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N=1/2+1/22+...+1/210
2N=1+1/2+...+1/29
2N-N=1-1/210=1-1/1024=1023/1024
Giải:
N=1/2+1/22+1/23+...+1/29+1/210
2N=1+1/2+1/22+...+1/28+1/29
2N-N=(1+1/2+1/22+...+1/28+1/29)-(1/2+1/22+1/23+...+1/29+1/210)
N=1-1/210=1023/1024
Chúc bạn học tốt!
\(C=5\frac{9}{10}:\frac{3}{2}-\left(2\frac{1}{3}.4\frac{1}{2}-2.2\frac{1}{3}\right):\frac{7}{4}\)
\(=\frac{59}{10}:\frac{3}{2}-\left(\frac{7}{3}.\frac{9}{2}-2.\frac{7}{3}\right):\frac{7}{4}\)
\(=\frac{59}{15}-\left[\frac{7}{3}\left(\frac{9}{2}-2\right)\right]:\frac{7}{4}\)
\(=\frac{59}{15}-\frac{35}{6}:\frac{7}{4}\)
\(=\frac{59}{15}-\frac{10}{3}\)
\(=\frac{9}{15}=\frac{3}{5}\)
\(\cdot62,87+35,14+4,13+8,35+4,86+5,65\)
\(=\left(62,87+4,13\right)+\left(35,14+4,86\right)+\left(8,35+5,65\right)\)
\(=67+40+14\)
\(=121\)
\(A=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{9}{10!}\)
\(A=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{10-1}{10!}\)
\(A=\frac{2}{2!}-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+...+\frac{10}{10!}-\frac{1}{10!}\)
\(A=\frac{1}{1!}-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}+...+\frac{1}{9!}-\frac{1}{10!}\)
\(A=1-\frac{1}{10!}\)
\(\Rightarrow A< 1\left(đpcm\right)\)
Ta có:
\(\frac{A}{B}=\frac{\frac{2000}{1}+\frac{1999}{2}+\frac{1998}{3}+...+\frac{1}{2000}+2000}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}\)
\(\Leftrightarrow\frac{A}{B}=\frac{\left(\frac{2000}{1}+1\right)+\left(\frac{1999}{2}+1\right)+\left(\frac{1998}{3}+1\right)+...+\left(\frac{1}{2000}+1\right)+2000+1}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}\)
\(\Leftrightarrow\frac{A}{B}=\frac{\frac{2001}{1}+\frac{2001}{2}+\frac{2001}{3}+...+\frac{2001}{2000}+2001}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}\)
\(\Leftrightarrow\frac{A}{B}=\frac{2001\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}\right)}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}\)
\(\Leftrightarrow\frac{A}{B}=2001\)
bn cộng trên tử rồi thì phải trừ đi chứ ko phân số sẽ thay đổi
A=[(1-22)/22][(1-32)/32]...[(1-20152)/20152]
A=[(1+2)(1-2)/22][(1-3)(1+3)/32]...[(1-2015)(1+2015)/20152]
=[(-1).3/2.2][(-2).4/3.3]...[-2014.2016/2015.2015]
=[(-1)(-2)(-3)...(-2013)(-2014).3.4.5...2015]/(2.2.3.3.4.4....2015.2015)
=[2(-3)...(-2014)]/(2.2.3.4.5....2015)
=(-3)(-4)...(-2014)/2.3.4.5....2015
=[-(3.4.5.6....2014)]/(2.3.4...2015)
=-1/1.2015=-1/2015
a) \(\frac{x-1}{6}=\frac{2x+3}{7}\)
\(\Leftrightarrow7\left(x-1\right)=6\left(2x+3\right)\)
\(\Leftrightarrow7x-7=12x+18\)
\(\Leftrightarrow5x+18=-7\)
\(\Leftrightarrow5x=-25\)
\(\Leftrightarrow x=-5\)
b) \(\left(2x^2-\frac{1}{2}x\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow x\left(2x-\frac{1}{2}\right)\left(x^2+1\right)=0\)
Vì \(x^2+1>0\)nên \(\orbr{\begin{cases}x=0\\2x-\frac{1}{2}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{4}\end{cases}}\)
\(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{10^2}\right)\)
= \(\dfrac{2^2-1}{2^2}.\dfrac{3^2-1}{3^2}.\dfrac{4^2-1}{4^2}...\dfrac{10^2-1}{10^2}\)
= \(\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}...\dfrac{9.11}{10^2}\)
= \(\dfrac{\left(1.2.3...9\right).\left(3.4.5...11\right)}{\left(2.3.4...10\right)\left(2.3.4...10\right)}\)
= \(\dfrac{1.11}{10.10}=\dfrac{11}{100}\)
quá hay