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3x + 2 . 5y = 45x
=> 3x . 32 . 5y = 3x . 5x . 3x
=> 32 . 5y = 5x . 3x
=> 32 - x = 5x - y
<=>\(\orbr{\begin{cases}3^{2-x}=1\\5^{x-y}=1\end{cases}}\)<=>\(\orbr{\begin{cases}2-x=0\\x=y\end{cases}}\)<=> x = y = 2
Vậy cặp số nguyên ( x ; y ) thỏa mãn là ( 2 ; 2 )
\(\left(7x-11\right)^3=2^5.5^2+200\)
\(\left(7x-11\right)^3=1000\)
\(=>7x-11=10\)
\(x=\frac{10+11}{7}=3\)
**** cho mik nhé!
\(\left(7x-11\right)^3=2^5\times5^2+200\)
\(\Rightarrow\left(7x-11\right)^3=1000=10^3\)
\(\Rightarrow7x-11=10\)
\(7x=10+11\)
\(7x=21\)
\(x=21\div7\)
\(x=3\)
\(A=\frac{25^3.5^5}{6.5^{10}}\)
\(A=\frac{\left(5^2\right)^3.5^5}{6.5^{10}}\)
\(A=\frac{5^6.5^5}{6.5^{10}}\)
\(A=\frac{5^{11}}{6.5^{10}}\)
\(A=\frac{5}{6}\)
(Dùng phương pháp giảm ước)
\(=\frac{\left(5^2\right)^3.5^5}{6.5^{10}}\)
\(=\frac{5^6.5^5}{6.5^{10}}\)
\(=\frac{5^{11}}{6.5^{10}}\)
\(=\frac{5}{6}\)
VẬY \(A=\frac{5}{6}\)
Ta có: B-A=1x3+2x4+3x5+4x6+...+100x102-(1x2+2x3+3x4+4x5+...+100x101)
=1x3+2x4+3x5+4x6+...100x102-1x2-2x3-3x4-4x5-...-100x101
=1+2+3+4+...+100
=((100-1):1+1)x((100-1):2)
=100x(101:2)
=5050
Có: A=\(\dfrac{3}{1.5}+\dfrac{3}{5.10}+...+\dfrac{3}{100.105}\)
=> A=\(3.\dfrac{5}{5}\left(\dfrac{1}{1.5}+\dfrac{1}{5.10}+...+\dfrac{1}{100.105}\right)\)
=> A= \(3.\dfrac{1}{5}\left(\dfrac{5}{1.5}+\dfrac{5}{5.10}+...+\dfrac{5}{100.105}\right)\)
=> A=\(\dfrac{3}{5}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{105}\right)\)
=> A= \(\dfrac{3}{5}\left(1-\dfrac{1}{105}\right)\)=\(\dfrac{3}{5}.\dfrac{104}{105}=\dfrac{312}{525}\)
Ta có:
\(A=\frac{3}{1\cdot5}+\frac{3}{5\cdot10}+...+\frac{3}{100\cdot105}\)
\(=\frac{3}{5}\cdot\left(\frac{5}{1\cdot5}+\frac{5}{5\cdot10}+...+\frac{5}{100\cdot105}\right)\)
\(=\frac{3}{5}\cdot\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{105}\right)\)
\(=\frac{3}{5}\left(1-\frac{1}{105}\right)=\frac{3}{5}\cdot\frac{104}{105}=\frac{312}{525}\)
Đề sai chắc luôn đoạn kìa là `3xx5^{x-2}` mới đúng
`3xx5^{x-2}+4xx5^{x-3}=19xx5^10`
`=>3xx5^{x-3+1}+4xx5^{x-3}=19xx5^10`
`=>3xx5xx5^{x-3}+4xx5^{x-3}=19xx5^10`
`=>15xx5^{x-3}+4xx5^{x-3}=19xx5^10`
`=>19xx5^{x-3}=19xx5^10`
`=>5^{x-3}=5^10`
`=>x-3=10`
`=>x=13`
Vậy `x=13`