tính tổng 1 cách hợp lí
-1-2-3-4-...-2009-2010
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Lời giải:
\(B=\frac{4.\frac{17}{4}}{\frac{119}{36}.\frac{16}{7}}=\frac{17}{\frac{68}{9}}=17.\frac{9}{68}=\frac{9}{4}\)
a)2.3x+1=10.312+8.312
=>2.3x+1=312.18
=>3x+1=312.9
=>3x+1=312.32
=>3x+1=314
=>x+1=14
=>x=13
b)2x+2x+3=144
=>2x+2x.23=144
=>2x(1+8)=144
=>2x.9=144
=>2x=16
=>2x=24
=>x=4
\(PA=2PB\Rightarrow\dfrac{PA}{AB}=\dfrac{2}{3};QA=2QC\Rightarrow\dfrac{QA}{AC}=\dfrac{2}{3}\)
Hai tg APK và tg ABK có chung đường cao từ K->AB nên
\(\dfrac{S_{APK}}{S_{ABK}}=\dfrac{PA}{AB}=\dfrac{2}{3}\Rightarrow S_{APK}=\dfrac{2}{3}xS_{ABK}\)
Hai tg AQK và tg ACK có chung đường cao từ K->AC nên
\(\dfrac{S_{AQK}}{S_{ACK}}=\dfrac{QA}{AC}=\dfrac{2}{3}\Rightarrow S_{AQK}=\dfrac{2}{3}xS_{ACK}\)
\(\Rightarrow S_{APKQ}=S_{APK}+S_{AQK}=\dfrac{2}{3}x\left(S_{ABK}+S_{ACK}\right)=\dfrac{2}{3}xS_{ABC}=\dfrac{2}{3}x360=240cm^2\)
Ta có:
\(S_{APM}=\dfrac{2}{3}\times S_{ABM}\) (chung đường cao hạ từ \(M\), \(AP=\dfrac{2}{3}\times AB\))
\(S_{APK}=\dfrac{1}{2}\times S_{APM}\) (chung đường cao hạ từ \(P\), \(AK=\dfrac{1}{2}\times AM\))
\(=\dfrac{1}{2}\times\dfrac{2}{3}\times S_{ABM}=\dfrac{1}{3}\times S_{ABM}\)
Tương tự \(S_{AQK}=\dfrac{1}{3}\times S_{ACM}\)
suy ra \(S_{APKQ}=S_{APK}+S_{AQK}=\dfrac{1}{3}\times S_{ABM}+\dfrac{1}{3}\times S_{ACM}\)
\(=\dfrac{1}{3}\times\left(S_{ABM}+S_{ACM}\right)=\dfrac{1}{3}\times S_{ABC}=\dfrac{1}{3}\times360=120\left(cm^2\right)\)
\(\left(5x-4\right)^{10}=\left(5x-4\right)^7\)
\(\left(5x-4\right)^{10}-\left(5x-4\right)^7=0\)
\(\left(5x-4\right)^7\left\{\left(5x-4\right)^3-1\right\}=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(5x-4\right)^7=0\\\left(5x-4\right)^3-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x-4=0\\\left(5x-4\right)^3=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\5x-4=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=1\end{matrix}\right.\)
`4.(x-3)-7.7=1`
`4.(x-3)-49=1`
`4.(x-3)=1+49=50`
`x-3=50:4=25/2`
`x=25/2+3=31/2`
_____________________________________________________
`9.(x+4)-25=5.4`
`9.(x+4)-25=20`
`9.(x+4)=20+25=45`
`x+4=45:9=5`
`x=5-4=1`
\(4.\left(x-3\right)-7.7=1\)
\(4.\left(x-3\right)-49=1\)
\(4.\left(x-3\right)=1+49\)
\(4.\left(x-3\right)=50\)
\(\left(x-3\right)=50\div4\)
\(x-3=12,5\)
\(x=12,5+3\)
\(x=15,5\)
\(9.\left(x+4\right)-25=5.4\)
\(9.\left(x+4\right)-25=20\)
\(9.\left(x+4\right)=20+25\)
\(9.\left(x+4\right)=45\)
\(\left(x+4\right)=45\div9\)
\(x+4=5\)
\(x=5-4\)
\(x=1\)
A = \(\dfrac{1}{9}\) + \(\dfrac{1}{18}\) + \(\dfrac{1}{30}\) + ......+ \(\dfrac{1}{135}\)
A = \(\dfrac{1}{3}\) ( \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) +........+ \(\dfrac{1}{45}\))
A = \(\dfrac{2}{3}\) ( \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\) +.....+\(\dfrac{1}{90}\))
A = \(\dfrac{2}{3}\) ( \(\dfrac{1}{2\times3}\) + \(\dfrac{1}{3\times4}\) + \(\dfrac{1}{4\times5}\)+......+\(\dfrac{1}{9\times10}\))
A = \(\dfrac{2}{3}\)( \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)+ \(\dfrac{1}{3}\)- \(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\)- \(\dfrac{1}{5}\)+........+ \(\dfrac{1}{9}\)- \(\dfrac{1}{10}\))
A = \(\dfrac{2}{3}\)( \(\dfrac{1}{2}\)- \(\dfrac{1}{10}\))
A = \(\dfrac{2}{3}\) x( \(\dfrac{5}{10}\) - \(\dfrac{1}{10}\))
A = \(\dfrac{2}{3}\)x \(\dfrac{4}{10}\)
A = \(\dfrac{4}{15}\)
Giảm đường kính đi 20% thì bán kính cũng giảm đi 20%
Bán kính của hình tròn mới là 100% – 20%= 80%
Diện tích hình tròn có bán kính 80% là:80%x80% = 64%
Diên tích hình tròn cũ hơn hình tròn mới là: 100%x100%-64%= 36%
36%=113,04cm2 => diện tích hình tròn ban đầu là 113,04: 36x 100 = 314cm2.
Chúc e học tốt nha!☘
A = -1 -2 - 3 -4 -......-2010
A = - (1+2+3+4+......2010)
A = - (2010 +1)[ (2010 -1):1 +1] : 2
A = - 2011.2010:2
A = -2021055