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Bài 1:
\(a)\dfrac{20^5.5^{10}}{100^5}=\dfrac{20^5.5^5.5^5}{100^5}=\dfrac{100^5.3125}{100^5}=3125\)
2.
a)A có 36 sô hạng , chia A thành 18 nhóm , mỗi nhóm có 2 số hạng .
Ta có : A = \(\left(3+3^2\right)+\left(3^3+3^4\right)+....+\left(3^{35}+3^{36}\right)\)
\(A=3.\left(1+3\right)+3^3.\left(1+3\right)+...+3^{35}.\left(1+3\right)\)
\(A=3.4+3^3.4+...+3^{35}.4\)
\(A=4.\left(3+3^3+...+3^{35}\right)\)
Vậy A chia hết cho 4 .
b)Chia A thành 13 nhóm mỗi nhóm có 3 số hạng
Ta có : \(A=\left(3+3^2+3^3\right)+...+\left(3^{34}+3^{35}+3^{36}\right)\)
\(A=3.\left(1+3+9\right)+...+3^{34}.\left(1+3+9\right)\)
A=\(3.13+...+3^{34}.13\)
A= \(13.\left(3+..+3^{34}\right)\)
Vậy A chia hết cho 13
c) Tương tự như câu a và câu b
f) \(\left(1:\frac{1}{7}\right)^2\left[\left(2^2\right)^3:2^5\right]\cdot\frac{1}{49}\)
\(=\left(1\cdot7\right)^2:\left(2^6:2^5\right)\cdot\frac{1}{49}=7^2\cdot\frac{1}{2}\cdot\frac{1}{49}=49\cdot\frac{1}{49}\cdot\frac{1}{2}=\frac{1}{2}\)
g) \(\frac{4^6\cdot3^5-2^{12}\cdot3^6}{2^{12}\cdot9^3+8^4\cdot3^5}=\frac{\left(2^2\right)^6\cdot3^5-2^{12}\cdot3^6}{2^{12}\cdot\left(3^2\right)^3+\left(2^3\right)^4\cdot3^5}\)
\(=\frac{2^{12}\cdot3^5-2^{12}\cdot3^6}{2^{12}\cdot3^6+2^{12}\cdot3^5}=\frac{2^{12}\left(3^5-3^6\right)}{2^{12}\left(3^6+3^5\right)}=\frac{2^{12}\left[3^5\left(1-3\right)\right]}{2^{12}\left[3^5\left(3+1\right)\right]}=\frac{2^{12}\cdot3^5\cdot\left(-2\right)}{2^{12}\cdot3^5\cdot4}=\frac{-2}{4}=-\frac{1}{2}\)
Bài giải
\(f,\text{ }\left(1\text{ : }\frac{1}{7}\right)^2\left[\left(2^2\right)^3\text{ : }2^5\right]\cdot\frac{1}{49}\)
\(=7^2\left(2^6\text{ : }2^5\right)\cdot\frac{1}{7^2}\)
\(=2\)
\(g,\text{ }\frac{4^6\cdot3^5-2^{12}\cdot3^6}{2^{12}\cdot9^3+8^4\cdot3^5}=\frac{2^{12}\cdot3^5-2^{12}\cdot3^6}{2^{12}\cdot3^6+2^{12}\cdot3^5}=\frac{2^{12}\cdot3^5\cdot\left(1-3\right)}{2^{12}\cdot3^5\cdot\left(3+1\right)}=-\frac{2}{4}=-\frac{1}{2}\)
\(\frac{131.145+100}{45-132.140}=\frac{132.145-45}{45-132.140}=-1\)
\(\frac{49^6.5-7^{11}}{\left(-7\right)^{10}.5+2.49^5}=\frac{7^{11}.7-7^{11}.1}{7^{10}.5+2.7^{10}}=\frac{7^{11}.\left(7-1\right)}{7^{10}.\left(5+2\right)}=\frac{7^{11}.6}{7^{11}}=6\)
\(M=\frac{131.145+100}{45-132.145}\)
\(=\frac{131-\frac{100}{145}}{\frac{45}{145}-132}\)
\(=\frac{131-\frac{20}{29}}{\frac{9}{29}-132}\)
\(=\frac{131\frac{-20}{29}}{-132\frac{9}{29}}\)