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Đặt \(P=\frac{5-3}{7}+\frac{5^2-3^2}{7^2}+...+\frac{5^n-3^n}{7^n}=\frac{5}{7}+\left(\frac{5}{7}\right)^2+...+\left(\frac{5}{7}\right)^n-\frac{3}{7}-\left(\frac{3}{7}\right)^2-...-\left(\frac{3}{7}\right)^n=A-B\)
\(A=\frac{5}{7}+\left(\frac{5}{7}\right)^2+...+\left(\frac{5}{7}\right)^n\) là tổng CSN với \(u_1=\frac{5}{7};q=\frac{5}{7};n=n\)
\(\Rightarrow A=\frac{5}{7}.\frac{1-\left(\frac{5}{7}\right)^{n+1}}{1-\frac{5}{7}}=\frac{5}{2}-\frac{5}{2}.\left(\frac{5}{7}\right)^{n+1}\)
\(B=\frac{3}{7}+\left(\frac{3}{7}\right)^2+...+\left(\frac{3}{7}\right)^n\) là tổng CSN với \(u_1=\frac{3}{7};q=\frac{3}{7}\)
\(\Rightarrow B=\frac{3}{7}.\frac{1-\left(\frac{3}{7}\right)^{n+1}}{1-\frac{3}{7}}=\frac{3}{4}-\frac{3}{4}.\left(\frac{3}{7}\right)^{n+1}\)
\(\Rightarrow limP=lim\left(\frac{5}{2}-\frac{5}{2}\left(\frac{5}{7}\right)^{n+1}-\frac{3}{4}+\frac{3}{4}\left(\frac{3}{7}\right)^{n+1}\right)=\frac{5}{2}-\frac{3}{4}=\frac{7}{4}\)
Lời giải:
a.
\(\lim\limits_{x\to 0}\frac{\sqrt{x+1}-\sqrt{x^2+x+1}}{x^2-2x+1}=\lim\limits_{x\to 0}\frac{\sqrt{0+1}-\sqrt{0^2+0+1}}{0^2-2.0+1}=0\)
b.
\(\lim\limits_{x\to 7}\frac{\sqrt{x-3}-2}{49-x^2}=\lim\limits_{x\to 7}\frac{(x-3)-2^2}{(49-x^2)(\sqrt{x-3}+2)}\)
\(=\lim\limits_{x\to 7}\frac{x-7}{-(x-7)(x+7)(\sqrt{x-3}+2)}=\lim\limits_{x\to 7}\frac{1}{-(x+7)(\sqrt{x-3}+2)}=\frac{1}{-(7+7)(\sqrt{7-3}+2)}=\frac{-1}{56}\)
a) \(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {x + 1} - \sqrt {{x^2} + x + 1} }}{{{x^2} - 2x + 1}} = \frac{{\sqrt {0 + 1} - \sqrt {{0^2} + 0 + 1} }}{{{0^2} - 2.0 + 1}} = 0\)
b) \(\mathop {\lim }\limits_{x \to 7} \frac{{\sqrt {x - 3} - 2}}{{49 - {x^2}}} = \mathop {\lim }\limits_{x \to 7} \frac{{x - 3 - {2^2}}}{{\left( {7 - x} \right)\left( {7 + x} \right)\left( {\sqrt {x - 3} + 2} \right)}} = \mathop {\lim }\limits_{x \to 7} \frac{{ - 1}}{{\left( {7 + x} \right)\left( {\sqrt {x - 3} + 2} \right)}} = \frac{{ - 1}}{{56}}\)
\(y'=\dfrac{1}{2\sqrt{x-1}}+\dfrac{1}{\sqrt{2x+1}}\)
\(\Rightarrow y'\left(3\right)=\dfrac{1}{2\sqrt{2}}+\dfrac{1}{\sqrt{7}}\Rightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=1\end{matrix}\right.\Rightarrow a+b=\dfrac{3}{2}\)
