Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Đặt \(A=\frac{7^{15}}{1+7+7^2+...+7^{14}}\)
Đặt \(B=1+7+7^2+...+7^{14}\)
\(\Rightarrow7B=7+7^2+...+7^{15}\)
\(\Rightarrow7B-B=6B=7^{15}-1\)
\(\Rightarrow B=\frac{7^{15}-1}{6}\)
\(\Rightarrow A=\frac{7^{15}-1+1}{\frac{7^{15}-1}{6}}=\left(7^{15}-1\right).\frac{6}{7^{15}-1}+\frac{6}{7^{15}-1}=6+\frac{6}{7^{15}-1}\)
Tự làm tiếp nha
b, Ta có:\(\dfrac{1+3+3^2+.....+3^{10}}{1+3+3^2+.....+3^9}\) \(=\dfrac{1}{1+3+3^2+...+3^9}+\dfrac{3+3^2+...+3^{10}}{1+3+3^2+...+3^9}\)\(=\dfrac{1}{1+3+3^2+...+3^9}+\dfrac{3.\left(1+3+3^2+...+3^9\right)}{1+3+3^2+...+3^9}\)
\(=\dfrac{1}{1+3+3^2+...+3^9}+3< 4\)
\(\Rightarrow\) \(\dfrac{1+3+3^2+...+3^{10}}{1+3+3^2+...+3^9}< 4\) \(\left(1\right)\)
Ta có :\(\dfrac{1+5+5^2+...+5^{10}}{1+5+5^2+...+5^9}\)
\(=\dfrac{1}{1+5+5^2+...+5^9}+\dfrac{5+5^2+...+5^{10}}{1+5+5^2+....+5^9}\)
\(=\dfrac{1}{1+5+5^2+...+5^9}+\dfrac{5.\left(1+5+5^2+...+5^9\right)}{1+5+5^2+...+5^9}\)
\(=\dfrac{1}{1+5+5^2+...+5^9}+5>5\)
\(\Rightarrow\) \(\dfrac{1+5+5^2+...+5^{10}}{1+5+5^2+...+5^9}>5\) \(\left(2\right)\)
Từ \(\left(1\right)và\left(2\right)\)
\(\Rightarrow\dfrac{1+3+3^2+...+3^{10}}{1+3+3^2+...+3^9}< \dfrac{1+5+5^2+...+5^{10}}{1+5+5^2+...+5^9}\)
Vậy \(\dfrac{1+3+3^2+...+3^{10}}{1+3+3^2+...+3^9}< \dfrac{1+5+5^2+...+5^{10}}{1+5+5^2+...+5^9}\)
a, Đặt \(A\)\(=\dfrac{7^{15}}{1+7+7^2+...+7^{14}}\)
\(\Rightarrow\) \(\dfrac{1}{A}\) \(=\dfrac{1+7+7^2+...+7^{14}}{7^{15}}=\dfrac{1}{7^{15}}+\dfrac{7}{7^{15}}+\dfrac{7^2}{7^{15}}+...+\dfrac{7^{14}}{7^{15}}\)
\(=\dfrac{1}{7^{15}}+\dfrac{1}{7^{14}}+\dfrac{1}{7^{13}}+....+\dfrac{1}{7}\)
Đặt \(B=\dfrac{9^{15}}{1+9+9^2+...+9^{14}}\)
\(\Rightarrow\dfrac{1}{B}=\dfrac{1+9+9^2+...+9^{14}}{9^{15}}=\dfrac{1}{9^{15}}+\dfrac{9}{9^{15}}+\dfrac{9^2}{9^{15}}+...+\dfrac{9^{14}}{9^{15}}\)
\(=\dfrac{1}{9^{15}}+\dfrac{1}{9^{14}}+\dfrac{1}{9^{13}}+...+\dfrac{1}{9}\)
Mà \(\dfrac{1}{7^{15}}>\dfrac{1}{9^{15}};\dfrac{1}{7^{14}}>\dfrac{1}{9^{14}};\dfrac{1}{7^{13}}>\dfrac{1}{9^{13}};....;\dfrac{1}{7}>\dfrac{1}{9}\)
\(\Rightarrow\dfrac{1}{A}>\dfrac{1}{B}\) \(\Rightarrow A< B\)
Vậy\(\dfrac{7^{15}}{1+7+7^2+...+7^{14}}>\dfrac{9^{15}}{1+9+9^2+....+9^{14}}\)
Bài 1:
a: -8/12<0<-3/-4
b: -56/24<0<7/3
c: 4/25<1<15/13
=>-4/25>-15/13
Bài 2:
a: =-60/45=-4/3
b: =4/15-3/2-8/5=8/30-45/30-48/30=-85/30=-17/6
\(M=\frac{\left(-7\right).15.9.15.14}{9.49.7.15}=\frac{-15.2}{7}=\frac{-30}{7}.\)
\(N=\frac{200}{189}+\frac{1}{14}=\)1.12962962963
\(M=\left(\frac{-7}{9}\cdot\frac{9}{7}\right)\cdot\left(\frac{15}{49}\cdot\frac{14}{15}\right)\cdot15\)
\(M=\left(-1\right)\cdot\frac{2}{7}\cdot15\)
\(M=\frac{-30}{7}\)
\(N=\frac{5}{9}\cdot\frac{4}{7}\cdot\frac{10}{3}+\frac{3}{9}\cdot\frac{3}{7}\cdot\frac{1}{2}\)
\(N=\frac{200\cdot2}{189\cdot2}+\frac{9\cdot3}{126\cdot3}\)
\(N=\frac{400}{378}+\frac{27}{378}\)
\(N=\frac{61}{51}\)
T i ck nha
\(a,\frac{7}{12}\cdot\frac{6}{11}+\frac{7}{12}\cdot\frac{5}{11}+2\frac{7}{12}\)
\(=\frac{7}{12}\cdot\left(\frac{6}{11}+\frac{5}{11}\right)+2\frac{7}{12}\)
\(=\frac{7}{12}+\frac{31}{12}\)
\(=\frac{38}{12}=\frac{19}{6}\)
\(b,\frac{-5}{9}\cdot\frac{-6}{13}+\frac{5}{-9}\cdot\frac{-5}{13}-\frac{5}{9}\)
\(=\frac{-5}{9}\cdot\frac{-6}{13}+\frac{-5}{9}\cdot\frac{-5}{13}+\frac{-5}{9}\cdot1\)
\(=\frac{-5}{9}\cdot\left(\frac{-6}{13}+\frac{-5}{13}+1\right)\)
\(=\frac{-5}{9}\cdot\left(\frac{-11}{13}+1\right)\)
\(=\frac{-5}{9}\cdot\frac{2}{13}\)
\(=\frac{-10}{117}\)
\(c,\)\(0,8\cdot\frac{-15}{14}-\frac{4}{5}\cdot\frac{13}{14}-1\frac{2}{5}\)
\(=\frac{4}{5}\cdot\frac{-15}{14}-\frac{4}{5}\cdot\frac{13}{14}-\frac{7}{5}\)
\(=\frac{4}{5}\cdot\left(\frac{-15}{14}-\frac{13}{14}\right)-\frac{7}{5}\)
\(=\frac{4}{5}\cdot\left(-2\right)-\frac{7}{5}\)
\(=\frac{-8}{5}-\frac{7}{5}\)
\(=-3\)
\(d,\)\(75\%\cdot\frac{6}{7}+5\%\cdot\frac{6}{7}+\frac{7}{10}\cdot1\frac{1}{7}\)
\(=\frac{3}{4}\cdot\frac{6}{7}+\frac{1}{20}\cdot\frac{6}{7}+\frac{7}{10}\cdot\frac{8}{7}\)
\(=\left(\frac{3}{4}+\frac{1}{20}\right)\cdot\frac{6}{7}+\frac{7}{10}\cdot\frac{8}{7}\)
\(=\frac{4}{5}\cdot\frac{6}{7}+\frac{4}{5}\cdot1\)
\(=\frac{4}{5}\cdot\left(\frac{6}{7}+1\right)\)
\(=\frac{4}{5}\cdot\frac{13}{7}\)
\(=\frac{52}{35}\)
a)7/12.6/11+7/12.5/11-2.7/12
=7/12(6/11+5/11-2)
=7/12(1-2)
=7/12.(-1)
=-7/12
a) Ta có: \(12 = 2^2 . 3; 15 = 3.5\)
\(BCNN(12, 15) = 2^2.3.5 = 60\) nên chọn mẫu chung là 60.
\(\begin{array}{l}\frac{9}{{12}} = \frac{{9.5}}{{12.5}} = \frac{{45}}{{60}}\\\frac{7}{{15}} = \frac{{7.4}}{{15.4}} = \frac{{28}}{{60}}\end{array}\)
b) Ta có: \(10 = 2.5; 4 = 2^2; 14=2.7\)
\(BCNN(10, 4, 14) =2^2.5.7= 140\) nên chọn mẫu chung là 140.
\(\begin{array}{l}\frac{7}{{10}} = \frac{{7.14}}{{10.14}} = \frac{{98}}{{140}}\\\frac{3}{4} = \frac{{3.35}}{{4.35}} = \frac{{105}}{{140}}\\\frac{9}{{14}} = \frac{{9.10}}{{14.10}} = \frac{{90}}{{140}}\end{array}\)
