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=-1/2.3-1/3.4-1/4.5-1/5.6
=-(1/2.3+1/3.4+1/4.5+1/5.6)
=-(1/2-1/3+1/3-1/4+1/5-1/6)
=-(1/2-1/6)
=1/6-1/2=-1/3
1.1+3.1+5.1+....+99.1
= 1+3+5+...+99
= (99+1) .[(99-1):2+1] : 2
= 2500
mik nha !!
a, \(A=-\dfrac{1}{20}-\left(\dfrac{1}{20\cdot19}+\dfrac{1}{19\cdot18}+...+\dfrac{1}{2\cdot1}\right)\\ \Rightarrow A=-\dfrac{1}{20}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)\\ \Rightarrow A=-\dfrac{1}{20}-1+\dfrac{1}{20}=-1\)
b, \(B=\dfrac{1}{99}-\dfrac{1}{99\cdot97}-\dfrac{1}{97\cdot95}-...-\dfrac{1}{3\cdot1}\\ \Rightarrow B=\dfrac{1}{99}-\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{97\cdot99}\right)\\ \Rightarrow B=\dfrac{1}{99}-\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\\ \Rightarrow B=\dfrac{1}{99}-\dfrac{1}{2}+\dfrac{1}{2\cdot99}=-\dfrac{16}{33}\)
\(A=-\dfrac{1}{2.3}-\dfrac{1}{3.4}-\dfrac{1}{4.5}-...-\dfrac{1}{9.10}\)
\(\Rightarrow-A=\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+\dfrac{5-4}{4.5}+...+\dfrac{10-9}{9.10}=\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}=\)
\(=\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{2}{5}\Rightarrow A=-\dfrac{2}{5}\)
Giải
Tìm x:
a)\(\left(x-2\right)^2=1\Leftrightarrow\left(x-2\right)^2=1^2.\)
\(\Rightarrow\orbr{\begin{cases}x-2=1\Rightarrow x=1+2=3\\x-2=-1\Rightarrow x=-1+2=1\end{cases}}\)
=> Vậy \(x=\orbr{\begin{cases}3\\1\end{cases}}\)
b) \(\left(2x-1\right)^3=-8\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Rightarrow\left(2x-1\right)=-2\Rightarrow2x=-2+1=-1\)
\(\Rightarrow x=-1:2=-\frac{1}{2}\)
Vậy \(x=-\frac{1}{2}\)
c) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\Leftrightarrow\left(x+\frac{1}{2}\right)^2=\orbr{\begin{cases}\left(-\frac{1}{4}\right)^2\\\left(\frac{1}{4}\right)^2\end{cases}}\)
\(\Rightarrow\left(x+\frac{1}{2}\right)=\orbr{\begin{cases}-\frac{1}{4}\\\frac{1}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}x+\frac{1}{2}=-\frac{1}{4}\Rightarrow x=-\frac{1}{4}-\frac{1}{2}=-\frac{3}{4}\\x+\frac{1}{2}=\frac{1}{4}\Rightarrow x=\frac{1}{4}-\frac{1}{2}=-\frac{1}{4}\end{cases}}\)
Vậy \(x=-\frac{3}{4};-\frac{1}{4}\)
BT2:
Giải
a) \(9.3^3.\frac{1}{81}.3^2=3^2.3^3.\left(\frac{1}{3}\right)^4.3^2=\left(3^2.3^3.3^2\right).\left(\frac{1}{3}\right)^4\)
\(=3^{2+3+2}.\left(\frac{1}{3}\right)^4=3^7.\left(\frac{1}{3}\right)^4=\frac{3^7.1^4}{1.3^4}=3^3\)
b) \(4.2^5:\left(2^3.\frac{1}{16}\right)=2^2.2^5:\left(2^3.\left(\frac{1}{2}\right)^4\right)=2^{2+5}:\left(\frac{2^3.1^4}{2^4}\right)\)
\(=2^7:\left(\frac{1}{2}\right)=2^7.\frac{2}{1}=2^8\)
c) Chị đang nghĩ...