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Lời giải:
a)
$H=\frac{(x^2+y^2)(x+y)-x^2(x+1)-y^2(y-1)}{(x+1)(y-1)(x+y)}$
$=\frac{x^2y+xy^2-x^2+y^2}{(x+1)(y-1)(x+y)}$
$=\frac{xy(x+y)-(x-y)(x+y)}{(x+1)(y-1)(x+y)}=\frac{(x+y)(xy-x+y)}{(x+1)(y-1)(x+y)}$
$=\frac{xy-x+y}{(x+1)(y-1)}=\frac{xy-x+y}{xy-x+y-1}=1+\frac{1}{(x+1)(y-1)}$
b)
$H=6\Leftrightarrow \frac{1}{(x+1)(y-1)}=5$
$\Leftrightarrow (x+1)(y-1)=\frac{1}{5}$ (vô lý với mọi $x,y$ nguyên.
\(A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{n^2}\right)\)
\(=\left(\frac{2^2-1}{2^2}\right)\left(\frac{3^2-1}{3^2}\right)\left(\frac{4^2-1}{4^2}\right)...\left(\frac{n^2-1}{n^2}\right)\)
\(=\text{[}\frac{\left(2-1\right)\left(2+1\right)}{2^2}\text{]}.\text{[}\frac{\left(3-1\right)\left(3+1\right)}{3^2}\text{]}.\text{[}\frac{\left(4-1\right)\left(4+1\right)}{4^2}\text{]}...\text{[}\frac{\left(n-1\right)\left(n+1\right)}{n^2}\text{]}\)
\(=\left(\frac{1.3}{2^2}\right).\left(\frac{2.4}{3^2}\right).\left(\frac{3.5}{4^2}\right)...\text{[}\frac{\left(n-1\right)\left(n+1\right)}{n^2}\text{]}\)
\(=\frac{\text{[}1.2.3...\left(n-1\right)\text{]}.\text{[}3.4.5...\left(n+1\right)\text{]}}{\text{[}2.3.4...n\text{]}.\text{[}2.3.4...n\text{]}}\)
\(=\frac{1}{n}.\frac{n+1}{2}\)
\(=\frac{n+1}{2n}\)
a, \(I=s\left(s^2-t\right)+\left(t^2+s\right)=s^3-st+t^2+s\)
Thay t = -1 và s = 1 vào biểu thức trên ta được :
\(1+1+1+1=4\)
b, \(N=u^2\left(u-v\right)-v\left(v^2-u^2\right)=u^2\left(u-v\right)+v\left(u+v\right)\left(u-v\right)\)
\(=\left(u-v\right)\left(u^2+v\left(u+v\right)\right)\)
Thay \(u=0,5=\frac{1}{2};v=-\frac{1}{2}\)
\(=\left(\frac{1}{2}+\frac{1}{2}\right).\frac{1}{4}=\frac{1}{4}\)
mk thấy bn Quán quân hơn