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Ta có :
\(\frac{21^2.14.125}{35^3.6}=\frac{3^2.7^2.2.7.5^3}{5^3.7^3.2.3}=\frac{2.3^2.5^3.7^3}{2.3.5^3.7^3}=\frac{3}{1}=3\)
Vậy \(\frac{21^2.14.125}{35^3.6}=3\)
\(\frac{21^2.14.125}{35^3.6}\)= \(\frac{21^2.2.7.125}{42875.2.3}\)= \(\frac{21^2.7.125}{125.343.3}\)= \(\frac{21^2.7.125}{125.7.49.3}\)= \(\frac{21^2}{49.3}\)= \(\frac{441}{147}\)
Mình làm rồi nhưng bạn thử tính lại cho chắc nha
Chúc bạn học tốt!
Câu 1 : \(1,321338308x10^{-4}\)
Câu 2 : \(1316,572106\)
Câu 3 : \(1,641302619x10^{-13}\)
Ủng hộ nhé,tớ đang âm.
Ta có :
\(M=\frac{9^4.27^5.3^6.3^4}{3^8.81^4.23^4.8^2}\)
\(M=\frac{\left(3^2\right)^4.\left(3^3\right)^5.3^{10}}{3^8.\left(3^4\right)^4.23^4.8^2}\)
\(M=\frac{3^8.3^{15}.3^{10}}{3^8.3^{16}.23^4.8^2}\)
\(M=\frac{3^{33}}{3^{24}.23^4.8^2}\)
\(M=\frac{3^9}{23^4.8^2}\)
Bài 1
a) \(P=\frac{6n+5}{2n-4}=\frac{6n-12+7}{2n-4}=3+\frac{7}{2n-4}\)
Để P là phân số thì \(\hept{\begin{cases}2n-4\ne7\\2n-4\ne1\end{cases}}\Leftrightarrow\hept{\begin{cases}n\ne\frac{11}{2}\\n\ne\frac{5}{2}\end{cases}}\)
Vậy...
b) \(P=\frac{6n+5}{2n-4}=3+\frac{7}{2n-4}\)
Để \(P\in Z\)thì \(\orbr{\begin{cases}2n-4=7\\2n-4=1\end{cases}\Leftrightarrow\orbr{\begin{cases}n=\frac{11}{2}\notin Z\\n=\frac{5}{2}\notin Z\end{cases}}}\)
Vậy không có giá trị n nào thuộc Z để P thuộc Z.
c) \(\left|2n-3\right|=\frac{5}{3}\)
Trường hợp: \(2n-3=\frac{5}{3}\Rightarrow n=\frac{7}{3}\)
\(P=\frac{6.\frac{7}{3}+5}{2.\frac{7}{3}-4}=\frac{19}{\frac{2}{3}}=\frac{57}{2}\)
Trường hợp: \(2n-3=-\frac{5}{3}\Rightarrow n=\frac{2}{3}\)
\(P=\frac{6.\frac{2}{3}+5}{2.\frac{2}{3}-4}=\frac{9}{\frac{-8}{3}}=\frac{27}{-8}\)
Bài 2
\(N=\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\frac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^{10}.4.5}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)
\(=\frac{2^{12}.3^{10}+5.2^{12}.3^{10}}{2^{12}.3^{12}-6^{11}}=\frac{6.2^{12}.3^{10}}{6^{12}-6^{11}}\)
\(=\frac{2.3.2^{12}.3^{10}}{6.6^{11}-6^{11}}=\frac{2^{13}.3^{11}}{5.\left(2.3\right)^{11}}=\frac{2^{13}.3^{11}}{5.2^{11}.3^{11}}=\frac{4}{5}\)
d)\(\frac{2.3+4.6+14.21}{3.5+6.10+21.35}=\frac{2.3+2.2.6+2.7.21}{3.5+3.2.10+3.7.35}=\frac{2.3+2.12+2.147}{3.5+3.20+3.245}=\frac{2\left(3+12+147\right)}{3\left(5+20+245\right)}\)
\(=\frac{2.162}{3.270}=\frac{54}{135}=\frac{2}{5}\)
\(a.\frac{-2019.2018+1}{\left(-2017\right).\left(-2019\right)+2018}\)
\(=\frac{2019.\left(-2018\right)+1}{2019.2017+2018}\)
\(=\frac{2019.\left(-2018\right)+1}{2019.2018-1}\)
\(=-\frac{2018}{2018}\)
\(=-1\)
\(A=\frac{4^6.3^4.9^5}{6^{12}}=\frac{\left(2^2\right)^6.3^3.\left(3^2\right)^5}{6^{12}}\)
\(=\frac{2^{12}.3^3.3^{10}}{6^{12}}=3^{13}.3^{12}=3^{25}\)
\(A=\frac{4^6.3^4.9^5}{6^{12}}\)
\(A=\frac{2^6.2^6.3^4.3^5.3^5}{2^{12}.3^{12}}\)
\(A=\frac{3^3.3^5}{1}\)
\(A=3^8\)
\(B=\frac{21^2.14.125}{35^3.6}\)
\(B=\frac{3^2.7^2.2.7.5^3}{5^3.7^3.2.3}\)
\(B=\frac{3.1.1.1.1}{1.1.1.1}\)
\(B=3\)