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a)\(\frac{2}{3}+\frac{3}{4}+\frac{5}{6}\)
\(=\frac{8+9+10}{12}\)
\(=\frac{27}{12}=\frac{9}{4}\)
b)\(\frac{15}{8}-\frac{7}{12}+\frac{5}{6}\)
\(=\frac{45-14+20}{24}\)
\(=\frac{51}{24}=\frac{17}{8}\)
2)
a)\(\frac{2}{5}+\frac{7}{13}+\frac{3}{5}+\frac{1}{7}\)
\(=\frac{2}{5}+\frac{3}{5}+\frac{7}{13}+\frac{1}{7}\)
\(=1+\frac{7}{13}+\frac{1}{7}\)
\(=\frac{20}{13}+\frac{1}{7}\)
\(=\frac{153}{91}\)
Tí tớ trả lời tiếp
\(6\frac{5}{7}-\left(1\frac{3}{4}+2\frac{5}{7}\right)\)
\(=6\frac{5}{7}-1\frac{3}{4}-2\frac{5}{7}\)
\(=\left(6\frac{5}{7}-2\frac{5}{7}\right)-1\frac{3}{4}\)
\(=4-1\frac{3}{4}\)
\(=2\frac{1}{4}\)
\(6\frac{5}{7}-\left(1\frac{3}{4}+2\frac{5}{7}\right)\)
\(=6\frac{5}{7}-1\frac{3}{4}-2\frac{5}{7}\)
\(=\left(6\frac{5}{7}-2\frac{5}{7}\right)-1\frac{3}{4}\)
\(=4-\frac{7}{4}\)
\(=\frac{16}{4}-\frac{7}{4}\)
\(=\frac{9}{4}\)
= \(6\frac{5}{7}-1\frac{3}{4}-2\frac{5}{7}=6\frac{5}{7}-2\frac{5}{7}\)\(-1\frac{3}{4}\)= \(4-1\frac{3}{4}\)=\(\frac{9}{4}\)
47/7 - (7/4 + 19/7) =47/7 -125/28 =gomen =tao xin lỗi chịu rùi = <3
a) 1 - 2 + 3 - 4 + 5 - 6 + ... + 97 - 98 + 99 - 100 + 101 (có 101 số; 101 chia 2 dư 1)
= (1 - 2) + (3 - 4) + (5 - 6) + ... + (97 - 98) + (99 - 100) + 101 (có 50 nhóm và dư 1 số)
= -1 + (-1) + (-1) + ... + (-1) + (-1) + 101
= -1.50 + 101
= -50 + 101
= 51
2 câu c` lại cx chia thành nhóm r` lm tương tự
1 - 2 + 3 - 4 + ........ + 99 - 100 + 101
= ( 1 - 2) + ( 3 - 4 ) + ..... + ( 99 -100 ) + 101 ( 50 nhóm )
= - 1 + ( - 1 ) + ...... + ( -1) + 101 ( 50 số -1 )
= - 1 . 50 + 101
= 50 + 101
= 151
\(\frac{5}{16}+\frac{5}{7}+0,4+\frac{1}{6}-\frac{4}{35}\)
= \(\frac{5}{16}+\frac{25}{35}+\frac{2}{5}+\frac{1}{6}-\frac{4}{35}\)
= \(\frac{5}{16}+\frac{2}{5}+\frac{1}{6}+\frac{25}{35}-\frac{4}{35}\)
= \(\frac{5}{16}+\frac{2}{5}+\frac{1}{6}+\frac{2}{3}\)
= \(\frac{5}{16}+\frac{2}{5}+\frac{5}{6}\)
= \(\frac{5}{16}+\frac{37}{30}\)
= \(\frac{371}{240}\)
\(a)\frac{4^{17}.3^4.9^6}{6^{13}}=\frac{2^{34}.3^4.3^{12}}{2^{13}.3^{13}}=\frac{2^{21}.3^{16}}{3^{13}}=2^{21}.3^3\)
\(b)\frac{2^{16}+2^8}{2^{13}+2^5}=\frac{2^8\left(2^8+1\right)}{2^5\left(2^8+1\right)}=\frac{2^3\left(2^8+1\right)}{2^8+1}=2^3=8\)
\(a,\frac{4^{17}\times3^4\times9^6}{6^{13}}\)
\(=\frac{2^{34}\times3^4\times3^{12}}{2^{13}\times3^{13}}\)
\(=\frac{2^{34}\times3^{16}}{2^{13}\times3^{13}}\)
\(=2^{21}\times3^3\)
\(b,\frac{2^{16}+2^8}{2^{13}+2^5}\)
\(=\frac{2^8\left(2^8+2^0\right)}{2^5\left(2^8+2^0\right)}\)
\(=\frac{2^8}{2^5}\)
\(=8\)
~Study well~