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Ở câu a) số mũ lúc nào cug dương mà bạn ( 45-10 = 4510). Nếu số mũ là dương thì:
a)\(\frac{45^{10}.5^{20}}{75^{15}}\)
= \(\frac{\left(3^2.5\right)^{10}.5^{20}}{\left(3.5^2\right)^{15}}\)
= \(\frac{3^{20}.5^{10}.5^{20}}{3^{15}.5^{30}}\)
= \(\frac{3^{20}.5^{30}}{3^{15}.5^{30}}\)
= \(\frac{3^5.1}{1.1}\)
= \(\frac{243}{1}\)
= 243
b)\(\frac{2^{15}.9^4}{6^6.8^2}\)
= \(\frac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^2}\)
= \(\frac{2^{15}.3^8}{2^6.3^6.2^6}\)
= \(\frac{2^{15}.3^8}{2^{12}.3^6}\)
= \(\frac{2^3.3^2}{1.1}\)
= \(\frac{8.9}{1}\)
= \(\frac{72}{1}\)
= 72
Bài làm
\(A=\frac{45^{10}\cdot5^{20}}{75^{15}}\)
\(A=\frac{\left(3^2\right)^{10}\cdot5^{10}\cdot5^{20}}{3^{15}\cdot\left(5^2\right)^{15}}\)
\(A=\frac{3^{20}\cdot5^{30}}{3^{15}\cdot5^{30}}\)
\(A=3^5\)
Vậy \(A=3^5\)
\(B=\frac{2^{15}\cdot5^{20}}{6^6\cdot8^3}\)
\(B=\frac{2^{15}\cdot5^{20}}{2^6\cdot3^3\cdot\left(2^3\right)^3}\)
\(B=\frac{2^{15}\cdot5^{20}}{2^{15}\cdot3^3}\)
\(B=\frac{5^{20}}{3^3}\)
Vậy \(B=\frac{5^{20}}{3^3}\)
1)
a)
\(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)
\(\frac{-1}{1}.\frac{20}{5}< x< \frac{-1}{5}.\frac{3}{2}\)
\(\frac{-20}{5}< x< \frac{-3}{10}\)
\(\frac{-40}{10}< x< \frac{-3}{10}\)
\(\Rightarrow Z\in\left\{-4;-5;-6;-7;-8;-9;-10;...;-39\right\}\)
Bài 1:
a, \(\frac{1}{-16}-\frac{3}{45}=\frac{-1}{16}-\frac{1}{15}\)
\(=\frac{-15}{240}-\frac{16}{240}\)
\(=\frac{-31}{240}\)
b, \(=\frac{-10}{12}-\frac{-12}{12}\)
\(=\frac{2}{12}=\frac{1}{6}\)
c, \(=\frac{-30}{6}-\frac{1}{6}\)
\(=\frac{-31}{6}\)
Bài 2:
a, \(x=-\frac{1}{2}-\frac{3}{4}\)
\(x=-\frac{1}{4}\)
b, \(\frac{1}{2}+x=-\frac{11}{2}\)
\(x=-\frac{11}{2}-\frac{1}{2}\)
\(x=-6\)
Bạn nhớ k đúng và chọn câu trả lời này nhé!!!! Mình giải đúng và chính xác hết ^_^
\(A=\frac{1\cdot2+2\cdot4+3\cdot6+4\cdot8+5\cdot10+6\cdot12}{3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20+18\cdot24}\)
\(A=\frac{2\cdot3\left[1\cdot2\right]+2\cdot3\left[2\cdot4\right]+2\cdot3\left[3\cdot6\right]+2\cdot3\left[4\cdot8\right]+2\cdot3\left[5\cdot10\right]}{3\cdot4\left[3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20\right]}\)
\(A=\frac{\left[3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20\right]}{2\cdot3\left[3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20\right]}=\frac{1}{2\cdot3}=\frac{1}{6}\)
\(a,\left(\frac{31}{20}-\frac{26}{45}\right)\cdot\left(\frac{-36}{35}\right)< x< \left(\frac{51}{56}+\frac{8}{21}+\frac{1}{3}\right)\cdot\frac{8}{13}\)
\(taco:\left(\frac{31}{20}-\frac{26}{45}\right)\cdot\left(\frac{-36}{35}\right)=\frac{35}{36}\cdot\frac{-36}{35}=-1\)
\(\left(\frac{51}{56}+\frac{8}{21}+\frac{1}{3}\right)\cdot\frac{8}{13}=\frac{13}{8}\cdot\frac{8}{13}=1\)
\(=>x=0\)
\(b,\frac{-5}{6}+\frac{8}{3}+\frac{29}{-3}< x< \frac{-1}{2}+2+\frac{5}{2}\)(dau <co dau gach ngang o duoi nha)
\(taco:\frac{-5}{6}+\frac{8}{3}+\frac{29}{-3}=\frac{-5}{6}+\frac{8}{3}+\frac{-29}{3}=\frac{-5}{6}+\frac{16}{6}+\frac{-58}{6}=\frac{-47}{6}=-7,8\)
\(\frac{-1}{2}+2+\frac{5}{2}=\frac{3}{2}+\frac{5}{2}=4\)
tu do \(=>x=-7,8;...;0;1;2;3;4\)