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\(B=1+5+5^2+5^3+...+5^{2008}+5^{2009}\)
\(\Rightarrow 5B=5+5^2+5^3+5^4+...+5^{2009}+5^{2010}\)
Trừ theo vế:
\(5B-B=(5+5^2+5^3+5^4+...+5^{2009}+5^{2010})-(1+5+5^2+...+5^{2009})\)
\(4B=5^{2010}-1\)
\(B=\frac{5^{2010}-1}{4}\)
\(S=\frac{3^0+1}{2}+\frac{3^1+1}{2}+\frac{3^2+1}{2}+..+\frac{3^{n-1}+1}{2}\)
\(=\frac{3^0+3^1+3^2+...+3^{n-1}}{2}+\frac{\underbrace{1+1+...+1}_{n}}{2}\)
\(=\frac{3^0+3^1+3^2+..+3^{n-1}}{2}+\frac{n}{2}\)
Đặt \(X=3^0+3^1+3^2+..+3^{n-1}\)
\(\Rightarrow 3X=3^1+3^2+3^3+...+3^{n}\)
Trừ theo vế:
\(3X-X=3^n-3^0=3^n-1\)
\(\Rightarrow X=\frac{3^n-1}{2}\). Do đó \(S=\frac{3^n-1}{4}+\frac{n}{2}\)
1.Thực hiện phép tính:
a, \(\frac{1}{3}.(\frac{-4}{5}).\left(\frac{-1}{2}\right)^2:\frac{1}{2}\)
=\(\frac{1}{3}.\frac{-4}{5}.(\frac{-1}{2}).2\)
=\(\frac{1}{3}.\left(\frac{-4}{5}\right).\left(\frac{-1}{2}\right).2\)
=\(\frac{1}{3}.\frac{-4}{5}.\left(-1\right)\)
=\(\frac{1}{3}.\frac{4}{5}=\frac{4}{15}\)
b,\(\left(1+\frac{1}{3}-\frac{2}{5}\right)-\left(2-\frac{2}{3}-\frac{3}{5}\right)+\left(3-\frac{4}{3}-\frac{1}{5}\right)\)
=\(1+\frac{1}{3}-\frac{2}{5}-2+\frac{2}{3}+\frac{3}{5}+3-\frac{4}{3}-\frac{1}{5}\)
=(1+2+3)+\(\left(\frac{1}{3}+\frac{2}{3}-\frac{4}{3}\right)-\left(\frac{2}{5}-\frac{3}{5}+\frac{1}{5}\right)\)
=2+\(\frac{-1}{3}-0\)
=\(\frac{5}{3}-0=\frac{5}{3}\)
2.Tìm x, biết:
a,\(\frac{1}{x}.\frac{-2}{7}=\frac{3}{8}\)
\(\frac{1}{x}=\frac{3}{8}:\frac{-2}{7}\)
\(\frac{1}{x}=\frac{-21}{16}\)\(\Rightarrow x=\frac{-21}{16}\)
b,\(\left|x-\frac{1}{4}\right|=\frac{2}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{1}{4}=\frac{-2}{3}\\x-\frac{1}{4}=\frac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-5}{12}\\x=\frac{11}{12}\end{matrix}\right.\)
Vậy:\(x\in\left\{\frac{-5}{12};\frac{11}{12}\right\}\)
3,Tìm a,b\(\in Z\)
a,-19<a<\(\frac{-20}{3}\)
\(\Rightarrow\frac{-57}{3}< \frac{-3a}{3}< \frac{-20}{3}\)
\(\Rightarrow\) -57<-3a<-20
\(\Rightarrow\) a\(\in\)\(\left\{-7;-8;...;-16\right\}\)
b,\(\frac{1}{3}< \frac{4}{b}< \frac{1}{2}\)
Ta có:\(\frac{4}{12}< \frac{4}{b}< \frac{4}{8}\)
\(\Rightarrow12< b< 8\)
Vì quy đồng tử thì mẫu nào lớn hơn thì số đó bé hơn nên:
\(b\in\left\{9;10;11\right\}\)
CHÚC BẠN CÓ MỘT BÀI LÀM THẬT TỐT NHÉ!!!
Bn ơi câu a) bài 3 là -19/2<a<-20/3
Mk ghi sai đề bn làm lại mk vs ạ camơn
a) Ta có: \(A=\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right)...\left(\frac{1}{10}-1\right)=\frac{-1}{2}.\frac{-2}{3}...\frac{-9}{10}=\frac{-\left(1.2.3...9\right)}{2.3.4...10}=-\frac{1}{10}\)
b) Ta có : \(B=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)....\left(\frac{1}{100}-1\right)=\frac{-3}{4}.\frac{-8}{9}....\frac{-99}{100}=-\frac{3.8....99}{\left(2.3...10\right)\left(2.3...10\right)}\)
\(=-\frac{1.3.2.4...9.11}{\left(2.3....10\right)\left(2.3...10\right)}=\frac{\left(1.2.3...10\right).\left(3.4..10.11\right)}{\left(2.3...10\right).\left(2.3.4...10\right)}=\frac{11}{2}=5,5\)
c) Ta có : \(C=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{n+1}\right)=\frac{1}{2}.\frac{2}{3}...\frac{n}{n+1}=\frac{1.2...n}{2.3...\left(n+1\right)}=\frac{1}{n+1}\)