cho x^2+y^2=2 va xy=1 cmr x-x^3=y^3-y
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Từ giải thiết, ta suy ra được những điều sau :
\(\frac{x}{y^3-1}-\frac{y}{x^3-1}=\frac{x}{\left(y-1\right)\left(y^2+y+1\right)}-\frac{y}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{x}{\left[y-\left(x+y\right)\right]\left(y^2+y+1\right)}-\frac{y}{\left[x-\left(x+y\right)\right]\left(x^2+x+1\right)}\)
\(=\frac{x}{-x\left(y^2+y+1\right)}-\frac{y}{-y\left(x^2+x+1\right)}\)
\(=\frac{-1}{y^2+y+1}+\frac{1}{x^2+x+1}\) (1)
Và \(\left(x^2+x+1\right)\left(y^2+y+1\right)\)
\(=x^2y^2+x^2y+x^2+xy^2+xy+x+y^2+y+1\)
\(=x^2y^2+\left(x^2+xy\left(x+y\right)+xy+y^2\right)+\left(x+y\right)+1\)
\(=x^2y^2+\left(x^2+2xy+y^2\right)+1+1\)
\(=x^2y^2+\left(x+y\right)^2+2\)
\(=x^2y^2+3\) (2)
Từ (1) và (2) suy ra :
\(\frac{x}{y^3-1}-\frac{y}{x^3-1}+\frac{2\left(x-y\right)}{x^2y^2+3}\)
\(=\frac{-1}{y^2+y+1}+\frac{1}{x^2+x+1}+\frac{2\left(x-y\right)}{\left(y^2+y+1\right)\left(x^2+x+1\right)}\)
\(=\frac{-x^2-x-1+y^2+y+1+2x-2y}{\left(y^2+y+1\right)\left(x^2+x+1\right)}\)
\(=\frac{-x^2+y^2+x-y}{\left(y^2+y+1\right)\left(x^2+x+1\right)}\)
\(=\frac{\left(x+y\right)\left(y-x\right)+x-y}{\left(y^2+y+1\right)\left(x^2+x+1\right)}\)
\(=\frac{y-x+x-y}{\left(y^2+y+1\right)\left(x^2+x+1\right)}\)
\(=0\)(ĐPCM)
Biến đổi
\(\frac{x}{y^3-1}-\frac{y}{x^3-1}=\frac{x^4-x-y^4+y}{\left(x^3-1\right)\left(y^3-1\right)}=\frac{\left(x^4-y^4\right)-\left(x-y\right)}{xy\left(y^2+y+1\right)\left(x^2+x+1\right)}\)
(do x+y=1 => y-1=-x và x-1=-y)
\(=\frac{\left(x-y\right)\left(x+y\right)\left(x^3+y^3\right)-\left(x-y\right)}{xy\left(x^2y^2+y^2x+y^2+yx^2+xy+y+x^2+x+1\right)}\)
\(=\frac{\left(x-y\right)\left(x^2+y^2-1\right)}{xy\left[x^2y^2+xy\left(x+y\right)+x^2+y^2+xy+2\right]}\)
\(=\frac{\left(x-y\right)\left(x^2-x+y^2-y\right)}{xy\left[x^2y^2+\left(x+y\right)^2+2\right]}=\frac{\left(x-y\right)\left[x\left(x-1\right)+y\left(y-1\right)\right]}{xy\left(x^2y^2+3\right)}\)
\(=\frac{\left(x-y\right)\left[x\left(-y\right)+y\left(-x\right)\right]}{xy\left(x^2y^2+3\right)}=\frac{\left(x-y\right)\left(-2xy\right)}{xy\left(x^2y^2+1\right)}=\frac{-2\left(x-y\right)}{x^2y^2+3}\)
=> ĐPCM
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Trả lời nhanh nha các bn, mik đang cần gấp, cảm ơn nhiều.
Kết hợp với giả thiết nêu ra ở đề bài, ta có vài biến đổi sau:
\(\frac{x}{y^3-1}=\frac{x}{\left(y-1\right)\left(y^2+y+1\right)}=\frac{x}{\left[y-\left(x+y\right)\right]\left(y^2+y+1\right)}=-\frac{1}{y^2+y+1}\) \(\left(1\right)\)
\(\frac{y}{x^3-1}=\frac{y}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{y}{\left[x-\left(x+y\right)\right]\left(x^2+x+1\right)}=-\frac{1}{x^2+x+1}\) \(\left(2\right)\)
Mặt khác, ta lại có: \(\left(x^2+x+1\right)\left(y^2+y+1\right)=x^2y^2+xy^2+y^2+x^2y+xy+y+x^2+x+1\)
\(=x^2y^2+\left[x^2+xy\left(x+y\right)+xy+y^2\right]+\left(x+y\right)+1=x^2y^2+\left(x+y\right)^2+2=x^2y^2+3\)
Khi đó, trừ đẳng thức \(\left(1\right)\) cho đẳng thức \(\left(2\right)\) vế theo vế, ta được:
\(\frac{x}{y^3-1}-\frac{y}{x^3-1}=\frac{1}{x^2+x+1}-\frac{1}{y^2+y+1}=\frac{\left(y-x\right)\left(x+y+1\right)}{\left(x^2+x+1\right)\left(y^2+y+1\right)}=\frac{-2\left(x-y\right)}{x^2y^2+3}\)
Vậy, \(\frac{x}{y^3-1}-\frac{y}{x^3-1}+\frac{2\left(x-y\right)}{x^2y^2+3}=-\frac{2\left(x-y\right)}{x^2y^2+3}+\frac{2\left(x-y\right)}{x^2y^2+3}=0\)
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\(xy\ne0,x,y\ne1\)
\(A=\dfrac{x^{ }}{y^3-1}-\dfrac{y}{x^3-1}+\dfrac{2\left(x+y\right)}{x^2y^2+3}\)
\(xét:\dfrac{2\left(x+y\right)}{x^2y^2+3}=\dfrac{2}{x^2y^2+3}\left(1\right)\)
\(\dfrac{x^{ }}{y^3-1}-\dfrac{y}{x^3-1}=\dfrac{x^4-x-y^4+y}{\left(x^3-1\right)\left(y^3-1\right)}\left(2\right)\)
\(xét:\) \(x^4-x-y^4+y=\left(x-y\right)\left(x^3+x^2y+xy^2+y^3-1\right)\)
\(=\left(x-y\right)\left[\left(x+y\right)^3-3xy\left(x+y\right)+xy\left(x+y\right)-1\right]\)
\(=\left(x-y\right)\left(1-3xy+xy-1\right)\)
\(=\left(x-y\right)\left(-2xy\right)=-2xy\left(x-y\right)=2xy\)
\(xét\) \(\left(y^3-1\right)\left(x^3-1\right)=x^3y^3-\left[\left(x+y\right)^3-3xy\left(x+y\right)\right]+1\)
\(=x^3y^3-\left(1-3xy\right)+1=x^3y^3+3xy=xy\left(x^2y^2+3\right)\)
\(\Rightarrow\left(2\right)\Leftrightarrow\dfrac{-2\left(x-y\right)}{x^2y^2+3}\)
\(\left(1\right)\left(2\right)\Rightarrow A=\dfrac{2}{x^2y^2+3}-\dfrac{2\left(x-y\right)}{x^2y^2+3}=\dfrac{2-2x+2y}{x^2y^2+3}\ne0\left(đề-sai\right)\)
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2.
\(4n^3+n+3=4n^3+2n^2+2n-2n^2-n-1+4=2n\left(2n^2+n+1\right)-\left(2n^2+n+1\right)+4\)-Để \(\left(4n^3+n+3\right)⋮\left(2n^2+n+1\right)\) thì \(4⋮\left(2n^2+n+1\right)\)
\(\Leftrightarrow2n^2+n+1\in\left\{1;-1;2;-2;4;-4\right\}\) (do n là số nguyên)
*\(2n^2+n+1=1\Leftrightarrow n\left(2n+1\right)=0\Leftrightarrow n=0\) (loại) hay \(n=\dfrac{-1}{2}\) (loại)
*\(2n^2+n+1=-1\Leftrightarrow2n^2+n+2=0\) (phương trình vô nghiệm)
\(2n^2+n+1=2\Leftrightarrow2n^2+n-1=0\Leftrightarrow n^2+n+n^2-1=0\Leftrightarrow n\left(n+1\right)+\left(n+1\right)\left(n-1\right)=0\Leftrightarrow\left(n+1\right)\left(2n-1\right)=0\)
\(\Leftrightarrow n=-1\) (loại) hay \(n=\dfrac{1}{2}\) (loại)
\(2n^2+n+1=-2\Leftrightarrow2n^2+n+3=0\) (phương trình vô nghiệm)
\(2n^2+n+1=4\Leftrightarrow2n^2+n-3=0\Leftrightarrow2n^2-2n+3n-3=0\Leftrightarrow2n\left(n-1\right)+3\left(n-1\right)=0\Leftrightarrow\left(n-1\right)\left(2n+3\right)=0\)\(\Leftrightarrow n=1\left(nhận\right)\) hay \(n=\dfrac{-3}{2}\left(loại\right)\)
-Vậy \(n=1\)
1. \(x^2+y^2=z^2\)
\(\Rightarrow x^2+y^2-z^2=0\)
\(\Rightarrow\left(x-z\right)\left(x+z\right)+y^2=0\)
-TH1: y lẻ \(\Rightarrow x-z;x+z\) đều lẻ.
\(x+3z-y=x+z-y+2x\) chia hết cho 2. \(\Rightarrow\)Hợp số.
-TH2: y chẵn \(\Rightarrow\)1 trong hai biểu thức \(x-z;x+z\) chia hết cho 2.
*Xét \(\left(x-z\right)⋮2\):
\(x+3z-y=x-z+4z-y\) chia hết cho 2. \(\Rightarrow\)Hợp số.
*Xét \(\left(x+z\right)⋮2\):
\(x+3z-y=x+z+2z-y\) chia hết cho 2 \(\Rightarrow\)Hợp số.
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Đề là CMR $x^4-x^3y+x^2y^2-xy^3+y^4> x^2+y^2$ thì đúng hơn bạn ạ.
Lời giải:
Ta có:
$\text{VT}=(x^4+y^4-x^3y-xy^3)+x^2y^2$
$=(x-y)^2(x^2+xy+y^2)+x^2y^2\geq x^2y^2$
Mà:
$x^2y^2=\frac{x^2y^2}{2}+\frac{x^2y^2}{2}> \frac{x^2.2}{2}+\frac{2.y^2}{2}=x^2+y^2$ do $x^2> 2, y^2>2$
Do đó: $\text{VT}> x^2+y^2$ (đpcm)