Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1 :
\(\left(\frac{1}{8}+\frac{1}{8.15}+\frac{1}{15.22}+...+\frac{1}{43.50}\right)\frac{4-3-5-7-...-49}{217}\)
\(=\frac{1}{7}\left(1-\frac{1}{8}+\frac{1}{8}-\frac{1}{15}+\frac{1}{15}-\frac{1}{22}+...+\frac{1}{43}-\frac{1}{50}\right).\frac{5-\left(1+3+5+7+...+49\right)}{217}\)
\(=\frac{1}{7}\left(1-\frac{1}{50}\right).\frac{5-\left(12.50\right)+25}{217}\)
\(=\frac{1}{7}.\frac{49}{50}.\frac{5-625}{217}\)
\(=\frac{-2}{5}\)
Bài 2 :
\(B=\frac{x^2+17}{x^2+7}=\frac{\left(x^2+7\right)+10}{x^2+7}=1+\frac{10}{x^2+7}\)
Ta có : \(x^2\ge0\). Dấu '' = '' xảy ra khi :
\(x=0\Rightarrow x^2+7\ge7\)( 2 vế dương )
\(\Rightarrow\frac{10}{x^2+7}\le\frac{10}{7}\)
\(\Rightarrow1+\frac{10}{x^2+7}\le1+\frac{10}{7}\)
\(\Rightarrow B\le\frac{17}{7}\)
Dấu '' = '' xảy ra < = > x = 0
Vậy Max \(B=\frac{17}{7}\Leftrightarrow x=0\)
\(\left(\frac{-1}{4}+\frac{7}{33}-\frac{5}{3}\right)-\left(\frac{-5}{4}+\frac{6}{11}-\frac{48}{49}\right)=\left(\frac{-1}{4}-\frac{16}{11}\right)-\left(-\frac{31}{44}-\frac{48}{49}\right)=-\frac{1}{4}-\frac{16}{11}+\frac{31}{44}+\frac{48}{49}=-\frac{1}{49}\)
ta có
1/5(5/36+5/126+...+5/44*49)1-3-5-7-9-...-49/89
=1/5(1/4-1/9+1/9-1/14+...+1/44-1/49)-623/89
=1/5*-7(1/4-1/49)
=-7/5*45/196
=-9/128
\(\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+....+\frac{1}{44.49}\right)\cdot\frac{1-3-5-7-....-49}{89}\)
\(\text{Đặt }:\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+...+\frac{1}{44.49}\right)\)là \(A\)
\(\frac{1-3-5-7-...-49}{89}\)là \(B\);ta có :
\(A=\frac{1}{5}\cdot\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}\right)\)
\(A=\frac{1}{5}\cdot\left(\frac{1}{4}-\frac{1}{49}\right)=\frac{1}{5}\cdot\frac{45}{196}=\frac{9}{196}\)
\(B=\frac{1-3-5-7-....-49}{89}=\frac{1-\left(3+5+7+...+49\right)}{89}\)
Tổng của \(3+5+7+...+49\)là:
\(\frac{\left(3+49\right).24}{2}=624\)
\(\Rightarrow\frac{1-624}{89}=\frac{-623}{89}=-7\)
\(\Rightarrow\left(\frac{1}{4.9}+\frac{1}{9.14}+...+\frac{1}{44.49}\right)\cdot\frac{1-3-5-7-...-49}{89}=A.B=\frac{9}{196}\cdot-7=-\frac{9}{28}\)
mk ko viết lại đề đâu
=\(\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}\right)\)\(.\frac{1-\left(3+5+...+49\right)}{89}\)
=\(\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right).\frac{\left(1-\frac{\left(49+3\right).24}{2}\right)}{89}\)
=\(\frac{1}{5}.\frac{45}{196}.\frac{1-\left(\frac{52.24}{2}\right)}{89}\)
=\(\frac{9}{196}.\left(1-\frac{624}{89}\right)=\frac{9}{196}.\left(\frac{-623}{89}\right)\)
=\(\frac{-9}{28}\)
\(\left(\frac{-1}{4}+\frac{7}{33}-\frac{5}{3}\right)\) \(-\left(-\frac{15}{12}+\frac{6}{11}-\frac{48}{49}\right)\)
\(=\left(\frac{-1}{4}+\frac{7}{33}-\frac{55}{33}\right)\)\(+\frac{15}{12}-\frac{6}{11}+\frac{48}{49}\)
\(=\left(\frac{-1}{4}-\frac{48}{33}\right)\)\(+\frac{8085}{6468}\)\(-\)\(\frac{3528}{6468}\)\(+\frac{6603}{6468}\)
\(=\frac{-75}{44}\)\(+\frac{930}{539}\)\(=\frac{45}{2156}\)
Bước đến nhà em bóng xế tà
Đứng chờ năm phút bố em ra
Lơ thơ phía trước vài con chó
Lác đác đằng sau chiếc chổi chà
Sợ quá anh chuồn quên đôi dép
Bố nàng ngoác mỏ đứng chửi cha
Phen này nhất quyết thuê cây kiếm
Trở về chém ổng đứt làm ba
a) \(4\frac{5}{9}:\left(-\frac{5}{7}\right)+\frac{49}{9}:\left(-\frac{5}{7}\right)=\frac{41}{9}:\left(-\frac{5}{7}\right)+\frac{49}{9}:\left(-\frac{5}{7}\right)\)
\(=\frac{41}{9}\cdot\left(-\frac{7}{5}\right)+\frac{49}{9}\cdot\left(-\frac{7}{5}\right)=\left(\frac{41}{9}+\frac{49}{9}\right)\cdot\left(-\frac{7}{5}\right)=10\cdot\left(-\frac{7}{5}\right)=-14\)
b) \(\left(\frac{-3}{5}+\frac{4}{9}\right):\frac{7}{11}+\left(\frac{-2}{5}+\frac{5}{9}\right):\frac{7}{11}\)
\(=\left(\frac{-3}{5}+\frac{4}{9}+\frac{-2}{5}+\frac{5}{9}\right):\frac{7}{11}\)
\(=\left(\frac{-3}{5}+\frac{-2}{5}+\frac{4}{9}+\frac{5}{9}\right):\frac{7}{11}\)
\(=\left(-1+1\right):\frac{7}{11}=0\cdot\frac{11}{7}=0\)
c) \(\left(\frac{3}{4}\right)^4\cdot\left(\frac{8}{9}\right)^2=\left(\frac{3}{4}\right)^2\cdot\left(\frac{3}{4}\right)^2\cdot\left(\frac{8}{9}\right)^2=\left(\frac{3}{4}\cdot\frac{3}{4}\cdot\frac{8}{9}\right)^2\)
\(=\left(\frac{1}{2}\right)^2=\frac{1}{4}\)
d) \(\left(-\frac{3}{5}\right)^6\cdot\left(-\frac{5}{3}\right)^5=\left(-\frac{3}{5}\right)^5\cdot\left(-\frac{3}{5}\right)\cdot\left(-\frac{5}{3}\right)^5=\left[\left(-\frac{3}{5}\right)\cdot\left(-\frac{5}{3}\right)\right]^5\cdot\left(-\frac{3}{5}\right)\)
\(=1^5\cdot\left(-\frac{3}{5}\right)=1\cdot\left(-\frac{3}{5}\right)=-\frac{3}{5}\)
e) \(\frac{8^{14}}{4^4\cdot64^5}=\frac{\left(2^3\right)^{14}}{\left(2^2\right)^4\cdot\left(2^6\right)^5}=\frac{2^{42}}{2^8\cdot2^{30}}=\frac{2^{42}}{2^{38}}=2^4=16\)
f) \(\frac{9^{10}\cdot27^7}{81^7\cdot3^{15}}=\frac{\left(3^2\right)^{10}\cdot\left(3^3\right)^7}{\left(3^4\right)^7\cdot3^{15}}=\frac{3^{20}\cdot3^{21}}{3^{28}\cdot3^{15}}=\frac{3^{41}}{3^{43}}=3^{-2}=\frac{1}{3^2}=\frac{1}{9}\)