2.x-49=5.9
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Ta có : \(\left(x-1\right)^2+\dfrac{1}{5.9}+\dfrac{1}{9.13}+...+\dfrac{1}{41.45}=\dfrac{49}{900}\)
\(\Leftrightarrow\left(x-1\right)^2+\dfrac{1}{4}.\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}\right)=\dfrac{49}{900}\)
\(\Leftrightarrow\left(x-1\right)^2+\dfrac{1}{4}\left(\dfrac{1}{5}-\dfrac{1}{45}\right)=\dfrac{49}{900}\)
\(\Leftrightarrow\left(x-1\right)^2=\dfrac{1}{100}\) \(\Leftrightarrow\left[{}\begin{matrix}x-1=\dfrac{1}{10}\\x-1=-\dfrac{1}{10}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{10}\\x=\dfrac{9}{10}\end{matrix}\right.\)
Vậy ...
a, ta có A.5 = 5 ( 1+5 +52 +...........+549 +550)
5A = 5 +52 +53 +............... + 550 +551
5A-A = (5 +52 +53 +............+ 551) - (1+5+52 +......+550)
4A = 551 -1
A =\(\dfrac{5^{51}-1}{4}\)
vậy A =
b, B= \(\dfrac{4^5.9^4-2.6^9}{2^{10}.3+6^8.20}\)
= \(\dfrac{\left(2^2\right)^5.\left(3^3\right)^4-2.6^9}{2^{10}.3+6^8.20}\)
=\(\dfrac{2^{10}.3^{12}-2.6^9}{2^{10}.3+6^8.20}\)
= \(\dfrac{3^{11}-6}{10}\)
\(9^x=5\cdot9^7+5\cdot9^7\)
\(\Rightarrow9^x=9^7\cdot\left(5+4\right)\)
\(\Rightarrow9^x=9^7\cdot9\)
\(\Rightarrow9^x=9^8\)
\(\Rightarrow x=8\)
`2.x-49=5.9`
`=> 2.x-49=45`
`=>2.x=45+49`
`=>2.x=94`
`=>x=94:2`
`=>x=47`
2 . x - 49 = 45
2 . x = 49 + 45
2 . x = 94
x = 94 : 2
x = 47