Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)38:3^4+2^2.2^3=3^8-4+2^2+3=3^4+2^5=81+32=113
b)3.4^2-2.3^2=3(4^2-2.3)=3(16-6)=3.10=30
c)4^6.3^4.9^5/6^12=(2^2)^6.3^4.(3^2)^5/(2.3)^12=2^12.3^4.3^10/2^12.3^12=2^12.3^14/2^12.3^12=3^14/3^12=3^2=9
e)45^3.20^4.18^2/180^5=(5.3^2)^3.(2^2.5)^4.(2.3^2)^2/(2^2.3^2.5)^5=5^3.3^6.2^8.5^4.2^2.3^4/2^10.3^10.5^5=5^9.3^10.2^10/2^10.3^10.5^5
=5^4/1=5^4=625
d)21^2.14.125/35^3.6=(2.7)^2.2.7.5^3/(5.7)^3.2.3=2^2.7^2.2.7.5^3/5^3.7^3.2.3=2^3.7^3.5^3/5^3.7^3.2.3=2^3/2.3=8/6=4/3
g)2^13+2^5/2^10+2^2=2^5(2^8+1)/2^2(2^8+1)=2^5/2^2=32/4=8
Chúc bạn học tốt !!!!
Câu 1 : \(1,321338308x10^{-4}\)
Câu 2 : \(1316,572106\)
Câu 3 : \(1,641302619x10^{-13}\)
Ủng hộ nhé,tớ đang âm.
\(A=\frac{4^6.3^4.9^5}{6^{12}}=\frac{\left(2^2\right)^6.3^3.\left(3^2\right)^5}{6^{12}}\)
\(=\frac{2^{12}.3^3.3^{10}}{6^{12}}=3^{13}.3^{12}=3^{25}\)
\(A=\frac{4^6.3^4.9^5}{6^{12}}\)
\(A=\frac{2^6.2^6.3^4.3^5.3^5}{2^{12}.3^{12}}\)
\(A=\frac{3^3.3^5}{1}\)
\(A=3^8\)
\(B=\frac{21^2.14.125}{35^3.6}\)
\(B=\frac{3^2.7^2.2.7.5^3}{5^3.7^3.2.3}\)
\(B=\frac{3.1.1.1.1}{1.1.1.1}\)
\(B=3\)
a, =\(3^4+2^5=81+32=113\)
b, =\(3.\left(4^2-2.3\right)=3.\left(16-6\right)=3.10=30\)
c, =\(\dfrac{2^{12}.3^4.3^{10}}{2^{12}.3^{12}}=\dfrac{2^{12}.3^{14}}{2^{12}.3^{12}}=3^2=9\)
d, =\(\dfrac{3^2.7^2.2.7.5^3}{5^3.7^3.2.3}=3\)
e, =\(\dfrac{3^6.5^3.2^8.5^4.2^2.3^4}{2^{10}.3^{10}.5^5}=\dfrac{3^{10}.2^{10}.5^7}{2^{10}.3^{10}.5^5}=5^2=25\)
g, =\(\dfrac{2^5.\left(2^8+1\right)}{2^2.\left(2^8+1\right)}=\dfrac{2^5}{2^2}=2^3=8\)
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
\(N=\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\frac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^{10}.4.5}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)
\(=\frac{2^{12}.3^{10}+5.2^{12}.3^{10}}{2^{12}.3^{12}-6^{11}}=\frac{6.2^{12}.3^{10}}{6^{12}-6^{11}}\)
\(=\frac{2.3.2^{12}.3^{10}}{6.6^{11}-6^{11}}=\frac{2^{13}.3^{11}}{5.\left(2.3\right)^{11}}=\frac{2^{13}.3^{11}}{5.2^{11}.3^{11}}=\frac{4}{5}\)