a: ĐKXĐ: \(x\notin\left\{3;-3;-1\right\}\)

\(P=\left(\dfrac{2x}{x+3}+\dfrac{x}{x-3}+\dfrac{3x^2+3}{9-x^2}\right):\left(\dfrac{2x-2}{x-3}-1\right)\)

\(=\dfrac{2x\left(x-3\right)+x\left(x+3\right)-3x^2-3}{\left(x-3\right)\left(x+3\right)}:\dfrac{2x-2-x+3}{x-3}\)

\(=\dfrac{2x^2-6x+x^2+3x-3x^2-3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x-3}{x+1}\)

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

b: |x-2|=1

=>\(\left[{}\begin{matrix}x-2=-1\\x-2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=3\left(loại\right)\end{matrix}\right.\)

Khi x=1 thì \(P=\dfrac{3}{1+3}=\dfrac{3}{4}\)

c: Để P nguyên thì \(-3⋮x+3\)

=>\(x+3\in\left\{1;-1;3;-3\right\}\)

=>\(x\in\left\{-2;-4;0;-6\right\}\)

\(16^x:2^x=6^4\\ \Rightarrow\left(16:2\right)^x=6^4\\ \Rightarrow8^x=6^4\)
Vì \(8^x⋮̸6,\forall x\) và \(6^4⋮6\) nên:
Không có giá trị x thỏa mãn đề bài
Vậy...
P/s: Bạn kiểm tra lại đề bài nhé

Cửa hàng giảm giá số tiền là:

      \(20000-15000=5000\) (đồng)

Cửa hàng giảm giá số phần trăm là:

       \(5000:20000=0,25=25\%\)

               Đ/s: \(25\%\)

Số tiền mà của hàng giảm giá cho bìa sách là: 

`20000 - 15000 = 5000` (đồng)

Của hàng đã giảm giá số `%` là: 

`5000 : 20000` x `100 = 25%` (Giá vốn)

Đáp số: ...

\(x^3+\left(1+x\right)^3-\left(2x+1\right)\left(x+1\right)=0\)

=>\(\left(x+x+1\right)\left[x^2-x\left(x+1\right)+\left(x+1\right)^2\right]-\left(2x+1\right)\left(x+1\right)=0\)

=>\(\left(2x+1\right)\left(x^2-x^2-x+x^2+2x+1-x-1\right)=0\)

=>\(\left(2x+1\right)\left(x^2\right)=0\)

=>\(\left[{}\begin{matrix}x^2=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)

`123 : x - 23 : x = 20`

`=> (123 - 23) : x = 20`

`=> 100 : x = 20`

`=> x = 100 : 20`

`=> x = 5`

Vậy `x = 5`

      \(123\div\)\(x-\)\(23\div\)\(x=\)\(20\)

     \(\left(123-23\right)\)\(\div\)\(x\)  \(=\)\(20\)

       \(100\div x\)          \(=\)\(20\)

                 \(x\)          \(=\)\(100\div20\)

                 \(x\)          \(=\)\(5\)

`(3x + 1)^4 =` \(\dfrac{1}{16}\)

`=> (3x + 1)^4 =` \(\left(\dfrac{1}{2}\right)^4\)

`=> 3x + 1 =` \(\dfrac{1}{2}\) hoặc `3x + 1 =` \(-\dfrac{1}{2}\)

`=> 3x =` \(-\dfrac{1}{2}\) hoặc `3x =` \(-\dfrac{3}{2}\)

`=> x =` \(-\dfrac{1}{6}\) hoặc `x =` \(-\dfrac{1}{2}\)

Vậy `x =` \(-\dfrac{1}{6}\) hoặc `x =` \(-\dfrac{1}{2}\)

`13 . (y + 3) - 9.( y+5) + (y+7) = 100`

`=> 13y + 39 - 9y - 45 + y+7 = 100`

`=> (13y - 9y  + y)+ 39 - 45+7 = 100`

`=> 5y + 1 = 100`

`=> 5y = 100 - 1`

`=> 5y = 99`

`=> y =` \(\dfrac{99}{5}\)

`A = 142 - [50-(2^3 - 10 - 2^3 . 5)]`

`A = 142 - [50-(8 - 10 - 8 . 5)]`

`A = 142 - [50-(-2 - 40)]`

`A = 142 - [50-(-42)]`

`A = 142 - (50+42)`

`A = 142 - 50-42`

`A = (142 - 42) - 50`

`A = 100 - 50`

`A = 50`

`(7 + 5y) : 2 + 13 = 19`

`=> (7 + 5y) : 2 = 19 - 13`

`=> (7 + 5y) : 2 = 6`

`=> 7 + 5y = 6 . 2`

`=> 7 + 5y = 12`

`=> 5y = 12 - 7`

`=> 5y = 5`

`=> y = 5 : 5`

`=> y = 1`

Ta có:

\(A=\dfrac{2022}{2023}+\dfrac{2023}{2024}+\dfrac{2024}{2025}\\ =\left(1-\dfrac{1}{2023}\right)+\left(1-\dfrac{1}{2024}\right)+\left(1-\dfrac{1}{2025}\right)\\ =1-\dfrac{1}{2023}+1-\dfrac{1}{2024}+1-\dfrac{1}{2025}\\=\left(1+1+\right)-\left(\dfrac{1}{2023}+\dfrac{1}{2024}+\dfrac{1}{2025}\right)\\ =3-\left(\dfrac{1}{2023}+\dfrac{1}{2024}+\dfrac{1}{2025}\right)< 3\)

A = \(\dfrac{2022}{2023}\) + \(\dfrac{2023}{2024}\) + \(\dfrac{2024}{2025}\) Vì \(\dfrac{2022}{2023}\) < 1; \(\dfrac{2023}{2024}\) < 1; \(\dfrac{2024}{2025}\) < 1

Vậy A = \(\dfrac{2022}{2023}+\dfrac{2023}{2024}+\dfrac{2024}{2025}\) < 1 + 1 + 1  = 2 + 1 = 3

Vậy A < 3