\(I=\lim\limits\dfrac{1+a+a^2+...+a^n}{1+b+b^2+...+b^n}\)

Xet tren tu la 1 csc voi : \(\left\{{}\begin{matrix}u_1=1\\q=a\end{matrix}\right.\Rightarrow S_a=1.\dfrac{a^{n+1}-1}{a-1}\)

Tuong tu cho mau so: \(S_b=1.\dfrac{b^{n+1}-1}{b-1}\)

\(\Rightarrow.....=\lim\limits\dfrac{\dfrac{a^{n+1}-1}{a-1}}{\dfrac{b^{n+1}-1}{b-1}}=\dfrac{\dfrac{1}{a-1}}{\dfrac{1}{b-1}}=\dfrac{1-b}{1-a}\)

\(B=\lim\limits\left(\sqrt{2n^2+1}-n\right)?\)

\(B=\lim\limits\left[n\left(\sqrt{\dfrac{2n^2}{n^2}+\dfrac{1}{n^2}}-\dfrac{n}{n}\right)\right]=\lim\limits\left[n\left(\sqrt{2}-1\right)\right]=+\infty\)

                     bài giải

diện tích hình vuông là 

    12*12/2=72(cm)

        đáp số 72 cm

\(B=\lim\limits\dfrac{4n^2+3n+1}{\left(3n-1\right)^2}=\lim\limits\dfrac{\dfrac{4n^2}{n^2}+\dfrac{3n}{n^2}+\dfrac{1}{n^2}}{\dfrac{9n^2}{n^2}-\dfrac{6n}{n^2}+1}=\dfrac{4}{9}\)

\(C=\lim\limits\dfrac{\sqrt{n^2+1}}{n+1}?\)

\(C=\lim\limits\dfrac{\sqrt{\dfrac{n^2}{n^2}+\dfrac{1}{n^2}}}{\dfrac{n}{n}+\dfrac{1}{n}}=1\)

Xai cai nay go cong thuc di ban :v Doc ko hieu

Vay thi tinh moi nhiet luong no tang tu -10 do-> 0 do c la duoc :v

sai cho nao ban? Neu nhu minh hieu ko sai de bai thi co nghia la nhiet luong de da tan chay o 0 do c bao gom ca nhiet luong de no tang len den 0 do c va nhiet luong de no tan chay ma van giu nguyen nhiet do do. Ko hieu ban nhin vo loi giai, hay nhin vo dap an ma bao sai?