\(\dfrac{8.9-4.15}{12.7-180}\)
\(\dfrac{^{2^3}.3}{2^2.3^2.5}\)
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a) \(\dfrac{2727-101}{3.303+404}=\dfrac{2626}{909+404}=\dfrac{2626}{1313}=2\)
b) \(\dfrac{8.9-4.15}{12.7-180}=\dfrac{72-60}{84-180}=\dfrac{12}{-96}=\dfrac{-1}{8}\)
c) \(\dfrac{-19}{3^2.7.11}=\dfrac{-19}{9.7.11}=\dfrac{-19}{63.11}=\dfrac{-19}{693}\)
d) \(\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\dfrac{2^{12}.3^{10}+120.6^9}{2^{12}.3^{12}-6^{11}}=\dfrac{2^2.6^{10}+20.6.6^9}{6^{12}-6^{11}}=\dfrac{4.6^{10}+20.6^{10}}{6^{11}\left(6-1\right)}=\dfrac{\left(4+20\right).6^{10}}{5.6^{11}}=\dfrac{24}{30}=\dfrac{4}{5}\)
\(a,\frac{5\cdot6+5\cdot7}{5\cdot8+20}=\frac{\left(5\cdot6+5\cdot7\right)\div2}{\left(5.8+20\right)\div2}=\frac{13}{12}\)
\(\frac{8.9-4.15}{12.7-180}=\frac{12.6-12.5}{12.7-12.15}=\frac{1}{-8}=\frac{-1}{8}\)
\(Ta\)\(co\)\(\frac{13}{12}=\frac{13.2}{12.2}=\frac{26}{24}\)
\(\frac{-1}{8}=\frac{3\left(-1\right)}{3.8}=\frac{-3}{24}\)
Phần b bạn tính ra rồi làm tương tự phần a nha chúc bạn học giỏi!!!
\(a,\dfrac{-5}{8};\dfrac{7}{10};\dfrac{16}{19};\dfrac{20}{23}\)
\(\Rightarrow\dfrac{20}{23}>\dfrac{16}{19}>\dfrac{7}{10}>\dfrac{-5}{8}\)
Vậy \(\dfrac{20}{23}>\dfrac{16}{19}>\dfrac{7}{10}>\dfrac{-5}{8}\)
\(b,\dfrac{5\times6+6\times7}{5\times5+20}\)và \(\dfrac{8\times9-4\times15}{12\times7-180}\)
Xét : \(\dfrac{5\times6+6\times7}{5\times5+20}\)
\(=\dfrac{\left(5+7\right)\times6}{5\times5+5\times4}\)
\(=\dfrac{12\times6}{\left(5+4\right)\times5}\)
\(=\dfrac{72}{9\times5}\)
\(=\dfrac{72}{45}\)
\(=\dfrac{8}{5}\)
Xét : \(\dfrac{8\times9-4\times15}{12\times7-180}\)
\(=\dfrac{72-60}{84-180}\)
\(=\dfrac{12}{-96}\)
\(=\dfrac{-1}{8}\)
Quy đồng \(\dfrac{8}{5}=\dfrac{8\times8}{5\times8}=\dfrac{64}{40}\\ \dfrac{-1}{8}=\dfrac{\left(-1\right)\times5}{8\times5}=\dfrac{-5}{40}\)
Vì \(\dfrac{64}{40}>\dfrac{-5}{40}\)
\(\Rightarrow\dfrac{8}{5}>\dfrac{-1}{8}\)
\(\Rightarrow\dfrac{5\times6+6\times7}{5\times5+20}>\dfrac{8\times9-4\times15}{12\times7-180}\)
Vậy \(\dfrac{5\times6+6\times7}{5\times5+20}>\dfrac{8\times9-4\times15}{12\times7-180}\)
\(\frac{5.6+5.7}{5.8+20}\)và \(\frac{8.9-4.15}{12.7-180}\)
Ta có : \(\frac{5.6+5.7}{5.8+20}=\frac{30+35}{40+20}=\frac{65}{60}=\frac{13}{12}\)
\(\frac{8.9-4.15}{12.7-180}=\frac{72-60}{84-180}=\frac{12}{-96}=\frac{-1}{8}\)
+) Tìm \(12=2^2.3\)
\(8=2^3\)
\(MC=BCNN\left(12,18\right)=2^3.3=24\)
Khi đó : \(\frac{13}{12}=\frac{13.2}{12.2}=\frac{26}{24}\)
\(\frac{-1}{8}=\frac{-1.3}{8.3}=\frac{-3}{24}\)
p/s : kham khảo
Ta có: \(\frac{2^5.7+2^5}{2^5.5^2-2^5.3}\) = \(\frac{2^5.\left(7+1\right)}{2^5.\left(5^2-3\right)}\) = \(\frac{2^5.8}{2^5.22}\) = \(\frac{8}{22}\) =\(\frac{56}{154}\)
\(\frac{3^4.5-3^6}{3^4.13+3^4}\) = \(\frac{3^4.\left(5-3^2\right)}{3^4.\left(13+1\right)}\) = \(\frac{3^4.\left(-4\right)}{3^4.14}\) = \(\frac{-4}{14}\)= \(\frac{-44}{154}\)
tA CÓ :
\(\frac{5.6+5.7}{5.8+20}=\frac{5\left(6+7\right)}{5.\left(8+4\right)}=\frac{5.13}{5.12}=\frac{13}{12}\)
\(\frac{8.9-4.15}{12.7-180}=\frac{3.4\left(2.3-5\right)}{12\left(7-15\right)}=\frac{12\left(6-5\right)}{12\left(7-15\right)}=\frac{6-5}{7-15}=\frac{1}{-8}\)
a) Ta có: \(\dfrac{8\cdot9-4\cdot15}{12\cdot7-180}\)
\(=\dfrac{4\cdot18-4\cdot15}{12\cdot7-12\cdot15}\)
\(=\dfrac{4\cdot\left(18-15\right)}{12\cdot\left(7-15\right)}\)
\(=\dfrac{4\cdot3}{12\cdot\left(-8\right)}\)
\(=\dfrac{12}{12\cdot\left(-8\right)}=\dfrac{-1}{8}\)
b) Ta có: \(\dfrac{2^3\cdot3}{2^2\cdot3^2\cdot5}\)
\(=\dfrac{2^2\cdot2\cdot3}{2^2\cdot3\cdot3\cdot5}\)
\(=\dfrac{2}{3\cdot5}=\dfrac{2}{15}\)
a)8⋅9−4⋅1512⋅7−1808⋅9−4⋅1512⋅7−180=4⋅18−4⋅1512⋅7−12⋅15=4⋅18−4⋅1512⋅7−12⋅15=4⋅(18−15)12⋅(7−15)=4⋅(18−15)12⋅(7−15)=4⋅312⋅(−8)=4⋅312⋅(−8)=1212⋅(−8)=−18=1212⋅(−8)=−18
b)23⋅322⋅32⋅523⋅322⋅32⋅5=2