a) tính : \(B=\dfrac{2.1+1}{\left(1\left(1+1\right)\right)^2}+\dfrac{2.2+1}{\left(2\left(2+1\right)\right)^2}+....+\dfrac{2.99+1}{\left(99.\left(99+1\right)\right)^2}\)
b) cho \(3a^2+b^2=4ab\). tính giá trị biểu thức \(P=\dfrac{a-b}{a+b}\)
c)cho \(N=0,7.\left(2007^{2009}-2013^{1999}\right)\). CMR N là 1 số nguyên
b) Giải:
ĐK: \(a\ne-b\)
Ta có:
\(3a^2+b^2=4ab\)
\(\Leftrightarrow4a^2-4ab+b^2-a^2=0\)
\(\Leftrightarrow\left(2a-b\right)^2-a^2=0\)
\(\Leftrightarrow\left(3a-b\right)\left(a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3a-b=0\\a-b=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}a=\dfrac{b}{3}\\a=b\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a=\dfrac{b}{3}\Leftrightarrow P=\dfrac{\dfrac{b}{3}-b}{\dfrac{b}{3}+b}=\dfrac{-1}{2}\\a=b\Leftrightarrow P=\dfrac{a-a}{a+a}=\dfrac{0}{2a}=0\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}P=\dfrac{-1}{2}\\P=0\end{matrix}\right.\)
