Số                    Căn bậc hai số học

25                                   5

144                                12

30                                 \(\sqrt{30}\)

155                               \(\sqrt{155}\)\

0                                    0

-36                                 Ko có

0,64                                0,8

\(\dfrac{49}{121}\)                               \(\dfrac{7}{11}\)

-4,41                                Ko có

\(\dfrac{4}{5}\)                                      \(\dfrac{2\sqrt{5}}{5}\)

Cậu có ghi nhầm đề không ạ .-.?

=> mình làm luôn nhưng ko viết lại đề bài ạ.

Có: \(C=\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2023^2}\)

= \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2023^2}\)

Ta thấy: \(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

...

\(\dfrac{1}{2023^2}< \dfrac{1}{2022.2023}\)

=> \(\dfrac{1}{2^2}+\dfrac{ 1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2023^2}\)

\(< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2022.2023}\)

=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2021}-\dfrac{1}{2022}+\dfrac{1}{2022}-\dfrac{1}{2023}\)

=\(1-\dfrac{1}{2023}\)

=\(\dfrac{2022}{2023}\)<\(\dfrac{3}{4}\)

hay C<3/4

b) \(B=\dfrac{2^{10}.\left(2^2.13\right)+2^{12}.\left(5.13\right)}{2^{11}.\left(2^2.13\right)}+\dfrac{3^{10}.11+3^9\left(11+4\right)}{3^8.2^4.3}\)

= \(\dfrac{2^{12}.13+2^{12}.5.13}{2^{13}.13}+\dfrac{3^{10}.11+3^9.11.4}{3^9.2^4}\)

= \(\dfrac{2^{12}.13\left(1+5\right)}{2^{12}.13.2}+\dfrac{3^9.11\left(3+4\right)}{3^9.2^4}\)

= \(\dfrac{6}{2}+\dfrac{11.7}{16}\)

= 3 + 77/16 = 125/16

Vậy B = 125/16

\(\dfrac{1}{2}:\dfrac{1}{2}-\dfrac{1}{4}:\dfrac{1}{4}=\dfrac{1}{8}:\dfrac{1}{8}+2023\)

1 - 1 = 1 + 2023

0 = 2024 (?)

Mong bn xem lại đề bài ạ.

a) => (2x - 15)⁵ - (2x - 15)³ = 0

(2x-15)³ [(2x-15)²-1]=0

\(\Rightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2=1\end{matrix}\right.\)  \(\Rightarrow\left[{}\begin{matrix}2x=15\\2x-15=1\\2x-15=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right.\)

b) x + 2x + 3x + ... + 2022x = 2022.2023

x(1 + 2 + 3 +...+ 2022)= 2022.2023 (1)

Đặt A = 1+2+3+...+2022

Số số hạng trong A là: (2022 - 1): 1 + 1 = 2022 (số)

Tổng A bằng: \(\dfrac{\left(2022+1\right).2022}{2}=\dfrac{2022.2023}{2}\)

Thay A vào (1) , ta  được:

x.\(\dfrac{2022.2023}{2}=2022.2023\)

=> x = 2