1)/(y^2+y+1)(x^2+x+1)=2(x-y)/(x^2y^2+3)
ta có x^2+x+1-y^2-y-1=x^2-y^2+x-y=(x-y)(x+y)+x-y=(x-y)(x+y+1)=2(x-y)
từ đó suy ra 2(x-y)/(y^2+y+1)(x^2+x+1)=2(x-y)/(x^2y^2+3)
suy ra (y^2+y+1)(x^2+x+1)=x^2+y^2+3
x^2y^2+xy^2+y^2+x^2y+xy+y+x^2+x+1=x^2y^2+3
x^2y^2+(xy^2+y^2+x^2y+xy+x^2)+x+y+1=x^2y^2+3
x^2y^2+(xy^2+y^2+x^2y+xy+x^2)+2=x^2y^2+3
ta có xy^2+y^2+x^2y+xy+x^2
=xy(x+y)+xy+y^2+x^2
=x^2+2xy+y^2
=(x+y)^2
=1^2
=1
thế vào ta được
x^2y^2+3=x^2y^2+3
vậy pt trên đúng khi x+y=1
HC
26 tháng 3 2017
Tk mình đi mọi người mình bị âm nè!
Ai tk mình mình tk lại cho!!
21 tháng 4 2017
Ta có:
\(\left(y^2+y+1\right)\left(x^2+x+1\right)\)
\(=x^2y^2+xy\left(x+y\right)+x^2+y^2+xy+x+y+1\)
\(=x^2y^2+x^2+y^2+2xy+2=x^2y^2+3\)
Ta lại có:
\(\left(y^2+y+1\right)-\left(x^2+x+1\right)=\left(y^2-x^2\right)+\left(y-x\right)\)
\(=\left(y-x\right)\left(x+y+1\right)=-2\left(x-y\right)\)
Theo đề bài ta có: (sửa đề luôn)
\(\frac{x}{y^3-1}-\frac{y}{x^3-1}+\frac{2\left(x-y\right)}{x^2y^2+3}\)
\(=\frac{x}{\left(y-1\right)\left(y^2+y+1\right)}-\frac{y}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2\left(x-y\right)}{x^2y^2+3}\)
\(=\frac{-1}{y^2+y+1}+\frac{1}{x^2+x+1}+\frac{2\left(x-y\right)}{x^2y^2+3}\)
\(=\frac{\left(y^2+y+1\right)-\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(y^2+y+1\right)}+\frac{2\left(x-y\right)}{x^2y^2+3}\)
\(=-\frac{2\left(x-y\right)}{x^2y^2+3}+\frac{2\left(x-y\right)}{x^2y^2+3}=0\)
22 tháng 2 2020
a, Ta có : \(\frac{3y}{4}=\frac{3y}{4}.1=\frac{3y}{4}.\frac{2x}{2x}=\frac{6xy}{8x}\) ( đpcm )
b, Ta có : \(6x^2y=6x^2y\)
=> \(3x^2.2y=\left(-3x^2\right).\left(-2y\right)\)
=> \(\frac{-3x^2}{2y}=\frac{3x^2}{-2y}\) ( đpcm )
c, Ta có : \(6x-6y=6x-6y\)
=> \(6x-6y=-6y+6x\)
=> \(6\left(x-y\right)=-6\left(y-x\right)\)
=> \(2\left(x-y\right).3=-2\left(y-x\right).3\)
=> \(\frac{2\left(x-y\right)}{3\left(y-x\right)}=\frac{-2}{3}\) ( đpcm )
13 tháng 5 2020
Biến đổi \(\frac{x}{y^3-1}-\frac{y}{x^3-1}=\frac{x^4-x-y^4+y}{\left(y^3-1\right)\left(x^3-1\right)}=\frac{\left(x^4-y^4\right)-\left(x-y\right)}{xy\left(y^2+y+1\right)\left(x^2+x+1\right)}\)
(Do x+y=1 => \(\hept{\begin{cases}y-1=-x\\x-1=-y\end{cases}}\))
\(=\frac{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)-\left(x-y\right)}{xy\left(x^2y^2+y^2x+y^2+yx^2+xy+y+x^2+x+1\right)}\)
\(=\frac{\left(x-y\right)\left(x^3+y^3-1\right)}{xy\left[x^2y^2+xy\left(x+y\right)+x^2+y^2+xy+2\right]}\)
\(=\frac{\left(x-y\right)\left(x^2-x+y^2-y\right)}{xy\left[x^2y^2+\left(x+y\right)^2+2\right]}=\frac{\left(x-y\right)\left[x\left(x-1\right)+y\left(y-1\right)\right]}{xy\left(x^2y^2+3\right)}\)
\(=\frac{\left(x-y\right)\left[x\left(-y\right)+y\left(-x\right)\right]}{xy\left(x^2y^2+3\right)}=\frac{\left(x-y\right)\left(-2xy\right)}{xy\left(x^2y^2+3\right)}=\frac{-2\left(x-y\right)}{x^2y^2+3}\)
\(\Rightarrow\frac{x}{y^3-1}-\frac{y}{x^3-1}+\frac{2\left(x-y\right)}{x^2y^2+3}=0\left(đpcm\right)\)
Y
17 tháng 4 2019
\(\frac{x}{y^3-1}-\frac{y}{x^3-1}+\frac{2\left(x-y\right)}{x^2y^2+3}\)
\(=\frac{-\left(y-1\right)}{\left(y-1\right)\left(y^2+y+1\right)}+\frac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}\) \(+\frac{2\left(x-y\right)}{x^2y^2+3}\)
\(=-\frac{1}{y^2+y+1}+\frac{1}{x^2+x+1}+\frac{2\left(x-y\right)}{x^2y^2+3}\)
\(=\frac{-\left(x^2+x+1\right)+y^2+y+1}{\left(y^2+y+1\right)\left(x^2+x+1\right)}+\frac{2\left(x-y\right)}{x^2y^2+3}\)
\(=\frac{-\left(x^2-y^2\right)-\left(x-y\right)}{x^2y^2+x^2y+xy^2+x^2+y^2+xy+x+y+1}+\frac{2\left(x-y\right)}{x^2y^2+3}\)
\(=\frac{-\left(x-y\right)\left(x+y\right)-\left(x-y\right)}{x^2y^2+xy\left(x+y\right)+xy+x^2+y^2+2}+\frac{2\left(x-y\right)}{x^2y^2+3}\)
\(=\frac{-\left(x-y\right)\left(x+y+1\right)}{x^2y^2+2xy+x^2+y^2+2}+\frac{2\left(x-y\right)}{x^2y^2+3}\)
\(=\frac{-2\left(x-y\right)}{x^2y^2+\left(x+y\right)^2+2}+\frac{2\left(x-y\right)}{x^2y^2+3}\)
\(=\frac{-2\left(x-y\right)}{x^2y^2+3}+\frac{2\left(x-y\right)}{x^2y^2+3}=0\)
Từ \(x+y=1\Rightarrow x=1-y\)
\(\Rightarrow y=1-x\)
Biến đổi \(\frac{y}{x^3-1}=\frac{y}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{\left(1-x\right)}{\left(x-1\right)\left(x^2+x+1\right)}=-\frac{\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=-\frac{1}{x^2+x+1}\)
\(\frac{x}{y^3-1}=\frac{x}{\left(y-1\right)\left(y^2+y+1\right)}=\frac{\left(1-y\right)}{\left(y-1\right)\left(y^2+y+1\right)}=-\frac{\left(y-1\right)}{\left(y-1\right)\left(y^2+y+1\right)}=-\frac{1}{y^2+y+1}\)
Ta có \(\frac{y}{x^3-1}-\frac{x}{y^3-1}=\frac{-1}{x^2+x+1}-\frac{-1}{y^2+y+1}=\frac{-y^2-y-1+x^2+x+1}{\left(x^2+x+1\right)\left(y^2+y+1\right)}\)
\(=\frac{\left(x^2-y^2\right)+\left(x-y\right)}{\left(x^2+x+1\right)\left(y^2+y+1\right)}=\frac{\left(x-y\right)\left(x+y\right)+\left(x-y\right)}{\left(x^2+x+1\right)\left(y^2+y+1\right)}=\frac{\left(x-y\right)\left(x+y+1\right)}{\left(x^2+x+1\right)\left(y^2+y+1\right)}\)
\(=\frac{\left(x-y\right)\left(1+1\right)}{x^2y^2+xy^2+y^2+xy+y+x^2+x+1}\)
\(=\frac{2\left(x-y\right)}{\left(x^2y^2+x+y+1\right)+\left(x^2+y^2\right)+\left(x^2y+xy^2+xy\right)}\)
\(=\frac{2\left(x-y\right)}{\left(x^2y^2+2\right)+\left(x^2+y^2\right)+xy\left(x+y+1\right)}\)
\(=\frac{2\left(x-y\right)}{\left(x^2y^2+2\right)+\left(x^2+y^2\right)+xy\left(1+1\right)}\)
\(=\frac{2\left(x-y\right)}{\left(x^2y^2+2\right)+x^2+y^2+2xy}\)
\(=\frac{2\left(x-y\right)}{\left(x^2y^2+2\right)+\left(x+y\right)^2}\)
\(=\frac{2\left(x-y\right)}{x^2y^2+2+1}\)
\(=\frac{2\left(x-y\right)}{x^2y^2+3}\)
Vậy \(\frac{x}{y^3-1}-\frac{y}{x^3-1}=\frac{2\left(x-y\right)}{x^2y^2+3}\) với \(x+y=1\&xy\ne0\)