Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A = (1- 2) \(\times\) ( 4 - 3) \(\times\) (5 - 6) \(\times\) (8 - 7) \(\times\) (9 - 10) \(\times\) (12 - 11) \(\times\)(13 - 14)
A = (-1) \(\times\) 1 \(\times\) (-1) \(\times\) 1 \(\times\) (-1) \(\times\) 1 \(\times\) (-1)
A = 1
a: \(=\left(\dfrac{-48}{12}+\dfrac{-8}{12}+\dfrac{21}{12}\right)\cdot\dfrac{-12}{13}\)
\(=\dfrac{-35}{12}\cdot\dfrac{-12}{13}=\dfrac{35}{13}\)
b: \(=\dfrac{-3}{6}+\dfrac{5}{6}-\dfrac{312}{100}+\dfrac{51}{10}\)
\(=\dfrac{1}{3}-\dfrac{312}{100}+\dfrac{51}{10}=\dfrac{347}{150}\)
c: \(=\left(\dfrac{48}{300}+\dfrac{175}{300}-\dfrac{135}{100}\right)\cdot\dfrac{5}{2}+\dfrac{1}{4}\)
\(=\dfrac{88}{300}\cdot\dfrac{5}{2}+\dfrac{1}{4}=\dfrac{59}{60}\)
Ta có:3/5+3/7-3/11=3.(1/5+1/7-1/11)
4/5+4/7-4/11=4.(1/5+1/7-1/11)
=>M=[3.(1/5+1/7-1/11)]/[4.(1/5+1/7-1/11)]=3/4
M = \(\frac{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{11}}=\frac{3\left(\frac{1}{5}+\frac{3}{7}-\frac{3}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}=\frac{3}{4}\)
\(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-3^{11}\cdot2^{11}}=\dfrac{2^{12}\cdot3^{10}\cdot6}{2^{11}\cdot3^{11}\cdot5}=\dfrac{2}{3}\cdot\dfrac{6}{5}=\dfrac{4}{5}\)
\(Q=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
\(Q=\left(\frac{1+12+123}{999}\right)\left(\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right)\)
\(Q=\left(\frac{136}{999}\right)\left(\frac{0}{6}\right)\)
\(Q=0\)
\(y=ax^2+bx-7\)đi qua điểm \(A\left(-1,-6\right)\)nên \(a-b-7=-6\Leftrightarrow a-b=1\)(1)
\(y=ax^2+bx-7\)có trục đối xứng \(x=-\frac{1}{3}\)nên \(\frac{-b}{2a}=-\frac{1}{3}\Leftrightarrow2a-3b=0\)(2)
Từ (1) và (2) suy ra \(\hept{\begin{cases}a=3\\b=2\end{cases}}\)
\(a^2-b^2=3^2-2^2=5\).
\(B=\left[-\dfrac{1}{6}+\dfrac{5}{12}\right]+\dfrac{7}{12}\)
\(=-\dfrac{1}{6}+\dfrac{5}{12}+\dfrac{7}{12}\)
\(=-\dfrac{1}{6}+1=\dfrac{5}{6}\)
B=[−61+125]+127
\(= - \frac{1}{6} + \frac{5}{12} + \frac{7}{12}\)
\(= - \frac{1}{6} + 1 = \frac{5}{6}\)