Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
- 1/4+17/50 +31/100=25/100+34/100+31/100 =90/100=0,9
- 3/4 +15/40+69/300=0,75+0,375+0,23=1,355
- 42/700+45/120+17/80=0,06+0,375+0,2125 = 0,6475
k cho mk vs
A = \(\dfrac{13}{50}\) + 9% + 41/100 + 0,24
A = 0,26 + 0,09 + 0,41 + 0,24
A = (0,26 + 0,24) + ( 0,09 + 0,41)
A = 0,5 + 0,5
A = 1
=26/100+9%+41%+24%
=26%+9%+41%+24%
=26%+24%+9%+41%=50%+50%=100%=1
26 % + 9 % + 41 % + 24 %
= ( 26 % + 24% ) + ( 41 % + 9 % )
=50 % + 50 %
=100 % =1
\(\frac{13}{50}+9\%+\frac{41}{100}+0,24\)
\(=\frac{13}{50}+\frac{9}{100}+\frac{41}{100}+\frac{24}{100}\)
\(=\frac{13}{50}+\left[\frac{9}{100}+\frac{41}{100}+\frac{24}{100}\right]\)
\(=\frac{13}{50}+\frac{74}{100}=\frac{13}{50}+\frac{37}{50}=1\)
đễ
=26%+9%+41%+24%
=(26%+24%)+(9%+41%)
=50%+50%
=100%
= 13/50 + 9/100 + 41/100 + 24/100
= 13/50 + ( 9/100 + 41/100 + 24/100 )
= 13/50 + 74/100
= 26/100 + 74/100
= 100/100 = 1
13/50 + 9% + 41/100 + 0,24
\(=\frac{26}{100}+\frac{9}{100}+\frac{41}{100}+\frac{24}{100}\)
\(=\frac{26+9+41+24}{100}\)
\(=\frac{\left(26+24\right)+\left(9+41\right)}{100}\)
\(=\frac{50+50}{100}\)
\(=\frac{100}{100}\)
\(=1\)
OK CHÚC BẠN HỌC TỐT !!!
THANK YOU TRƯỚC NHÉ !
13/50 + 9% + 41/100 + 0,24
= 0, 26 + 0, 09 + 0, 41 + 0, 24
= (0, 26 + 0, 24) + ( 0, 09 + 0, 41)
=0, 5 +0, 5
= 1
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)
\(\left(a\right)360-357+354-351+348-345+...+312-309+306-303+300-297\)
\(=\left(360-357\right)+\left(354-351\right)+\left(348-345\right)+...+\left(312-309\right)+\left(306-303\right)+\)\(\left(300-297\right)\)
\(=3+3+3+3+3+3+3+3+3+3+3=33\)
\(\left(b\right)2006-1-2-3-4-...-47-48-49-50\)
\(=2006-\left(1+2+3+4+...+47+48+49+50\right)\)
\(=2006-\frac{\left(50+1\right)\left[\left(50-1\right)+1\right]}{2}\)
\(=2006-1275=731\)
\(\left(c\right)280-276+272-268+264-260+...+216-212+208-204+200-196\)
\(=\left(280-276\right)+\left(272-268\right)+\left(264-260\right)+...+\left(216-212\right)+\left(208-204\right)+\)\(\left(200-196\right)\)
\(=4+4+4+4+4+4+4+4+4+4+4=44\)
\(\frac{13}{50}+74\%+\frac{41}{100}+0,59\)
\(=\frac{26}{100}+\frac{74}{100}+\frac{41}{100}+\frac{59}{100}\)
\(=\frac{\left(26+74\right)+\left(41+59\right)}{100}\)
\(=2\)
\(\frac{14}{5}+\frac{9}{13}+\frac{17}{13}-\frac{8}{9}+\frac{17}{9}-\frac{4}{5}\)
\(=\left(\frac{14}{5}-\frac{4}{5}\right)+\left(\frac{9}{13}+\frac{17}{13}\right)+\left(\frac{17}{9}-\frac{8}{9}\right)\)
\(=2+2+1\)
\(=5\)
\(\frac{14}{5}+\frac{9}{13}+\frac{17}{13}-\frac{8}{9}+\frac{17}{9}-\frac{4}{5}\)
\(=\left(\frac{14}{5}-\frac{4}{5}\right)+\left(\frac{9}{13}+\frac{17}{13}\right)-\left(\frac{8}{9}-\frac{17}{9}\right)\)
\(=\frac{10}{5}+\frac{26}{13}-\left(-1\right)\)
\(=2+2+1\)
\(=5\)
0,34+0,13+0,37+0,16
=(0,34+0,16)+(0,37+0,13)
=0,5+0,5=1
=0,34+0,13+0,37+0,16
=(0,34+0,16)+(0,37+0,13)
=0,5+0,5
=1