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hé hé bạn mik ớ ngân giới tính rất linh hoạt
P/s : đầu óc bạn thì ko đc linh hoạt bởi tên ngân còn hỏi là trai hay gái
\(-\dfrac{628628}{942942}=-\dfrac{2.314314}{3.314314}=-\dfrac{2}{3}\)
>> Mình không chép lại đề bài nhé ! <<
Cách 1 :
\(A=\left(\dfrac{36-4+3}{6}\right)-\left(\dfrac{30+10-9}{6}\right)-\left(\dfrac{18-14+15}{6}\right)=\dfrac{35}{6}-\dfrac{31}{6}-\dfrac{19}{6}=-\dfrac{15}{6}=-\dfrac{5}{2}\)
Cách 2 :
\(A=6-\dfrac{2}{3}+\dfrac{1}{2}-5+\dfrac{5}{3}-\dfrac{3}{2}-3-\dfrac{7}{3}+\dfrac{5}{2}\)
\(A=\left(6-5-3\right)-\left(\dfrac{2}{3}+\dfrac{5}{3}-\dfrac{7}{3}\right)+\left(\dfrac{1}{2}+\dfrac{3}{2}-\dfrac{5}{2}\right)\)
\(A=-2-0-\dfrac{1}{2}=-\dfrac{5}{2}\)
Cách 1 :
\(\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)
\(=\left(\dfrac{36}{6}-\dfrac{4}{6}+\dfrac{3}{6}\right)-\left(\dfrac{30}{6}+\dfrac{10}{6}-\dfrac{9}{6}\right)-\left(\dfrac{18}{6}-\dfrac{14}{6}+\dfrac{15}{6}\right)\)
\(=\dfrac{35}{6}-\dfrac{31}{6}-\dfrac{19}{6}\)
\(=-\dfrac{5}{2}\)
Cách 2 :
\(\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)
\(=6-\dfrac{2}{3}+\dfrac{1}{2}-5-\dfrac{5}{3}+\dfrac{3}{2}-3+\dfrac{7}{3}-\dfrac{5}{2}\)
\(=\left(6-5-3\right)+\left(\dfrac{-2}{3}+\dfrac{-5}{3}+\dfrac{7}{3}\right)+\left(\dfrac{1}{2}+\dfrac{3}{2}+\dfrac{-5}{2}\right)\)
\(=\left(-2\right)+0+\dfrac{-1}{2}\)
\(=\dfrac{-5}{2}\)
Ta có :
\(S=1.2+2.3+...+49.50\)
\(\Leftrightarrow3S=1.2.\left(3-0\right)+2.3.\left(4-1\right)+...+49.50.\left(51-48\right)\)
\(\Leftrightarrow3S=1.2.3-0.1.2+2.3.4-1.2.3+...+49.50.51-48.49.50\)
\(\Leftrightarrow3S=49.50.51\)
\(\Leftrightarrow S=\frac{49.50.51}{3}=41650\)
S=1 . 2 + 2.3+3.4+.....+49.100
3S=1.2.3+2.3.3+3.4.3+....+49.50.3
3S=1.2.3+2.3.(4-1)+3.4(5-2)+....+49.50(51-48)
3S=1.2.3-2.3.4+2.3.4-2.3.1+......+48.49.50+49.50.51
3S=49.50.51
S=49.50.51 / 3
S=41650
Ta có : \(\left\{\begin{matrix}Q=-\left(x-7\right)^2-6\\-\left(x-7\right)^2\le0\\-6=-6\end{matrix}\right.\)
\(\Rightarrow Q=-\left(x-7\right)^2-6\le0-6=-6\)
Vậy GTLN của \(Q=-\left(x-7\right)^2-6\) là \(-6\)
\(\left(\dfrac{-5}{13}\right)^{2017}\cdot\left(\dfrac{13}{5}\right)^{2016}=\left(\dfrac{-5}{13}\right)\cdot\left(-\dfrac{5}{13}\right)^{2016}\cdot\left(\dfrac{13}{5}\right)^{2016}=\left(\dfrac{-5}{13}\right)\cdot\left(\dfrac{5}{13}\right)^{2016}\cdot\left(\dfrac{13}{5}\right)^{2016}=\left(-\dfrac{5}{13}\right)\cdot\left[\left(\dfrac{5}{13}\right)^{2016}\cdot\left(\dfrac{13}{5}\right)^{2016}\right]=\left(-\dfrac{5}{13}\right)\cdot1^{2016}=\left(-\dfrac{5}{13}\right)\cdot1=-\dfrac{5}{13}\)
\(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{81}\)
<=> \(\left\{{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{9}\\x+\dfrac{1}{2}=-\dfrac{1}{9}\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=-\dfrac{7}{18}\\x=-\dfrac{11}{18}\end{matrix}\right.\)
\(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{81}\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{9}\\x+\dfrac{1}{2}=-\dfrac{1}{9}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{7}{18}\\x=-\dfrac{11}{18}\end{matrix}\right.\)
Vậy \(x_1=-\dfrac{7}{18};x_2=-\dfrac{11}{18}\).