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= \(\dfrac{4^5.9^4-2.6^9}{2^{10}.3+6^8.20}\)

= \(\dfrac{\left(2^2\right)^5.\left(3^3\right)^4-2.6^9}{2^{10}.3+6^8.20}\)

=\(\dfrac{2^{10}.3^{12}-2.6^9}{2^{10}.3+6^8.20}\)

= \(\dfrac{3^{11}-6}{10}\)

\(A=1+5+5^2+5^3+5^4+...+5^{49}+5^{50}\)

\(\Rightarrow5A=5\left(1+5+5^2+5^3+5^4+...+5^{49}+5^{50}\right)\)

\(\Rightarrow5A=5+5^2+5^3+5^4+5^5+...+5^{50}+5^{51}\)

\(\Rightarrow5A-A=\left(5+5^2+5^3+5^4+5^5+...+5^{50}+5^{51}\right)-\left(1+5+5^2+5^3+5^4+...+5^{49}+5^{50}\right)\)\(\Rightarrow4A=5^{51}-1\)

\(\Rightarrow A=\dfrac{5^{51}-1}{4}\)

Đáp số:

a) x=45

b) x=31

c) x=2

d) x=94

\(9^x=5\cdot9^7+5\cdot9^7\)

\(\Rightarrow9^x=9^7\cdot\left(5+4\right)\)

\(\Rightarrow9^x=9^7\cdot9\)

\(\Rightarrow9^x=9^8\)

\(\Rightarrow x=8\)

x = 8

9^x=5.9⁷+4.9⁷

9^x=9.9⁷

9^x=9⁸

=>x=8