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\(x^4+y^4=\left(x^2+y^2\right)^2-2x^2y^2=\left[\left(x+y\right)^2-2xy\right]^2-2\left(xy\right)^2\)
\(=\left(a^2-2b\right)^2-2b^2=a^4-2.a^2.2b+4b^2-2b^2=a^4-4a^2b+2b^2\)
Ta có: \(x-y=4\Rightarrow\left(x-y\right)^2=16\)
\(\Rightarrow x^2-2xy+y^2=16\Rightarrow x^2+y^2=16+2xy=16+2.3=22\)
\(M=x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)=4.\left(22+3\right)=100\)
\(x^2+y^2=\left(x+y\right)^2-2xy=6\)
\(x^3+y^3=\left(x+y\right)^3-3xy.\left(x+y\right)=14\)
\(x^4+y^4=\left(x^2+y^2\right)^2-2\left(xy\right)^2=34\)
\(\Rightarrow x^7+y^7=\left(x^3+y^3\right)\left(x^4+y^4\right)-\left(xy\right)^3\left(x+y\right)=478\)
Đề sai rồi, không thể tồn tại x; y sao cho \(\left\{{}\begin{matrix}x+y=3\\xy=5\end{matrix}\right.\) được
Vì \(\left(x+y\right)^2\ge4xy;\forall x;y\) nên \(3^2>4.5\) là vô lý
a: \(x^2+y^2=\left(x+y\right)^2-2xy=3^2-2\cdot5=-1\)
b: \(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=3^3-3\cdot3\cdot5=-18\)
\(\frac{x^2+y^2}{xy}=\frac{10}{3}\Rightarrow3x^2+3y^2-10xy=0\)
\(\Rightarrow\left(3x^2-9xy\right)-\left(xy-3y^2\right)=0\Rightarrow3x\left(x-3y\right)-y\left(x-3y\right)=0\)
\(\Rightarrow\left(x-3y\right)\left(3x-y\right)=0\Rightarrow3x-y=0\left(y>x>0\Rightarrow x-3y< 0\right)\Rightarrow3x=y\)
\(M=\frac{x-y}{x+y}=\frac{x-3x}{x+3x}=\frac{-2x}{4x}=-\frac{1}{2}\)
\(x^4+y^4=\left(x^2+y^2\right)^2-2x^2y^2=\left[\left(x+y\right)^2-2xy\right]^2-2x^2y^2\)
\(=\left(m^2-2n\right)^2-2n^2=m^4-4m^2n+4n^2-2n^2=m^4-4m^2n+2n^2\)
Ta có:
A=x2-2xy+y2+4xy-4xy
=(x+y)2-4xy
=9-40
=-31
B=x2+y2+2xy-2xy
=(x+y)2-2xy
=9-20
=-11
C=x3+y3
=(x+y)(x2-xy+y2)
=3.(-21)
=-63
\(x^2+y^2=\left(x+y\right)^2-2xy=m^2-2n\\ x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=m^3-3mn\\ \Rightarrow x^5+y^5=\left(x^3+y^3\right)\left(x^2+y^2\right)-x^2y^2\left(x+y\right)=\left(m^3-3mn\right)\left(m^2-2n\right)-n^2m\\ \Rightarrow x^7+y^7=\left(x^2+y^2\right)\left(x^5+y^5\right)-x^2y^2\left(x^3+y^3\right)=.....\)
tick mik nha