Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có :
\(P=\frac{\frac{6}{8}+\frac{6}{10}+\frac{6}{14}+\frac{6}{26}}{\frac{11}{4}+\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}\)
\(\Rightarrow P=\frac{\frac{3}{4}+\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{11\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\)
\(\Rightarrow P=\frac{3\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{11\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\)
\(\Rightarrow P=\frac{3}{11}\)
Vậy \(P=\frac{3}{11}\)
\(P=\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}=\frac{3}{11}\)
đề bài của bn sai nên mk sửa luôn nha
\(A=\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{26}{7}-\frac{13}{11}+\frac{13}{23}}-\frac{\frac{3}{5}+\frac{3}{13}-0,9}{\frac{7}{91}+0,2-\frac{3}{10}}\)
\(A=\frac{155-5\left(\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}{403-13\left(\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}-\frac{\frac{3}{5}+\frac{3}{13}-\frac{9}{10}}{\frac{7}{91}+\frac{2}{10}-\frac{3}{10}}\)
\(A=\frac{155-5}{403-13}-\frac{3\left(\frac{1}{5}+\frac{1}{13}\right)-\frac{9}{10}}{\frac{7}{91}+\left(-\frac{1}{10}\right)}\)
\(A=\frac{5}{13}-\frac{\left(-\frac{9}{130}\right)}{\left(-\frac{3}{130}\right)}=\frac{5}{13}-\frac{\frac{9}{130}}{\frac{3}{130}}\)
\(A=\frac{5}{13}-\frac{9}{130}\cdot\frac{130}{3}\)
\(A=\frac{5}{13}-3=-\frac{34}{13}\)
\(B=\frac{30\cdot4^7\cdot3^{29}-5\cdot14^5\cdot2^{12}}{54\cdot6^{14}\cdot9^7-12\cdot8^5\cdot7^5}\)
\(B=\frac{30\cdot\left(2^2\right)^7\cdot3^{29}-5\cdot\left(2\cdot7\right)^5\cdot2^{12}}{54\cdot\left(2\cdot3\right)^{14}\cdot\left(3^2\right)^7-12\cdot\left(2^3\right)^5\cdot7^5}\)
\(B=\frac{30\cdot2^{14}\cdot3^{29}-5\cdot2^5\cdot7^5\cdot2^{12}}{54\cdot2^{14}\cdot3^{14}\cdot3^{14}-12\cdot2^{15}\cdot7^5}\)
\(B=\frac{30\cdot3^{29}-5\cdot2^{17}\cdot7^5}{54\cdot3^{28}-12\cdot2^{15}\cdot7^5}=\frac{30\cdot3-5\cdot2^2}{54-12}=\frac{5}{3}\)
a) \(\frac{x+1}{3}=\frac{x-2}{4}\)
=> (x+1).4 = (x - 2) . 3
=> 4x + 4 = 3x - 6
=> 4x - 3x = - 6 - 4
=> x = - 10
b) \(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)
\(\Rightarrow\left(\frac{x-6}{7}+1\right)+\left(\frac{x-7}{8}+1\right)+\left(\frac{x-8}{9}+1\right)=\left(\frac{x-9}{10}+1\right)+\left(\frac{x-10}{11}+1\right)+\left(\frac{x-11}{12}+1\right)\)
\(\Rightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}=\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}\)
\(\Rightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}-\frac{x+1}{12}\) = 0
\(\Rightarrow\left(x+1\right).\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)\)
Vì \(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\ne0\) nên x + 1 =0
=> x = -1
c) Xem lại đề
\(B=\frac{x}{y}+\frac{\frac{1}{4}+\frac{5}{8}-\frac{7}{13}}{-\frac{2}{12}-\frac{10}{24}+\frac{14}{39}}\)
\(B=\frac{x}{y}+\frac{\frac{1}{4}+\frac{5}{8}-\frac{7}{13}}{-\left(\frac{2}{12}+\frac{10}{24}-\frac{14}{39}\right)}\)
\(B=\frac{x}{y}+\frac{\frac{1}{4}+\frac{5}{8}-\frac{7}{13}}{-\frac{2}{3}\left(\frac{1}{4}+\frac{5}{8}-\frac{7}{13}\right)}\)
\(B=\frac{x}{y}+\frac{1}{-\frac{2}{3}}\)
\(B=\frac{x}{y}-\frac{3}{2}\)
Thế x = 0, 5 = 1/2 ; y = 3 ta được :
\(B=\frac{\frac{1}{2}}{3}-\frac{3}{2}=\frac{1}{6}-\frac{9}{6}=-\frac{8}{6}=-\frac{4}{3}\)
Ta có:\(B=\frac{x}{y}+\frac{\frac{1}{4}+\frac{5}{8}-\frac{7}{13}}{\frac{-2}{12}-\frac{10}{24}+\frac{14}{39}}\)
\(B=\frac{x}{y}+\frac{\frac{1}{4}+\frac{5}{8}-\frac{7}{13}}{-\left(\frac{2}{12}+\frac{10}{24}-\frac{14}{39}\right)}\)
\(B=\frac{x}{y}+\frac{\frac{1}{4}+\frac{5}{8}-\frac{7}{13}}{-\frac{2}{3}\left(\frac{1}{4}+\frac{5}{8}-\frac{7}{13}\right)}\)
\(B=\frac{x}{y}+\frac{1}{-\frac{2}{3}}\)(Do\(\frac{1}{4}+\frac{5}{8}-\frac{7}{13}\ne0\))
\(B=\frac{x}{y}-\frac{3}{2}\)
Thay x = 0,5; y = 3 vào B ta được:
\(B=\frac{0,5}{3}-\frac{3}{2}\)
\(B=\frac{1}{6}-\frac{3}{2}\)
\(B=\frac{1}{6}-\frac{9}{6}\)
\(B=-\frac{4}{3}\)
Vậy\(B=-\frac{4}{3}\)tại x = 0,5; y = 3
Linz
=1356,545272