Số sách ở thư viện thứ nhất:

\(\left(15000+3000\right):2=9000\) (cuốn sách)

Số sách ở thư viện thứ hai:

\(15000-9000=6000\) (cuốn sách)

Đáp số:....

2\(xy\) + 6\(x\) -  \(y\) = 6

2\(xy\)  + 6\(x\)      = 6 + \(y\) 

\(x\)(2\(y\) + 6)       = 6 + \(y\)

\(x\)                    = (6 + \(y\) ): (2\(y\)+6)

\(x\) \(\in\) Z ⇔ 6 + \(y\) ⋮ 2\(y\) + 6 ⇒ 2.(6+\(y\)) ⋮ 2\(y\) + 6 ⇒ 12 + 2\(y\) ⋮ 2\(y\) + 6

⇒ 2\(y\) + 6 + 6  ⋮  2\(y\) + 6 ⇒ 6 ⋮ 2\(y\) + 6 ⇒ 3 ⋮ y + 3 

Ư(3) = {-3; -1; 1; 3}

Lập bảng ta có:

Các cặp (\(x;y\)) thỏa mãn đề bài lần lượt là:

(\(x\); \(y\)) = (0; -6); (-1; -4); (2; -2) ; (1; 0)

Vì : \(\left(2x-5\right)^{2022}\ge0\forall x,\left(3y+4\right)^{2024}\ge0\forall y\\ =>\left(2x-5\right)^{2022}+\left(3y+4\right)^{2024}\ge0\)

Do đó đề bài xảy ra khi và chỉ khi :

\(\left\{{}\begin{matrix}\left(2x-5\right)^{2022}=0\\\left(3y+4\right)^{2024}=0\end{matrix}\right.\\ =>\left(x;y\right)=\left(\dfrac{5}{2};-\dfrac{4}{3}\right)\)

Mình ko biết cách để làm ra đc kết quả này, có thể giải thích cụ thể hơn ko ạ?

\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)

\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)

\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)

\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)

\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)

\(\left[\left(0,1\right)^2\right]^0+\left[\left(\dfrac{1}{7}\right)^{-1}\right]^2\cdot\dfrac{1}{49}\cdot\left[\left(2^2\right)^3:2^5\right]\)

\(=1+7^2\cdot\dfrac{1}{49}\cdot\left(2^6:2^5\right)\)

\(=1+49\cdot\dfrac{1}{49}\cdot2\)

\(=1+1\cdot2\)

\(=3\)

cảm ơn ạ

`(4*2^5) \div (2^3*1/6)`

`= (2^2*2^5) \div (8/6)`

`= 2^7 \div 4/3`

`= 96`

\(\left(4.2^5\right):\left(2^3.\dfrac{1}{6}\right)\\=\dfrac{4.2^3.2^2}{2^3.\dfrac{1}{6}}\\ =4.2^2:\dfrac{1}{6}\\ =4.4.6=96\)

\(\left(0,125\right)^3.512=\left(0,125\right)^3.8^3=\left(0,125.8\right)^3=1^3=1\)

\(\left(0,125\right)^3.512\)

\(=\left(0,125\right)^3.8^3\)

\(=\left(0,125\cdot8\right)^3\)

\(=1^3\)

\(=1\)

`2xy + 6x - y = 6`

`=>(2xy-y)+(6x-3)=3`

`=>y(2x-1)+3(2x-1)=3`

`=>(2x-1)(y+3)=3=3.1=(-3).(-1)`

Mà \(x;y \in Z\)

`=>(x;y)=`{`(2;-2),(1;0),(-1;-4),(0;-6)`}

a, \(xy=9\) ( \(x\) < y)

Ư(9) = { -9; -3; -1; 1; 3; 9}

Lập bảng ta có: 

Theo bảng trên ta có: Vì \(x\) < y

nên (\(x;y\)) = (-9; -1); (1; 9)

b, (\(x\) - 6)(y +2) =7

   Ư(7) = {-7; -1; 1; 7}

Lập bảng ta có:

Theo bảng trên ta có:

Các cặp \(x;y\) thỏa mãn đề bài là:

(\(x;y\)) = (-1; -3); (5; -9); (7; 5); (13; -1)

\(a,\left(-8.2+17\right).\left[x+\left(-2\right).3\right]=20\)

\(\Rightarrow\left(-16+17\right).\left(x-6\right)=20\)

\(\Rightarrow1.\left(x-6\right)=20\)

\(\Rightarrow x=20+6\)

\(\Rightarrow x=26\)

\(b,\left(-5\right).x+5=\left(-3\right).\left(-8\right)+6\)

\(\Rightarrow-5x+5=24+6\)

\(\Rightarrow-5x+5=30\)

\(\Rightarrow-5x=25\)

\(\Rightarrow x=25:\left(-5\right)\)

\(\Rightarrow x=-5\)

\(c,-152-\left(3x+1\right)=\left(-2\right).\left(-27\right)\)

\(\Rightarrow-152-3x-1=54\)

\(\Rightarrow-3x=54+152+1\)

\(\Rightarrow-3x=207\)

\(\Rightarrow x=-69\)

\(d,\left(x+6\right)\left(x-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+6=0\\x-4=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=4\end{matrix}\right.\)

a) \(\left(-8,2+17\right).\left(x+\left(-2\right).3\right)=20\)

\(\Rightarrow8,8\cdot\left(x-6\right)=20\)

\(\Rightarrow x-6=\dfrac{25}{11}\)

\(\Rightarrow x=\dfrac{91}{11}\)

b) \(\left(-5\right)x+5=\left(-3\right).\left(-8\right)+6\)

\(\Rightarrow\left(-5\right)x+5=30\)

\(\Rightarrow\left(-5\right)x=25\)

\(\Rightarrow x=-5\)

c) \(-152-\left(3x+1\right)=\left(-2\right).\left(-27\right)\)

\(\Rightarrow-152-\left(3x+1\right)=54\)

\(\Rightarrow-\left(3x+1\right)=206\)

\(\Rightarrow-3x-1=206\)

\(\Rightarrow-3x=207\)

\(\Rightarrow x=-69\)

d) \(\left(x+6\right)\left(x-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+6=0\\x-4=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=4\end{matrix}\right.\)