B thuộc bội của 7 và 10 bé hơn b bé hơn 45

9^{10 =} (3²)¹⁰ = 3²⁰

`@` `\text {Ans}`

`\downarrow`

`(7x - 11)^2 = 2^5 * 5^2 + 100?`

`=> (7x - 11)^2 = 32*25 + 100`

`=> (7x - 11)^2 = 800 + 100`

`=> (7x - 11)^2 = 900`

`=> (7x - 11)^2 = (+-30)^2`

`=>`\(\left[{}\begin{matrix}7x-11=30\\7x-11=-30\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}7x=41\\7x=-19\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=\dfrac{41}{7}\\x=-\dfrac{19}{7}\end{matrix}\right.\)

Vậy, `x \in`\(\left\{\dfrac{41}{7};-\dfrac{19}{7}\right\}\)

\(A=\dfrac{5}{4}-\left(\dfrac{3}{8}+\dfrac{1}{a}\right):\dfrac{4}{5}\) \(\left(a\ne0\right)\)

Tại a = 12 biểu thức \(A=\dfrac{5}{4}-\left(\dfrac{3}{8}+\dfrac{1}{12}\right):\dfrac{4}{5}=\dfrac{5}{4}-\dfrac{11}{24}:\dfrac{4}{5}=\dfrac{5}{4}-\dfrac{11}{24}.\dfrac{5}{4}=\dfrac{5}{4}-\dfrac{55}{96}=\dfrac{65}{96}\)

Để \(A=\dfrac{15}{23}< =>\dfrac{5}{4}-\left(\dfrac{3}{8}+\dfrac{1}{a}\right):\dfrac{4}{5}=\dfrac{15}{23}\)

\(\Leftrightarrow\left(\dfrac{3}{8}+\dfrac{1}{a}\right):\dfrac{4}{5}=\dfrac{55}{92}< =>\dfrac{3}{8}+\dfrac{1}{a}=\dfrac{11}{23}< =>\dfrac{1}{a}=\dfrac{19}{184}< =>a=\dfrac{184}{19}\)

Thay \(a=12\) vào A ta có:

\(A=\dfrac{5}{4}-\left(\dfrac{3}{8}+\dfrac{1}{12}\right):\dfrac{4}{5}=\dfrac{65}{96}\)

Vậy:

____________________

Ta có:

\(A=\dfrac{15}{23}\) khi \(\dfrac{5}{4}-\left(\dfrac{3}{8}+\dfrac{1}{a}\right):\dfrac{4}{5}=\dfrac{15}{23}\)

\(\Rightarrow\left(\dfrac{3}{8}+\dfrac{1}{a}\right)\cdot\dfrac{5}{4}=\dfrac{5}{4}-\dfrac{15}{23}\)

\(\Rightarrow\dfrac{3}{8}+\dfrac{1}{a}=\dfrac{55}{92}:\dfrac{5}{4}\)

\(\Rightarrow\dfrac{3}{8}+\dfrac{1}{a}=\dfrac{11}{23}\)

\(\Rightarrow\dfrac{1}{a}=\dfrac{11}{184}\)

\(\Rightarrow a=\dfrac{1\cdot184}{11}=\dfrac{184}{11}\)

chịu

giúp với

a, 5ⁿ⁺¹ - 5^n-1 = 125⁴.2³.3

5^n-1.(5² - 1) = 125⁴.24

5^n-1.24         = 125⁴.24

5^n-1             = 125⁴

5^n-1             = (5³)⁴

5^n-1            = 5¹²

n - 1           = 12

n                = 12 + 1

n                = 13

b,2^2n-1 + 2²ⁿ⁺² = 3.2¹¹

   2^2n-1.(1 + 2³) = 3.2¹¹

  2^2n-1.9 = 3.2¹¹

 2^2n-1      = 2¹¹: 3

2²ⁿ        = 2¹² : 3 (xem lại đề bài em nhá)

đáp án 11

11

Do a chia 24 dư 18 nên a = 24k + 18 (k ∈ ℕ)

*) Do 24 ⋮ 3 nên 24k ⋮ 3

18 ⋮ 3

⇒ (24k + 18) ⋮ 3

Vậy a ⋮ 3

*) Do 24 ⋮ 4 nên 24k ⋮ 4

18 không chia hết cho 4

⇒ (24k + 18) không chia hết cho 4

Vậy a không chia hết cho 4