Xin lỗi em mới lớp ... 4 thôi hà :( 

\(\lim\dfrac{3^n+2.6^n}{6^{n-1}+5.4^n}=\lim\dfrac{6^n\left[\left(\dfrac{3}{6}\right)^n+2\right]}{6^n\left[\dfrac{1}{6}+5\left(\dfrac{4}{6}\right)^n\right]}=\lim\dfrac{\left(\dfrac{3}{6}\right)^n+2}{\dfrac{1}{6}+5\left(\dfrac{4}{6}\right)^n}=\dfrac{0+2}{\dfrac{1}{6}+0}=12\)

\(\lim\left(\sqrt{n^2+9}-n\right)=\lim\dfrac{\left(\sqrt{n^2+9}-n\right)\left(\sqrt{n^2+9}+n\right)}{\sqrt{n^2+9}+n}=\lim\dfrac{9}{\sqrt{n^2+9}+n}\)

\(=\lim\dfrac{n\left(\dfrac{9}{n}\right)}{n\left(\sqrt{1+\dfrac{9}{n^2}}+1\right)}=\lim\dfrac{\dfrac{9}{n}}{\sqrt{1+\dfrac{9}{n^2}}+1}=\dfrac{0}{1+1}=0\)

\(\lim\dfrac{\sqrt{15+9n^2}-3}{5-n}=\lim\dfrac{n\sqrt{\dfrac{15}{n^2}+9}-3}{5-n}=\lim\dfrac{n\left(\sqrt{\dfrac{15}{n^2}+9}-\dfrac{3}{n}\right)}{n\left(\dfrac{5}{n}-1\right)}\)

\(=\lim\dfrac{\sqrt{\dfrac{15}{n^2}+9}-\dfrac{3}{n}}{\dfrac{5}{n}-1}=\dfrac{\sqrt{9}-0}{0-1}=-3\)

\(Cl_2+H_2\rightarrow2HCl\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(ZnCl_2+Ba\left(OH\right)_2\rightarrow Zn\left(OH\right)_2+BaCl_2\)

\(Zn\left(OH\right)_2+2HCl\rightarrow ZnCl_2+2H_2O\)

Câu hỏi của Solyver - Toán lớp 7 - Học toán với OnlineMath

theo mình thì câu trên: dưới mẫu trong căn bỏ n^2 ra làm nhân tử chung xong đặt nhân tử chung của cả mẫu là n^2 . câu dưới thì mình k biết!!

\(\lim\dfrac{-3n+2}{n-\sqrt{4n+n^2}}=\lim\dfrac{\left(-3n+2\right)\left(n+\sqrt{4n+n^2}\right)}{\left(n-\sqrt{4n+n^2}\right)\left(n+\sqrt{4n+n^2}\right)}\)

\(=\lim\dfrac{\left(-3n+2\right)\left(n+\sqrt{4n+n^2}\right)}{-4n}=\lim\dfrac{n\left(-3+\dfrac{2}{n}\right)n\left(1+\sqrt{\dfrac{4}{n}+1}\right)}{-4n}\)

\(=\lim n\dfrac{\left(-3+\dfrac{2}{n}\right)\left(1+\sqrt{\dfrac{4}{n}+1}\right)}{-4}\)

Do \(\lim\left(n\right)=+\infty\)

\(\lim\dfrac{\left(-3+\dfrac{2}{n}\right)\left(1+\sqrt{\dfrac{4}{n}+1}\right)}{-4}=\dfrac{\left(-3+0\right)\left(1+\sqrt{0+1}\right)}{-4}=\dfrac{3}{2}>0\)

\(\Rightarrow\lim n\dfrac{\left(-3+\dfrac{2}{n}\right)\left(1+\sqrt{\dfrac{4}{n}+1}\right)}{-4}=+\infty\)

\(\lim\left(\sqrt[3]{n^3+9n^2}-n\right)=\lim\dfrac{\left(\sqrt[3]{n^3+9n^2}-n\right)\left(\sqrt[3]{\left(n^3+9n^2\right)^2}+n\sqrt[3]{n^3+9n^2}+n^2\right)}{\sqrt[3]{\left(n^3+9n^2\right)}+n\sqrt[3]{n^3+9n^2}+n^2}\)

\(=\lim\dfrac{9n^2}{\sqrt[3]{\left(n^3+9n^2\right)^2}+n\sqrt[3]{n^3+9n^2}+n^2}\)

\(=\lim\dfrac{9n^2}{n^2\sqrt[3]{\left(1+\dfrac{9}{n}\right)^2}+n^2\sqrt[3]{1+\dfrac{9}{n}}+n^2}\)

\(=\lim\dfrac{9}{\sqrt[3]{\left(1+\dfrac{9}{n}\right)^2}+\sqrt[3]{1+\dfrac{9}{n}}+1}\)

\(=\dfrac{9}{\sqrt[3]{\left(1+0\right)^2}+\sqrt[3]{1+0}+1}=\dfrac{9}{3}=3\)

1.A

2.C

3.D

4.D

5.B

6.B

Bạn tham khảo nhé:3

3. a) A = {90 ; 180 ; 270 ; 360 ; 450}

Tính tổng : (90 + 450) . 5 : 2 = 1350

b) B = {-99 ; -98 ; -97 ; -96}

Tính tổng : (-99 - 96) . 4 : 2 = -390