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\(\frac{1}{3}^{101}.....\frac{1}{2}^{145}\)
Đề có sai ko bạn?
![](https://rs.olm.vn/images/avt/0.png?1311)
thay x=-1 ta có : \(\left(-x^2\right)+\left(-x^4\right)+\left(-x^6\right)+\left(-x^8\right)+....+\left(-x^{100}\right)\) =\(\left(-1^2\right)+\left(-1^4\right)+\left(-1^6\right)+\left(-1^8\right)+...+\left(-1^{100}\right)\) =1+1+1+1+...+1 = 50
![](https://rs.olm.vn/images/avt/0.png?1311)
\(-3^{150}=-9^{75}\)
\(-2^{225}=-8^{75}\)
mà -9<-8
nên \(-3^{150}< -2^{225}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(\left(\frac{1}{5}\right)^{10}.5^{20}=\left(\frac{1}{5}\right)^{10}.5^{10.2}=\left(\frac{1}{5}\right)^{10}.25^{10}=\left(\frac{1}{5}.5\right)^{10}=1^{10}=1\)
b)\(5^2.3^5.\left(\frac{3}{5}\right)^2=\left(\frac{3}{5}.5\right)^2.3^5=3^2.3^5=3^7\)
c)\(\left(\frac{1}{16}\right)^3:\left(\frac{1}{8}\right)^2=\left(\frac{1}{8}\right)^{2.3}:\left(\frac{1}{8}\right)^2=\left(\frac{1}{8}\right)^{6+2}=\left(\frac{1}{8}\right)^8\)
\(a.\left(\frac{1}{5}\right)^{10}.5^{20}=\left(\frac{1}{5}\right)^{10}.5^{10.2}=\left(\frac{1}{5}\right)^{10}.\left(5^2\right)^{10}=\left(\frac{1}{5}\right)^{10}.25^{10}=\left(\frac{1}{5}.25\right)^{10}=5^{10}.\)
\(b.5^2.3^5.\left(\frac{3}{5}\right)^2=\left[5^2.\left(\frac{3}{5}\right)^2\right].3^5=\left(5.\frac{3}{5}\right)^2.3^5=3^2.3^5=3^7\)\(c.\left(\frac{1}{16}\right)^3:\left(\frac{1}{8}\right)^2=\left[\left(\frac{1}{4}\right)^2\right]^3:\left[\left(\frac{1}{2}\right)^3\right]^2=\left(\frac{1}{4}\right)^6:\left(\frac{1}{2}\right)^6=\left(\frac{1}{4}:\frac{1}{2}\right)^6=\left(\frac{1}{2}\right)^6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
3^x*5^x-1=224
3^x*5^x/5=224
15^x=224*5
15^x=1120
=>ko tồn tại x thỏa mãn đề bài vị 15^x luôn có tận cùng bằng 5 (x khác 0 ) hoặc 1 ( x=0) ma 1120 co tận cùng bằng 0
![](https://rs.olm.vn/images/avt/0.png?1311)
\(9^{70}=\left(9^7\right)^{10}=\left(\left(3^2\right)^7\right)^{10}=\left(3^{14}\right)^{10}\)
\(8^{100}=\left(2^3\right)^{100}=\left(\left(2^3\right)^{10}\right)^{10}=\left(2^{30}\right)^{10}\)
\(2^{30}>\left(2^4\right)^7>\left(3^2\right)^7=3^{14}\)
\(\Rightarrow2^{30}>3^{14}\Leftrightarrow9^{70}< 8^{100}\)
\(9^{70}=\left(9^7\right)^{10}=4782969^{10}\)
\(8^{100}=\left(8^{10}\right)^{10}=1073741824^{10}\)
\(4782969< 1073741824\)
\(\Rightarrow9^{70}< 8^{100}\)