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1)\(A=\frac{b\left(2a\left(a+5b\right)+\left(a+5b\right)\right)}{a-3b}.\frac{a\left(a-3b\right)}{ab\left(a+5b\right)}=\frac{b\left(a+5b\right)\left(2a+1\right).a\left(a-3b\right)}{\left(a-3b\right).ab\left(a+5b\right)}\)
\(A=2a+1\)=>lẻ với mọi a thuộc z=> dpcm
2) từ: x+y+z=1=> xy+z=xy+1-x-y=x(y-1)-(y-1)=(y-1)(x-1)
tường tự: ta có tử của Q=(x-1)^2.(y-1)^2.(z-1)^2=[(x-1)(y-1)(z-1)]^2=[-(z+y).-(x+y).-(x+y)]^2=Mẫu=> Q=1
3) kiểm tra lại xem đề đã chuẩn chưa
Hướng dẫn :\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow\frac{xy+yz+zx}{xyz}=0\Rightarrow xy+yz+zx=0\)
Thay vào:\(x^2+2yz=x^2+yz+yz=x^2+yz-xy-zx=x\left(x-y\right)-z\left(x-y\right)=\left(x-y\right)\left(x-z\right)\)
Tương tự thay vào mà quy đồng
1/
\(x^2-xy-2y^2=0\Leftrightarrow x^2+xy-2xy-2y^2=0\)
\(\Leftrightarrow x\left(x+y\right)-2y\left(x+y\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(x-2y\right)=0\Rightarrow x=2y\) (do \(x+y\ne0\))
\(\Rightarrow P=\frac{2y-y}{2y+y}=\frac{y}{3y}=\frac{1}{3}\)
2/
\(x^4-30x^2+31x-30=0\)
\(\Leftrightarrow x^4+x-30x^2+30x-30=0\)
\(\Leftrightarrow x\left(x^3+1\right)-30\left(x^2-x+1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)=0\)
\(\Leftrightarrow\left(x^2+x-30\right)\left(x^2-x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x-30=0\\x^2-x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left(x-5\right)\left(x+6\right)=0\\\left(x-\frac{1}{2}\right)^2+\frac{3}{4}=0\left(vn\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=5\\x=-6\end{matrix}\right.\)
\(x+y=1\Rightarrow\left\{{}\begin{matrix}y-1=-x\\x-1=-y\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left(y-1\right)^2=x^2\\\left(x-1\right)^2=y^2\end{matrix}\right.\)
\(\frac{x}{y^3-1}-\frac{y}{x^3-1}+\frac{2\left(x-y\right)}{\left(xy\right)^2+3}=\frac{x}{\left(y-1\right)\left(y^2+y+1\right)}-\frac{y}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2\left(x-y\right)}{\left(xy\right)^2+3}\)
\(=\frac{-1}{y^2+y+1}+\frac{1}{x^2+x+1}+\frac{2\left(x-y\right)}{\left(xy\right)^2+3}=\frac{-1}{x^2+3y}+\frac{1}{y^2+3x}+\frac{2\left(x-y\right)}{\left(xy\right)^2+3}\)
\(=\frac{-y^2-3x+x^2+3y}{\left(xy\right)^2+3x^3+3y^3+9xy}+\frac{2\left(x-y\right)}{\left(xy\right)^2+3}=\frac{\left(x-y\right)\left(x+y\right)-3x+3y}{\left(xy\right)^2+3\left(x+y\right)\left(\left(x+y\right)^2-3xy\right)+9xy}+\frac{2\left(x-y\right)}{\left(xy\right)^2+3}\)
\(=\frac{-2\left(x-y\right)}{\left(xy\right)^2+3}+\frac{2\left(x-y\right)}{\left(xy\right)^2+3}=0\)
Bài 3:
Ta có:
\(81^8-1=\left(9^2\right)^8-1=\left[\left(3^2\right)^2\right]^8-1=3^{32}-1\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
Do đó:
\(A=3^4-1=80\)