x - 18 = 7281

x = 7281 + 18 

x = 7299

x - 18 = 7281

x = 7281 + 18

x = 7299

125.101=125.(100+1)

             = 125.100 + 125.1

             = 12500 + 125

             = 12625

\(125.101=125.100+125\)

             \(=12500+125=12625.\)

ai giúp mình với

a.

\(\dfrac{x}{9}=\dfrac{4}{x}\)

\(\Rightarrow x^2=4.9\)

\(\Rightarrow x^2=36\)

\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=6\end{matrix}\right.\)

b.

\(\dfrac{x+1}{3}=\dfrac{3}{x+1}\)

\(\Rightarrow\left(x+1\right)^2=3^2\)

\(\Rightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

\(\dfrac{x}{9}=\dfrac{4}{x}\)

\(x^2=4.9\)

\(x^2=36\)

\(x^2=6^2\)

\(\Rightarrow\left\{{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)

\(---------\)

\(\dfrac{x+1}{3}=\dfrac{3}{x+1}\)

\(\left(x+1\right)^2=3.3=3^2\)

\(\Rightarrow\left(1\right):x+1=3\)

            \(x=3-1\Rightarrow x=2.\)

\(\Rightarrow\left(2\right):x+1=-3\)

             \(x=-3-1\Rightarrow x=-4\)

Từ \(\left(1\right)\) và \(\left(2\right)\), ta suy ra:

\(\Rightarrow x\in\left\{{}\begin{matrix}2\\-4\end{matrix}\right.\)

Bài 1

a) S = 1 + 2 + 2² + 2³ + ... + 2²⁰²³

2S = 2 + 2² + 2³ + 2⁴ + ... + 2²⁰²⁴

S = 2S - S = (2 + 2² + 2³ + ... + 2²⁰²⁴) - (1 + 2 + 2² + 2³)

= 2²⁰²⁴ - 1

b) B = 2²⁰²⁴

B - 1 = 2²⁰²⁴ - 1 = S

B = S + 1

Vậy B > S

a,

\(S=1+2+2^2+...+2^{2023}\)

\(2S=2+2^2+2^3+...+2^{2024}\)

\(\Rightarrow S=2^{2024}-1\)

b.

Do \(2^{2024}-1< 2^{2024}\)

\(\Rightarrow S< B\)

2.

\(H=3+3^2+...+3^{2022}\)

\(\Rightarrow3H=3^2+3^3+...+3^{2023}\)

\(\Rightarrow3H-H=3^{2023}-3\)

\(\Rightarrow2H=3^{2023}-3\)

\(\Rightarrow H=\dfrac{3^{2023}-3}{2}\)

Mỗi tam giác có 3 cạnh mỗi cạnh chính là một đoạn thẳng \(\Rightarrow\)n-1 = 36 .3 = 108 \(\Rightarrow\)n= 108+1=109

      Vậy có 109 đoạn thẳng

a) \(\dfrac{2x+5}{2x+1}=\dfrac{2x+1+4}{2x+1}=\dfrac{2x+1}{2x+1}+\dfrac{4}{2x+1}=1+\dfrac{4}{2x+1}\)  

Để \(\dfrac{2x+5}{2x+1}\in Z\) thì \(\dfrac{4}{2x+1}\in Z\) 

\(\Rightarrow4\) ⋮ \(2x+1\)

\(\Rightarrow2x+1\inƯ\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)

\(\Rightarrow2x\in\left\{0;-2;1;-3;3;-5\right\}\)

\(\Rightarrow x\in\left\{0;-1;\dfrac{1}{2};-\dfrac{3}{2};\dfrac{3}{2};-\dfrac{5}{2}\right\}\)

Mà x nguyên \(\Rightarrow\text{x}\in\left\{0;-1\right\}\) 

b) \(\dfrac{3x+5}{x+1}=\dfrac{3x+3+2}{x+1}=\dfrac{3\left(x+1\right)+2}{x+1}=\dfrac{3\left(x+1\right)}{x+1}+\dfrac{2}{x+1}=3+\dfrac{2}{x+1}\) 

Để \(\dfrac{3x+5}{x+1}\in Z\) thì \(\dfrac{2}{x+1}\in Z\) 

\(\Rightarrow2\) ⋮ \(x+1\)

\(\Rightarrow x+1\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)

\(\Rightarrow x\in\left\{0;-2;1;-3\right\}\) 

c) \(\dfrac{3x+8}{x-1}=\dfrac{3x-3+11}{x-1}=\dfrac{3\left(x-1\right)+11}{x-1}=\dfrac{3\left(x-1\right)}{x-1}+\dfrac{11}{x-1}=3+\dfrac{11}{x-1}\)  

Để: \(\dfrac{3x+8}{x-1}\in Z\) thì \(\dfrac{11}{x-1}\in Z\)

\(\Rightarrow11\) ⋮ \(x-1\)

\(\Rightarrow x-1\inƯ\left(11\right)=\left\{1;-1;11;-11\right\}\)

\(\Rightarrow x\in\left\{2;0;12;-10\right\}\)

d) \(\dfrac{5x+12}{x-2}=\dfrac{5x-10+22}{x-2}=\dfrac{5\left(x-2\right)+22}{x-2}=\dfrac{5\left(x-2\right)}{x-2}+\dfrac{22}{x-2}=5+\dfrac{22}{x-2}\)

Để: \(\dfrac{5x+12}{x-2}\in Z\) thì \(\dfrac{22}{x-2}\in Z\)

\(\Rightarrow22\) ⋮ \(x-2\)

\(\Rightarrow x-2\inƯ\left(22\right)=\left\{1;-1;2;-2;11;-11;22;-22\right\}\)

\(\Rightarrow x\in\left\{3;1;4;0;13;-9;24;-20\right\}\)

e) \(\dfrac{7x-12}{x+16}=\dfrac{7x+112-124}{x+16}=\dfrac{7\left(x+16\right)-124}{x+16}=\dfrac{7\left(x+16\right)}{x+16}-\dfrac{124}{x+16}=7-\dfrac{124}{x+16}\)

Để \(\dfrac{7x-12}{x+16}\in Z\) thì \(\dfrac{124}{x+16}\in Z\) 

\(\Rightarrow124\) ⋮ \(x+16\)

\(\Rightarrow x+16\inƯ\left(124\right)=\left\{1;-1;2;-2;4;-4;31;-31;62;-62;124;-124\right\}\)

\(\Rightarrow x\in\left\{-15;-17;-14;-18;-12;-20;15;-47;46;-78;108;-140\right\}\)

Số tự nhiên lớn nhất có ba chữ số khác nhau chia cho 2 dư 1 là: \(987\).

Số tự nhiên nhỏ nhất có ba chữ số khác nhau chia cho 2 dư 1 là: \(103.\)

Với $x=0, y=1$ thì $6x+11y=11$ chia hết cho 11 nhưng $x+7y=7$ không chia hết cho 31 bạn nhé. Bạn xem lại đề.