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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{899}{30^2}\)
\(\Rightarrow A=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{29.31}{30.30}\)
\(\Rightarrow A=\frac{1.3.2.4.3.5...29.31}{2.2.3.3.4.4...30.30}\)
\(\Rightarrow A=\frac{\left(1.2.3...29\right).\left(3.4.5...31\right)}{\left(2.3.4...30\right).\left(2.3.4...30\right)}\)
\(\Rightarrow A=\frac{1}{30}.\frac{31}{2}=\frac{31}{60}\)
Vậy \(A=\frac{31}{60}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\frac{-8}{15}.\left(-30\right).\frac{15}{-8}.\frac{9}{10}\)
\(=-\left(\frac{8}{15}.\frac{15}{8}\right).\left(30.\frac{9}{10}\right)\)
\(=-1.27
=-27\)
\(b,2\frac{1}{18}.\frac{23}{24}.\frac{9}{37}.\frac{48}{-15}\)
\(=\frac{-37.23.9.48}{18.24.37.15}=\frac{23}{15}\)
c, chịu rồi
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(2^5+8\left[\left(-2\right)^3:\frac{1}{2}\right]^0-\left(\frac{1}{2}\right)^3\times2+\left(-2\right)^3\)
\(=32+8\times1-\frac{1}{8}\times2+\left(-8\right)\)
\(=32+8-\frac{1}{4}+\left(-8\right)\)
\(=40-\frac{1}{4}+\left(-8\right)\)
\(=39\frac{3}{4}+\left(-8\right)\)
\(=31\frac{3}{4}\)
b vaf c mai minhf lamf, ht
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,15^8*9^7/27^7*25^4
=3^8*5^8*3^14/3^21*5^8
=3^22*5^8/3^21*5^8
=3
b,9^3/(3^4-3^3)^2
=3^6/3^8-3^6
=3^6/3^6*(3^2-1)
=3^6/3^6*8
=1/8
c,(1/2-2/3+3/4-2/5):x=11/30
=> 11/60:x=11/30
=> x=11/60:11/30
=> x=1/2
d,-3/4*x+0,7*x=1,25:1/8
=> x*(-3/4+0,7)=10
=> x*(-1/20)=10
=> x=10:(-1/20)=200
Mình làm tắt lắm mong bạn thông cảm.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\begin{array}{l}a)\frac{{{3^{12}} + {3^{15}}}}{{1 + {3^3}}}\\ = \frac{{{3^{12}} + {3^{12}}{{.3}^3}}}{{1 + {3^3}}}\\ = \frac{{{3^{12}}.(1 + {3^3})}}{{1 + {3^3}}}\\ = {3^{12}}\\b)2:{\left( {\frac{1}{2} - \frac{2}{3}} \right)^2} + 0,{125^3}{.8^3} - {( - 12)^4}:{6^4}\\ = 2:{\left( {\frac{3}{6} - \frac{4}{6}} \right)^2} + {(0,125.8)^3} - {12^4}:{6^4}\\ = 2:{\left( {\frac{{ - 1}}{6}} \right)^2} + {1^3} - {(\frac{{12}}{6})^4}\\ = 2:\frac{1}{{36}} + 1 - {2^4}\\ = 2.36 + 1 - 16\\ = 72 + 1 - 16=57\end{array}\)
\(A=\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{899}{30^2}=\frac{1.3.2.4.3.5...29.31}{2.2.3.3.4.4...30.30}=\frac{\left(2.3...29.30\right).\left(3.4.5...29.31\right)}{\left(2.3...29.30\right).\left(2.3.4...29.30\right)}=\frac{31}{2.30}=\frac{31}{60}\)
Gjngkbb. Mk