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a)\(\left(7\frac{4}{9}+4\frac{7}{11}\right)-3\frac{4}{9}\)=\(7\frac{4}{9}+4\frac{7}{11}-3\frac{4}{9}\)=\(\left(7\frac{4}{9}-3\frac{4}{9}\right)+4\frac{7}{11}\)= 4+\(4\frac{7}{11}\)=\(8\frac{7}{11}\)
b)\(\frac{-7}{9}.\frac{4}{11}+\frac{-7}{9}.\frac{7}{11}+5\frac{7}{9}\)=\(\frac{-7}{9}.\left(\frac{4}{11}+\frac{7}{11}\right)+5+\frac{7}{9}\)=\(\frac{-7}{9}.1+5+\frac{7}{9}\)=\(\frac{-7}{9}+\frac{7}{9}+5\)=\(\left(\frac{-7}{9}+\frac{7}{9}\right)+5\)= 0+5=5
c)\(50\%.1\frac{1}{3}.10\frac{7}{35}.0,75\)= \(\frac{1}{2}.\frac{4}{3}.10\frac{1}{5}.\frac{3}{4}\)=\(\frac{1}{2}.\frac{4}{3}.\frac{51}{5}.\frac{3}{4}\)=\(\frac{1.4.51.3}{2.3.5.4}\)=\(\frac{51}{2.5}\)=\(\frac{51}{10}\)
d)\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}\)=\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)=\(\frac{43}{43}-\frac{1}{43}\)=\(\frac{42}{43}\)
a) \(\left(7\frac{4}{9}+4\frac{7}{11}\right)-3\frac{4}{9}\)
\(=\left(\frac{67}{9}+\frac{51}{11}\right)-\frac{31}{9}\)
\(=\left(\frac{67}{9}-\frac{31}{9}\right)+\frac{51}{11}\)
\(=\frac{36}{9}+\frac{51}{11}\)
\(=\frac{95}{11}=8\frac{7}{11}\)
b) \(-\frac{7}{9}.\frac{4}{11}+-\frac{7}{9}.\frac{7}{11}+5\frac{7}{9}\)
\(=-\frac{7}{9}.\frac{4}{11}+-\frac{7}{9}.\frac{7}{11}+\frac{52}{9}\)
\(=-\frac{7}{9}.\left(\frac{4}{11}+\frac{7}{11}\right)+\frac{52}{9}\)
\(=-\frac{7}{9}.1+\frac{52}{9}\)
\(=-\frac{7}{9}+\frac{52}{9}\)
= 5
d) \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}\)
\(=1.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\right)\)
\(=1.\left(1-\frac{1}{43}\right)\)
\(=1.\frac{42}{43}\)
\(=\frac{42}{43}\)
a)\(\frac{3^{10}.\left(-5\right)^{21}}{\left(-5\right)^{20}.3^{12}}=\frac{-5}{9}\)
b)\(\frac{\left(-11\right)^5.13^7}{11^5.13^8}=-\frac{1}{13}\)
c)\(\frac{2^{10}.3^{10}-2^{10}.3^9}{2^9.3^{10}}=\frac{2^{10}.3^9\left(3-1\right)}{2^9.3^{10}}=2\)
d(\(\frac{5^{11}.7^{12}+5^{11}.7^{11}}{5^{12}.7^{12}+9.5^{11}.7^{11}}=\frac{5^{11}.7^{11}\left(7+1\right)}{5^{11}.7^{11}\left(35+9\right)}=\frac{1}{6}\)
\(\frac{2^{10}\cdot3^{10}-2^{10}\cdot3^9}{2^9\cdot3^{10}}=\frac{2^{10}\cdot3^9\left(3-1\right)}{2^9\cdot3^{10}}=\frac{2^{11}\cdot3^9}{2^9\cdot3^{10}}=\frac{2^2}{3}=\frac{4}{3}\)
Câu 1 xem kỉ đề
\(B,\frac{49^6.5-7^{11}}{\left(-7\right)^{10}.5-2.49^5}=\frac{7^{12}.5-7^{11}}{7^{10}.5-2.7^{10}}=\frac{7^{11}.\left(7.5-1\right)}{7^{10}.\left(5-2\right)}=\frac{7.34}{3}=\frac{238}{3}\)
a) A=212.35-\(\frac{2^{12}.3^6}{2^{12}}\)+93+84.35
=212.35-36+36+212.35
=213.35
b)B=496.5-5.\(\frac{7^{11}}{\left(-7\right)^{10}}-2.49^5\)
=496.5-7.5-2.495
=712.5-7.5-2.710
A=7/19(8/11+3/11)+12/19
=7/19.1+12/19
=1
B=5/9(7/13+9/13-3/13)
=5/9.1
=5/9
ba đội công nhân có tất cả 192 người . số người đội 1 chiếm 1/4 tổng số. số người đội 2 bằng 125% đội 1. tính số người đội 3
a) Đặt: A = \(\frac{5.7^{12}}{7^{10}.5^{10}}=\frac{5.7^2.7^{10}}{7^{10}.5^{10}}=\frac{7^2}{5^9}\)
b) Đặt: B = \(\frac{5^8.7^{12}-5^9.7^{11}}{7^{11}.5^{10}+7^{12}.5^{11}}=\frac{5^8.7.7^{11}-5.5^8.7^{11}}{7^{11}.5^{10}+7.7^{11}.5.5^{10}}\\ =\frac{5^8.7^{11}.\left(7-5\right)}{7^{11}.5^{10}.\left(1+7.5\right)}=\frac{2}{5^2.36}=\frac{2}{25.36}=\frac{1}{450}\)
a)\(\frac{5.7^{12}}{7^{10}.5^{10}}=\frac{5.7^{10}.7^2}{7^{10}.5.5^9}=\frac{7^2}{5^9}\)