a, (x + 1)(2y - 1) = 12

=> x + 1 ; 2y - 1 \(\in\) Ư(12) = {\(\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\)}

Mà 2y - 1 lẻ => 2y - 1 \(\in\) {\(\pm1;\pm3\)}

=> x + 1 \(\in\) {\(\pm4;\pm12\)}

Ta có bảng:

Vậy các cặp (x;y) là (3;2) ; (-5;-1) ; (11;1) ; (-13;0)

b, (2x + 1)(y - 3) = 10

=> 2x + 1 ; y - 3 \(\in\) Ư(10) = {\(\pm1;\pm2;\pm5;\pm10\)}

Mà 2x + 1 lẻ => 2x + 1 \(\in\left\{\pm1;\pm5\right\}\)

=> y - 3 \(\in\left\{\pm2;\pm10\right\}\)

Ta có bảng:

Vậy các cặp (x;y) là (0;13) ; (-1;-7) ; (2;5) ; (-3;1)

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}.\frac{22}{45}x=\frac{23}{45}\)

\(\Rightarrow\frac{11}{45}x=\frac{23}{45}\)

\(\Rightarrow x=\frac{23}{45}:\frac{11}{45}\)

\(\Rightarrow x=\frac{23}{11}\)

\(S=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{99}\right)\)

\(=\frac{1}{2}.\frac{98}{99}=\frac{49}{99}\)

\(A=1\frac{1}{2}.1\frac{1}{3}....1\frac{1}{2015}\)

\(=\frac{3}{2}.\frac{4}{3}......\frac{2016}{2015}\)

\(=\frac{2016}{2}=1008\)

xy + 2x + y + 11 = 0

<=> x(y + 2) + y + 2 + 9 = 0

<=> x(y + 2) + (y + 2) = 0 - 9

<=> (x + 1)(y + 2) = -9

=> x + 1 và y + 2 thuộc Ư(-9) = {1;-1;3;-3;9;-9}

Ta có bảng

Vậy các cặp (x;y) là (0;-11) ; (-2;7) ; (2;-5) ; (-4;1) ; (8;-3) ; (-10;-1)

Bài 2:

Ta có: \(\left|x-3\right|\ge0\)

=> \(\left|x-3\right|+10\ge10\)

Dấu "=" xảy ra <=> x - 3 = 0 <=> x = 3

Vậy A_Max = 10 tại x = 3

Bài 3:

Ta có: (a + b) - (-a + b - c) + (c - a - b)

= a + b + a - b + c + c - a - b

= (a + a - a) - (b - b + b) + (c + c)

= a - b + 2c