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A = -1 + -2 + -3 + -4 + ... + -99 + -100
= - ( 1 + 2 +3 + ... + 100)
= - 5050
\(...\\ A=-\left(1+2+3+...+100\right)\\ A=-\left(\frac{\left(1+100\right).100}{2}\right)\\ A=-101.50=-5050\)
Chúc bạn học tốt!!!
A = 21+2+3+...+10
1 +2 + 3 + ...+ 10 = (1+ 10).10 : 2 = 55
=>A = 255
2 đồng dư với -1 mod 3 => 255 đồng dư với (-1)55 = - 1 ( mod 3)
=> A chia cho 3 dư -1
A không chia hết cho 3
b./ \(\Leftrightarrow\frac{x+1}{2009}+1+\frac{x+2}{2008}+1+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1.\)
\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)(b)
Mà \(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}< 0\)
(b) \(\Leftrightarrow x+2010=0\Leftrightarrow x=-2010\)
a./
\(\Leftrightarrow\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}-\frac{x+1}{5}-\frac{x+1}{6}=0.\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)=0\)(a)
Mà \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}>0\)
(a) \(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
\(\frac{11.3^{2.2}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\)
\(=\frac{11.3^4.3^7-\left(3^2\right)^{15}}{2^2.3^{28}}\)
\(=\frac{11.3^{11}-3^{30}}{2^2.3^{28}}\)
\(=\frac{3^{11}.\left(11-3^{19}\right)}{2^2.3^{28}}\)
\(=\frac{11-3^{19}}{2^2.3^{17}}\)
Bạn Châu làm đúng rồi !
Tuy nhiên bạn My lại đề bài nhé!
Tham khảo đề và bài làm:
\(\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}=\frac{11.3^{22+7}-\left(3^2\right)^{15}}{2^2.3^{14.2}}\)
\(=\frac{11.3^{29}-3^{30}}{2^2.3^{28}}=\frac{3^{29}\left(11-3\right)}{2^2.3^{28}}=\frac{3.8}{2^2}=6\)
a) \(\frac{2^{10}\left(2+3\right)}{\left(2^2\right)^5.5.2}=\frac{2^{10}.5}{2^{10}.5.2}=\frac{1}{2}\); b) \(=\frac{\left(3^2\right)^4.2-3^6}{3^6.34.3}=\frac{3^6\left(2.3^2-1\right)}{3^6.34.3}=\frac{3^6.17}{3^6.17.2.3}=\frac{1}{6}\)
Bài 5:Giải:
Ta có: \(\left\{{}\begin{matrix}a+3c=2016\left(1\right)\\a+2b=2017\left(2\right)\end{matrix}\right.\)
Từ \(\left(1\right)\Leftrightarrow a=2016-3c\)
Lấy \(\left(2\right)-\left(1\right)\) ta được:
\(2b-3c=1\Leftrightarrow b=\dfrac{1+3c}{2}\)
Khi đó:
\(P=a+b+c=\left(2016-3c\right)+\dfrac{1+3c}{2}\) \(+\) \(c\)
\(=\left(2016+\dfrac{1}{2}\right)+\dfrac{-6c+3c+2c}{2}\)
\(=2016\dfrac{1}{2}-\dfrac{c}{2}\) Vì \(a,b,c\ge0\) nên:
\(P=2016\dfrac{1}{2}-\dfrac{c}{2}\le2016\dfrac{1}{2}\)
Vậy \(P_{max}=2016\dfrac{1}{2}\Leftrightarrow c=0\)