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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)
\(\left(a\right)360-357+354-351+348-345+...+312-309+306-303+300-297\)
\(=\left(360-357\right)+\left(354-351\right)+\left(348-345\right)+...+\left(312-309\right)+\left(306-303\right)+\)\(\left(300-297\right)\)
\(=3+3+3+3+3+3+3+3+3+3+3=33\)
\(\left(b\right)2006-1-2-3-4-...-47-48-49-50\)
\(=2006-\left(1+2+3+4+...+47+48+49+50\right)\)
\(=2006-\frac{\left(50+1\right)\left[\left(50-1\right)+1\right]}{2}\)
\(=2006-1275=731\)
\(\left(c\right)280-276+272-268+264-260+...+216-212+208-204+200-196\)
\(=\left(280-276\right)+\left(272-268\right)+\left(264-260\right)+...+\left(216-212\right)+\left(208-204\right)+\)\(\left(200-196\right)\)
\(=4+4+4+4+4+4+4+4+4+4+4=44\)
(dấu . là dấu nhân )
a, \(\frac{3}{2}\cdot\frac{4}{7}+\frac{3}{7}\cdot\frac{3}{2}\)
= \(\frac{3}{2}\left(\frac{4}{7}+\frac{3}{7}\right)\)
=\(\frac{3}{2}\cdot\frac{7}{7}=\frac{3}{2}\)
b, \(\frac{12}{5}\cdot4-4\cdot\frac{7}{5}\)
=\(4\left(\frac{12}{5}-\frac{7}{5}\right)=4\cdot\frac{5}{5}=4\)
c, \(\frac{5}{11}:\frac{1}{2}+\frac{6}{11}:\frac{1}{2}\)
=\(2\left(\frac{5}{11}+\frac{6}{11}\right)=2\cdot\frac{11}{11}=2\)
a;3/2x(4/7+3/7)
=3/2x1
=3/2
b;12/5x4-4x7/5
=4x(12/5-7/5)
=4x1
=4
c;5/11:1/2+6/11:1/2
=1/2:(5/11+6/11)
=1/2:1
=1/2
\(1\frac{1}{2}+1\frac{1}{3}=1\frac{3}{6}+1\frac{2}{6}\)\(=2\frac{5}{6}\)
\(\frac{5}{6}:\frac{3}{8}.\frac{3}{4}=\frac{5}{6}.\frac{8}{3}.\frac{3}{4}\)\(=\frac{5.8.3}{6.3.4}=\frac{120}{72}=\frac{3}{5}\)
\(\text{Học tốt!!}\)
a, Ta có:
\(\left(5\frac{3}{7}+2\frac{4}{9}\right)-\left(1\frac{4}{9}+3\frac{3}{7}\right)\)\(=5\frac{3}{7}+2\frac{4}{9}-1\frac{4}{9}-3\frac{3}{7}\)\(=5\frac{3}{7}-3\frac{3}{7}+2\frac{4}{9}-1\frac{4}{9}\)
\(=2+1\)\(=3\)
b, Bạn làm tương tự nhé!!! bạn phá ngoặc ra rồi nhóm các số như mình làm ở trên nha!!!
a) \(\left(5\frac{3}{7}+2\frac{4}{9}\right)-\left(1\frac{4}{9}+\frac{3}{7}\right)\)
\(=5\frac{3}{7}+2\frac{4}{9}-1\frac{4}{9}-\frac{3}{7}\)
\(=\left(5\frac{3}{7}-\frac{3}{7}\right)+\left(1\frac{4}{9}-2\frac{4}{9}\right)\)
\(=5+\left(-1\right)\)
\(=4\)
b)\(\left(3\frac{4}{5}+5\frac{3}{4}\right)-\left(2\frac{3}{4}+1\frac{4}{5}\right)\)
\(=3\frac{4}{5}+5\frac{3}{4}-2\frac{3}{4}-1\frac{4}{5}\)
\(=\left(3\frac{4}{5}-1\frac{4}{5}\right)+\left(5\frac{3}{4}-2\frac{3}{4}\right)\)
\(=2+3=5\)