A = \(\frac{3}{12}\)+\(\frac{7}{23}\)+\(\frac{13}{34}\)+\(\frac{21}{45}\)+\(\frac{21}{45}\)+\(\frac{31}{56}\)+\(\frac{43}{67}\)
GIÚP MK NHA MỌI NGƯỜI
Tính bằng cách hợp lý nhất:
a, \(\frac{31}{23}-\left(\frac{7}{32}+\frac{8}{23}\right)\)
b, \(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
c, \(\frac{38}{45}-\left(\frac{8}{45}-\frac{17}{51}-\frac{3}{11}\right)\)
Các bạn ơi giúp mk với!*****
a) 31/23 - ( 7/32 + 8/22)
= 31/23 - 7/32 + 8/23
= ( 31/23 + 8/23 ) - 7/32
= 32/22 - 7/32
= 39/32
Ccá ý khác làm tương tự
=(31\23-8\23)+7\32
=23\23+7\32
=1+7\32
=39\32
\(\frac{\frac{3}{67}\left(17\frac{21}{56}-13\frac{21}{45}\right):\left(\frac{3}{5.22}+\frac{54}{44.65}+\frac{18}{65.72}\right)}{\left(29^3:100-29^3:0,47\right)}\)
tính :\(\frac{3}{2}+\frac{7}{6}+\frac{13}{12}+\frac{21}{20}+\frac{31}{30}+\frac{43}{42}+\frac{57}{56}+\frac{73}{72}+\frac{91}{90}\)
= \(\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{6}\right)+\left(1+\frac{1}{12}\right)+....+\left(1+\frac{1}{90}\right)\)
= \(\left(1+1+1+....+1\right)+\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{90}\right)\)(9 số 1)
= 9 + \(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{9.10}\right)\)
= \(9+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)
= \(9+\left(1-\frac{1}{10}\right)=9+\frac{9}{10}=\frac{90}{10}+\frac{9}{10}=\frac{99}{10}\)
3/2+7/6+13/12+21/20+31/30+43/42+57/56+73/72+91/90=99/10=9,9
Thực hiện tính giải bằng cách hợp lý:
\(A=\frac{\frac{3}{67}.\left(17\frac{21}{56}-13\frac{21}{45}\right):\left(\frac{3}{5.22}+\frac{54}{44.65}+\frac{18}{65.72}\right)}{29^3:100-29^3:0,47}\)
Tìm x : \(2x+\frac{7}{6}+\frac{13}{12}+\frac{21}{20}+\frac{31}{30}+\frac{43}{42}+\frac{57}{56}+\frac{73}{72}+\frac{91}{90}=10\)10
\(2x+\frac{7}{6}+\frac{13}{12}+\frac{21}{20}+\frac{31}{30}+\frac{43}{42}+\frac{57}{56}+\frac{73}{72}+\frac{91}{90}=10\)
=> \(2x+\frac{6+1}{6}+\frac{12+1}{12}+....+\frac{90+1}{90}=10\)
=> \(2x+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}+10=10\)
=> \(2x+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}=0\)
=>\(2x+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}=0\)
=>\(2x+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}=0\)
=> \(2x-\frac{1}{10}=0\)
=>2x=\(\frac{1}{10}\)=> x=1/20
mình có bị nhầm chỗ dấu suy ra thứ 3. đáng lẽ ra biểu thức đó cộng 8 chứ k phải cộng 10 do mình sơ ý nên bạn hãy sủa lại chỗ ấy
thục hiện dãy tính :
\(\frac{\frac{3}{67}.\left(17\frac{21}{56}-13\frac{21}{45}\right):\left(\frac{3}{5.22}+\frac{54}{44.65}+\frac{18}{65.72}\right)}{29^3:100-29^3:0,47}\)
help me
\(\frac{\frac{3}{67}.\left(17\frac{21}{56}-13\frac{21}{45}\right):\left(\frac{3}{5.22}+\frac{54}{44.65}+\frac{18}{65.72}\right)}{29^3:100-29^3:0,47}\)
\(=\frac{\frac{3}{67}\left(\frac{139}{8}-\frac{202}{15}\right):\left(\frac{3}{110}+\frac{54}{2860}+\frac{18}{4680}\right)}{29^3.\frac{1}{100}-29^3.\frac{47}{100}}\)
\(=\frac{\frac{3}{67}.\frac{469}{120}:\frac{1}{20}}{29^3\left(\frac{1}{100}-\frac{47}{100}\right)}\)
\(=\frac{\frac{7}{40}.20}{29^3.\left(-\frac{23}{50}\right)}\).
\(=\frac{\frac{7}{2}}{-11218,94}\)
\(=-\frac{175}{560947}\)
arigato gozaimasu
bài 20 tính bằng phương pháp hợp lí nhất
a \(\frac{31}{23}-\left(\frac{7}{32}+\frac{8}{23}\right)\)
b \(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
c\(\frac{38}{45}-\left(\frac{8}{45}-\frac{17}{51}-\frac{3}{11}\right)\)
\(\frac{3}{2}+\frac{7}{6}+\frac{13}{12}+\frac{21}{20}+\frac{31}{30}+\frac{43}{42}+\frac{57}{56}+\frac{73}{72}+\frac{91}{90}\)
ai lam di mk tick cho
chi tiết cách giải
\(\frac{3}{2}\)+ \(\frac{7}{6}\)+\(\frac{13}{12}\)+\(\frac{21}{20}\)+\(\frac{31}{30}\)+ \(\frac{43}{42}\)+\(\frac{57}{56}\)+\(\frac{73}{72}\)+\(\frac{91}{90}\)
=\(\frac{3780+2940+2730+2646+2604+2580+2565+2555+2548}{2520}\)
= \(\frac{24948}{2520}\)=\(\frac{99}{10}\)
HỌC TỐT
\(\frac{3}{67}\left(17\frac{21}{45}-13\frac{21}{45}\right)\)
\(\frac{3}{67}\left(17\frac{21}{45}-13\frac{21}{45}\right)\)
\(=\frac{3}{67}.4\)
\(=\frac{12}{67}\)
\(\frac{3}{67}\left(17\frac{21}{45}-13\frac{21}{45}\right)\)
\(=\frac{3}{67}\times4\)
\(=\frac{12}{67}\)