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1)
\(\left(a\right)37+397+3997+39997\)
\(=40-3+400-3+4000-3+40000-3\)
\(=\left(40+400+4000+40000\right)-\left(3+3+3+3\right)\)
\(=44440-12=44428\)
\(\left(b\right)298+2998+29998+299998\)
\(=300-2+3000-2+30000-2+300000-2\)
\(=\left(300+3000+30000+300000\right)-\left(2+2+2+2\right)\)
\(=333300-8=333296\)
\(\left(c\right)9+99+999+9999+99999\)
\(=10-1+100-1+1000-1+10000-1+100000-1\)
\(=\left(10+100+1000+10000+100000\right)-\left(1+1+1+1+1\right)\)
\(=111110-5=111105\)
2)
\(\left(a\right)\left(2+4+6+...+2002+2004+2006\right)-\left(1+3+5+...+2001+2003+2005\right)\)
\(=\left(2-1\right)+\left(4-3\right)+\left(6-5\right)+...+\left(2002-2001\right)+\left(2004-2003\right)+\left(2006-2005\right)\)
\(=1+1+1+...+1+1+1\)( 1003 số 1 )
\(=1003\)
\(\left(b\right)88-87+86-85+84-83+...+6-5+4-3+2-1\)
\(=\left(88-87\right)+\left(86-85\right)+\left(84-83\right)+...+\left(6-5\right)+\left(4-3\right)+\left(2-1\right)\)
\(=1+1+1+...+1+1+1\)( 44 số 1 )
\(=44\)
\(\left(c\right)100-98+96-94+92-90+...+12-10+8-6+4-2\)
\(=\left(100-98\right)+\left(96-94\right)+\left(92-90\right)+...+\left(12-10\right)+\left(8-6\right)+\left(4-2\right)\)
\(=2+2+2+...+2+2+2\) ( 25 số 2 )
\(=50\)
3)
\(\left(a\right)360-357+354-351+348-345+...+312-309+306-303+300-297\)
\(=\left(360-357\right)+\left(354-351\right)+\left(348-345\right)+...+\left(312-309\right)+\left(306-303\right)+\)\(\left(300-297\right)\)
\(=3+3+3+3+3+3+3+3+3+3+3=33\)
\(\left(b\right)2006-1-2-3-4-...-47-48-49-50\)
\(=2006-\left(1+2+3+4+...+47+48+49+50\right)\)
\(=2006-\frac{\left(50+1\right)\left[\left(50-1\right)+1\right]}{2}\)
\(=2006-1275=731\)
\(\left(c\right)280-276+272-268+264-260+...+216-212+208-204+200-196\)
\(=\left(280-276\right)+\left(272-268\right)+\left(264-260\right)+...+\left(216-212\right)+\left(208-204\right)+\)\(\left(200-196\right)\)
\(=4+4+4+4+4+4+4+4+4+4+4=44\)
a) \(\dfrac{2}{3}+\dfrac{3}{4}< x< 1\dfrac{1}{3}+\dfrac{4}{5}\)
\(\dfrac{2\times4}{3\times4}+\dfrac{3\times3}{4\times3}< x< \dfrac{\left(1\times3+1\right)\times5}{3\times5}+\dfrac{4\times3}{5\times3}\)
\(\dfrac{8}{12}+\dfrac{9}{12}< x< \dfrac{20}{15}+\dfrac{12}{15}\\ \dfrac{17}{12}< x< \dfrac{32}{15}\)
Ước tính: \(\dfrac{17}{12}=1,4\) và \(\dfrac{32}{15}=2,1\). Vậy số tự nhiên x = 2 sẽ thõa mãn 1,4 < x < 2,1
b)
\(\dfrac{5}{6}-\dfrac{1}{4}< x< 2\dfrac{1}{3}-\dfrac{2}{5}\\ \dfrac{5\times4}{6\times4}-\dfrac{1\times6}{4\times6}< x< \dfrac{\left(2\times3+1\right)\times5}{3\times5}-\dfrac{2\times3}{5\times3}\\ \dfrac{20}{24}-\dfrac{6}{24}< x< \dfrac{35}{15}-\dfrac{6}{15}\\ \dfrac{14}{24}< x< \dfrac{29}{15}\)
Ước tính \(\dfrac{14}{24}=0,5\) và \(\dfrac{29}{15}=1,9\)
Vậy với x là số tự nhiên x = 1 sẽ thõa mãn 0,5 < x < 1,9
a) \(M=\frac{2\times2}{1\times5}+\frac{2\times2}{5\times9}+\frac{2\times2}{9\times13}+...+\frac{2\times2}{45\times40}\)
\(M=\frac{4}{1\times5}+\frac{4}{5\times9}+\frac{4}{9\times13}+...+\frac{4}{45\times49}\)
\(M=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{45}-\frac{1}{49}\)
\(M=1-\frac{1}{49}\)
\(M=\frac{48}{49}\)
b) \(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+5+...+10}\)
= \(\frac{2}{2\times\left(1+2\right)}+\frac{2}{2\times\left(1+2+3\right)}+...+\frac{2}{2\times\left(1+2+3+...+10\right)}\)
\(=\frac{2}{6}+\frac{2}{12}+...+\frac{2}{110}\)
\(=\frac{2}{2\times3}+\frac{2}{3\times4}+...+\frac{2}{10\times11}\)
\(=2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(=2\times\left(\frac{1}{2}-\frac{1}{11}\right)\)
\(=2\times\frac{9}{22}\)
\(=\frac{9}{11}\)
Mình trả lời câu a nha M= 4/1*5+4/5*9+4/9*13+...+4/45*49 M=1-1/5+1/5-1/9+1/9-1/13+...+1/45-1/49 M=1-1/49=48/49
a) mk chỉnh đề
\(A=\left(1+\frac{1}{2005}\right)\left(1+\frac{1}{2006}\right)\left(1+\frac{1}{2019}\right)\)
\(=\frac{2006}{2005}.\frac{2007}{2006}.....\frac{2020}{2019}\)
\(=\frac{2020}{2005}\)
\(=\frac{404}{401}\)
\(B=\frac{3}{1}+\frac{3}{1+2}+\frac{3}{1+2+3}+....+\frac{3}{1+2+3+...+100}\)
\(=3+3\left(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+100}\right)\)
\(=3+3.\left(\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+....+\frac{1}{\frac{100.101}{2}}\right)\)
\(=3+3.\left(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{100.101}\right)\)
\(=3+6\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(=3+6\left(\frac{1}{2}-\frac{1}{101}\right)=3+6.\frac{99}{202}\)
\(=3+2\frac{95}{101}=5\frac{95}{101}\)
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