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`

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

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

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

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

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

\(=1^3\)

\(=1\)

\(0,25\cdot\dfrac{7}{15}-\dfrac{1}{4}\cdot\left(-\dfrac{8}{15}\right)+2,75\)

\(=\dfrac{1}{4}\cdot\left(\dfrac{7}{15}+\dfrac{8}{15}\right)+2,75\)

\(=\dfrac{1}{4}\cdot1+2,75\)

\(=0,25+2,75\\ =3\)

\(a,\) Ta có : \(\left\{{}\begin{matrix}AD\perp AB\left(gt\right)\\BC\perp AB\left(gt\right)\end{matrix}\right.\)

\(\Rightarrow AD//BC\) ( cùng vuông góc với \(AB\) )

\(b,\) Ta có tứ giác thì 4 góc là \(360^o\)

\(\Rightarrow\widehat{ADC}=360^o-\widehat{BAD}-\widehat{ABC}-\widehat{DCB}\)

\(=360^o-90^o-90^o-75^o=105^o\)

Vậy \(\widehat{ADC}=105^o\)

`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)`}

\(a,\left(7+3\dfrac{1}{4}-\dfrac{3}{5}\right)+\left(0,4-5\right)-\left(4\dfrac{1}{4}-1\right)\)

\(=\left(7+\dfrac{13}{4}-\dfrac{3}{5}\right)-\dfrac{23}{5}-\left(\dfrac{17}{4}-1\right)\)

\(=7+\dfrac{13}{4}-\dfrac{3}{5}-\dfrac{23}{5}-\dfrac{17}{4}+1\)

\(=\left(7+1\right)+\left(\dfrac{13}{4}-\dfrac{17}{4}\right)-\left(\dfrac{3}{5}+\dfrac{23}{5}\right)\)

\(=8-\dfrac{4}{4}-\dfrac{26}{5}\)

\(=7-\dfrac{26}{5}\)

\(=\dfrac{9}{5}\)

\(b,\dfrac{2}{3}-\left[\left(-\dfrac{7}{4}\right)-\left(\dfrac{1}{2}+\dfrac{3}{8}\right)\right]\)

\(=\dfrac{2}{3}-\left(-\dfrac{7}{4}-\dfrac{1}{2}-\dfrac{3}{8}\right)\)

\(=\dfrac{2}{3}-\left(-\dfrac{14}{8}-\dfrac{4}{8}-\dfrac{3}{8}\right)\)

\(=\dfrac{2}{3}-\left(-\dfrac{21}{8}\right)\)

\(=\dfrac{2}{3}+\dfrac{21}{8}\)

\(=\dfrac{79}{24}\)

\(c,\left(9-\dfrac{1}{2}-\dfrac{3}{4}\right):\left(7-\dfrac{1}{4}-\dfrac{5}{8}\right)\)

\(=\left(\dfrac{36}{4}-\dfrac{2}{4}-\dfrac{3}{4}\right):\left(\dfrac{56}{8}-\dfrac{2}{8}-\dfrac{5}{8}\right)\)

\(=\dfrac{31}{4}:\dfrac{49}{8}\)

\(=\dfrac{62}{49}\)

\(d,3-\dfrac{1-\dfrac{1}{7}}{1+\dfrac{1}{7}}=3-\dfrac{\dfrac{7}{7}-\dfrac{1}{7}}{\dfrac{7}{7}+\dfrac{1}{7}}=3-\left(\dfrac{6}{7}:\dfrac{8}{7}\right)=3-\dfrac{3}{4}=\dfrac{9}{4}\)

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)