Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)Để phân số thuộc Z
=>3 chia hết cho 2n-1
=>2n-1=Ư(3)=(-1,-3,1,3)
=>2n=(0,-2,2,4)
=>n=(0,-1,1,2)
Vậy n=0,-1,1,2
b)Để phân số thuộc Z
=>2n+3 chia hết cho 7
=>2n+3-7 chia hết cho 7
=>2n-4 chia hết cho 7
=>2n:7(dư 4)
=>2n đồng dư với 4(mod 7)
=>n đồng dư với 2(mod 7)
=>n:7(dư 2)
=>n-2 chia hết cho 7
=>n-2=7k
=>n=7k+2(k thuộc Z)
Vậy n=7k+2(k thuộc Z)
\(\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)
\(=\frac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(2^2.3\right)^{10}}\)
\(=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\)
\(=\frac{2^{19}.3^9+2^{19}.3^9.5}{2^{19}.3^9+2^{20}.3^{10}}\)
\(=\frac{2^{19}.3^9.\left(1+5\right)}{2^{19}.3^9\left(1+2.3\right)}\)
\(=\frac{6}{7}\)
\(H=\frac{4116-14}{10290-35}=\frac{14.294-14}{35.294-35}=\frac{14.\left(294-1\right)}{35.\left(294-1\right)}=\frac{14.293}{35.293}=\frac{2}{5}\)
\(K=\frac{29.101-101}{2.19.101+4.101}=\frac{101.\left(29-1\right)}{101.\left(38+4\right)}=\frac{28}{42}=\frac{2}{3}\)
\(I=\frac{1313-1717}{303}=\frac{13.101-17.101}{3.101}=\frac{101.\left(13-17\right)}{3.101}=\frac{-4}{3}\)
\(M=\frac{12-24.3}{1-35}=\frac{12-12.2.3}{-34}=\frac{12.\left(1-6\right)}{-34}=\frac{-60}{-34}=\frac{30}{17}\)
\(H\)\(=\) \(\frac{4116-14}{10290-35}\)
\(=\) \(\frac{4102}{10255}\)
\(=\) \(\frac{4102:2051}{10255:2051}\)
\(=\) \(\frac{2}{5}\)
\(K=\frac{2929-101}{2.1919+404}\)
\(=\) \(\frac{2828}{4242}\)
\(=\) \(\frac{2828:1414}{4242:1414}\)
\(=\) \(\frac{2}{3}\)
\(M=\frac{12-24.3}{1-35}\)
\(=\) \(\frac{-60}{-34}\)
\(=\) \(\frac{60}{34}\)
\(=\) \(\frac{30}{17}\)
:D
\(\text{Câu 1 :}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{12.13}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{12}-\frac{1}{13}\)
\(=\frac{1}{1}-\frac{1}{13}\)
\(=\frac{12}{13}\)
\(\text{Câu 2 :}\)
\(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)
\(=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\frac{100}{101}\)
\(=\frac{250}{101}\)
\(\frac{6^5+6^6.7}{6^3+6^6.4}=\frac{6^5.\left(1+6.7\right)}{6^3.\left(1+6^3.4\right)}\)
\(=\frac{6^2.43}{865}=\frac{36.43}{865}\)
\(=\frac{1568}{865}\)
+) \(\frac{-180}{270}=\frac{-2}{3}\) +) \(\frac{22}{-165}=\frac{-2}{15}\)
+) \(\frac{-39}{261}=\frac{-13}{87}\) +) \(\frac{4.26}{52.20}=\frac{4.13.2}{4.13.2.10}=\frac{1}{10}\)
+) \(\frac{7.\left(-32\right)}{16.35}=\frac{7.\left(-2\right).16}{16.7.5}=\frac{-2}{5}\)
Học tốt
\(\frac{560}{720}=\frac{560:80}{720:80}=\frac{7}{9}\)
\(\frac{189}{270}=\frac{189:27}{270:27}=\frac{7}{10}\)
k mk nha
thank you very much
cau 1 :7/9 cau 2 :7/10