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\(\frac{2^{35}.9^{25}.5^{25}.13^{22}.7^{16}.5^{16}}{9^{26}.13^{22}5.^{22}.7^{17}.4^{17}.5^{2.9}}=2^{\left(35-2.17\right)}.9^{\left(25-26\right)}.5^{\left(25+16\right)-\left(22+2.9\right)}.13^{22-22}.7^{16-17}.\)
\(2^1.9^{-1}.5^1.13^0.7^{-1}=\frac{2.5}{9.7}=\frac{10}{63}\)
Ta có:
\(\frac{2^{35}.45^{25}.13^{22}.35^{16}}{9^{26}.65^{22}.28^{17}.25^9}=\frac{2^{35}.\left(9.5\right)^{25}.13^{22}.\left(5.7\right)^{16}}{9^{26}.\left(5.13\right)^{22}.\left(7.4\right)^{17}.\left(5.5\right)^9}\)
\(=\frac{2^{35}.9^{25}.5^{25}.13^{22}.5^{16}.7^{16}}{9^{26}.13^{22}.5^{22}.7^{17}.4^{17}.5^9.5^9}\)
\(=\frac{2^{35}.5^3}{9.4^{17}.5^2.7}\)
\(=\frac{2^{35}.5}{9.4^{17}.7}\)
\(=\frac{2^{35}.5}{9.2^{34}.7}\)
\(=\frac{2.5}{9.7}\)
\(=\frac{10}{63}\)
p/s: - Ko chắc ~
\(A=\dfrac{2^{35}.45^{25}.13^{22}.35^{16}}{9^{26}.65^{22}.28^{17}.25^9}=\dfrac{2^{35}.3^{50}.5^{25}.13^{22}.5^{16}.7^{16}}{3^{52}.13^{22}.5^{22}.2^{34}.7^{17}.5^{18}}\)
\(=\dfrac{2.5^3}{3^2.5^2.7}=\dfrac{2.5}{3^2.7}=\dfrac{10}{63}\)
\(\frac{2^{23}\cdot45^{25}\cdot13^{22}\cdot35^{16}}{9^{26}\cdot65^{22}\cdot28^{17}\cdot25^9}=\frac{2^{23}\cdot5^{25}\cdot3^{50}\cdot13^{22}\cdot5^{16}\cdot7^{16}}{3^{52}\cdot5^{22}\cdot13^{22}\cdot7^{17}\cdot2^{54}\cdot5^{18}}=\frac{2^{23}\cdot3^{50}\cdot5^{31}\cdot7^{16}\cdot13^{22}}{2^{54}\cdot3^{52}\cdot5^{22}\cdot7^{17}\cdot13^{22}}=\frac{5^9}{2^{31}\cdot3^2\cdot7}\)
A=\(\frac{2^{35}.9^{25}.5^{25}.13^{22}.7^{16}.5^{16}}{4^{17}.7^{17}.9^{26}.13^{22}.5^{22}.5^9.5^9}=\frac{2^{35}.5^1}{4^{17}.7^1.9}=\frac{2^{35}.5}{2^{34}.7^1.9}\)= \(\frac{2.5}{7.9}=\frac{10}{63}\)
\(\frac{x^6y^{n+2}}{x^ny^4z^{n-3}}=x^{\left(6-n\right)}y^{n+2-4}z^{n-3}\Rightarrow\hept{\begin{cases}6-n\ge0\\n-2\ge0\\n-3\ge0\end{cases}}\)=>n={3,4,5,6}
Ta có:
a) ( 3 n + 1 ) 2 - 25 = 3(3n - 4)(n + 2) chia hết cho 3;
b) ( 4 n + 1 ) 2 - 9 = 8(2n - 1)(n +1) chia hết cho 8.
a: \(\dfrac{x^ny^6}{x^5y^{n-2}}=x^{n-5}y^{8-n}\)
Để đây là phép chia hết thì n-5>=0và 8-n>=0
=>5<=n<=8
b: \(\dfrac{x^6y^{n+2}}{x^ny^4z^{n-3}}=x^{6-n}y^{n-4}z^{3-n}\)
Để đây là phép chia hết thì \(\left\{{}\begin{matrix}6-n>=0\\n-4>=0\\3-n>=0\end{matrix}\right.\Leftrightarrow n\in\varnothing\)
c: \(\dfrac{\left(\dfrac{1}{2}x^5y^{7-n}\right)}{-2x^ny^3}=-\dfrac{1}{4}x^{5-n}y^{4-n}\)
Để đây là phép chia hết thì 5-n>=0 và 4-n>=0
=>n<=4