Câu 1 xem kỉ đề

\(B,\frac{49^6.5-7^{11}}{\left(-7\right)^{10}.5-2.49^5}=\frac{7^{12}.5-7^{11}}{7^{10}.5-2.7^{10}}=\frac{7^{11}.\left(7.5-1\right)}{7^{10}.\left(5-2\right)}=\frac{7.34}{3}=\frac{238}{3}\)

a) A=2¹².3⁵-\(\frac{2^{12}.3^6}{2^{12}}\)+9³+8⁴.3⁵

=2¹².3⁵-3⁶+3⁶+2¹².3⁵

=2¹³.3⁵

^b)B=49⁶.5-5.\(\frac{7^{11}}{\left(-7\right)^{10}}-2.49^5\)

=49⁶.5-7.5-2.49⁵

⁼7¹².5-7.5-2.7¹⁰

a. - \(\frac{-1}{2}\)

b. \(\frac{238}{3}\)

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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

Bài 1:

b: \(=\dfrac{7^{12}\cdot5-7^{11}}{7^{10}\cdot5-2\cdot7^{10}}=\dfrac{7^{11}\cdot\left(7\cdot5-1\right)}{7^{10}\cdot3}=7\cdot\dfrac{34}{3}=\dfrac{238}{3}\)

c: \(=\dfrac{3^3\cdot2^3+3^3\cdot2^2+3^3}{-13}=\dfrac{3^3\cdot13}{-13}=-27\)

d: \(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\dfrac{2^{13}\cdot3^{11}}{2^{11}\cdot3^{11}\cdot5}=\dfrac{4}{5}\)

\(\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^4.7^4}{5^9.7^3+5^92^3.7^3}\)\(=\frac{2^{12}\left(3^5-3^4\right)}{2^{12}\left(3^6+3^5\right)}-\frac{7^3\left(5^{10}-5^4.7^4\right)}{7^3\left(5^9+5^9.2^3\right)}\)

\(=\frac{3^5-3^4}{3^6+3^5}-\frac{5^{10}-5^4.7^4}{5^9+5^9.2^3}\)\(=\frac{3^4\left(3-1\right)}{3^4\left(3^2+3\right)}-\frac{5^4\left(5^6-7^4\right)}{5^4\left(5^5+5^5.2^3\right)}\)

\(=\frac{1}{6}-\frac{5^6-7^4}{5^5+5^5.2^3}\)cậu cứ phân tích từ từ

a) A = 110 - (-761) + 296 + 1454 - (-813 + 1077)

        = 110 + 761 + 296 + 1454 - 264

        = 871 + 1750 - 264

        = 2631 - 264

        = 2357

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

\(f,\text{ }\left(1\text{ : }\frac{1}{7}\right)^2\left[\left(2^2\right)^3\text{ : }2^5\right]\cdot\frac{1}{49}\)

\(=7^2\left(2^6\text{ : }2^5\right)\cdot\frac{1}{7^2}\)

\(=2\)

\(g,\text{ }\frac{4^6\cdot3^5-2^{12}\cdot3^6}{2^{12}\cdot9^3+8^4\cdot3^5}=\frac{2^{12}\cdot3^5-2^{12}\cdot3^6}{2^{12}\cdot3^6+2^{12}\cdot3^5}=\frac{2^{12}\cdot3^5\cdot\left(1-3\right)}{2^{12}\cdot3^5\cdot\left(3+1\right)}=-\frac{2}{4}=-\frac{1}{2}\)