|x|+|y|<=3

mà x,y nguyên

nên \(\left(\left|x\right|;\left|y\right|\right)\in\left\{\left(0;3\right);\left(0;1\right);\left(0;2\right);\left(0;0\right);\left(1;1\right);\left(1;2\right);\left(3;0\right);\left(1;0\right);\left(2;0\right);\left(2;1\right)\right\}\)

=>(x;y)\(\in\){(0;0);(0;1);(1;0);(0;-1);(-1;0);(0;2);(2;0);(0;-2);(-2;0);(0;3);(0;-3);(3;0);(-3;0);(1;1);(1;-1);(-1;1);(1;2);(2;1);(-1;-2);(-2;-1);(1;-2);(-2;1);(-1;2);(2;-1)}

Bài 2:

1: ĐKXĐ: \(x\ne-1\)

Để A là số nguyên thì \(x+5⋮x+1\)

=>\(x+1+4⋮x+1\)

=>\(4⋮x+1\)

=>\(x+1\in\left\{1;-1;2;-2;4;-4\right\}\)

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

2: ĐKXĐ: \(x\ne-3\)

Để B là số nguyên thì \(2x+4⋮x+3\)

=>\(2x+6-2⋮x+3\)

=>\(-2⋮x+3\)

=>\(x+3\in\left\{1;-1;2;-2\right\}\)

=>\(x\in\left\{-2;-4;-1;-5\right\}\)

3: ĐKXĐ: \(x\ne1\)

Để C nguyên thì \(3x+8⋮x-1\)

=>\(3x-3+11⋮x-1\)

=>\(11⋮x-1\)

=>\(x-1\in\left\{1;-1;11;-11\right\}\)

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

4: ĐKXĐ: \(x\ne1\)

Để D là số nguyên thì \(2x-3⋮x-1\)

=>\(2x-2-1⋮x-1\)

=>\(-1⋮x-1\)

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

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

5: ĐKXĐ: \(x\ne-5\)

Để E là số nguyên thì \(5x+9⋮x+5\)

=>\(5x+25-16⋮x+5\)

=>\(-16⋮x+5\)

=>\(x+5\in\left\{1;-1;2;-2;4;-4;8;-8;16;-16\right\}\)

=>\(x\in\left\{-4;-6;-3;-7;-1;-9;3;-13;11;-21\right\}\)

\(\left(7+3\dfrac{1}{4}-\dfrac{3}{5}\right)+\left(0,4-5\right)-\left(4\dfrac{1}{4}-1\right)\\ =7+3\dfrac{1}{4}-\dfrac{3}{5}+\dfrac{2}{5}-5-4\dfrac{1}{4}+1\\ =\left(7-5+1\right)+\left(3\dfrac{1}{4}-4\dfrac{1}{4}\right)+\left(\dfrac{-3}{5}+\dfrac{2}{5}\right)\\ =3-1+\dfrac{-1}{5}\\ =2+\dfrac{-1}{5}\\=\dfrac{9}{5}\)

\(\left(-8\dfrac{2}{5}\right):\left(-2\dfrac{4}{5}\right)=\dfrac{-42}{5}:\dfrac{-14}{5}=\dfrac{42}{14}=3\)

\(\dfrac{8^{11}\cdot3^{17}}{27^{10}\cdot9^{15}}=\dfrac{2^{33}\cdot3^{17}}{3^{30}\cdot3^{30}}=\dfrac{2^{33}\cdot3^{17}}{3^{60}}=\dfrac{2^{33}}{3^{43}}\)

\(\dfrac{8^{11}.3^{17}}{27^{10}.9^{15}}\\ =\dfrac{\left(2^3\right)^{11}.3^{17}}{\left(3^3\right)^{10}.\left(3^2\right)^{15}}\\ =\dfrac{2^{33}.3^{17}}{3^{30}.3^{30}}\\ =\dfrac{2^{33}.3^{17}}{3^{60}}\\ =\dfrac{2^{33}}{3^{43}}\)

a: \(-\dfrac{2}{3}\cdot x=\dfrac{4}{15}\)

=>\(x=\dfrac{4}{15}:\dfrac{-2}{3}=\dfrac{4}{15}\cdot\dfrac{-3}{2}=\dfrac{-12}{30}=-\dfrac{2}{5}\)

b: \(-\dfrac{7}{19}\cdot x=\dfrac{-13}{24}\)

=>\(x=\dfrac{13}{24}:\dfrac{7}{19}=\dfrac{13}{24}\cdot\dfrac{19}{7}=\dfrac{247}{168}\)

\(-\dfrac{2}{3}x=\dfrac{4}{15}\)

<=> \(x=\dfrac{4}{15}:\left(-\dfrac{2}{3}\right)\)

<=> \(x=\dfrac{4}{15}.\left(-\dfrac{3}{2}\right)\)

<=> \(x=-\dfrac{2}{5}\)

\(-\dfrac{7}{19}.x=-\dfrac{13}{24}\)

=> \(x=\left(-\dfrac{13}{24}\right):\left(-\dfrac{7}{19}\right)\)

=> \(x=\dfrac{13}{24}.\dfrac{19}{7}\)

=> \(x=\dfrac{247}{168}\)

`A(x) + B(x) = 6x^4 - 3x^2 - 5`

`A(x) - B(x) = 4x^4 - 6x^3 + 7x^2 + 8x - 9`

Áp dụng bài toán tổng hiệu ta có: 

`A(x) = [(6x^4 - 3x^2 - 5) + (4x^4 - 6x^3 + 7x^2 + 8x - 9)] : 2`

`= (6x^4 - 3x^2 - 5 + 4x^4 - 6x^3 + 7x^2 + 8x - 9) : 2`

`= (10x^4 - 6x^3 + 4x^2 + 8x - 14) : 2`

`= 5x^4 - 3x^3 + 2x^2 + 4x - 7`

`B(x) = (6x^4 - 3x^2 - 5) - (5x^4 - 3x^3 + 2x^2 + 4x - 7)`

`= 6x^4 - 3x^2 - 5 - 5x^4 + 3x^3 - 2x^2 - 4x + 7`

`= x^4 + 3x^3 - 5x^2 - 4x + 2`

Vậy ....

\(2A\left(x\right)=\left(6x^4-3x^2-5\right)+\left(4x^4-6x^3+7x^2+8x-9\right)\\ =\left(6x^4+4x^4\right)-6x^3+\left(-3x^2+7x^2\right)+8x+\left(-5-9\right)\\ =10x^4-6x^3+4x^2+8x-14\\ =>A\left(x\right)=5x^4-3x^3+2x^2+4x-7\)

\(=>B\left(x\right)=\left(6x^4-3x^2-5\right)-A\left(x\right)\\ =\left(6x^4-3x^2-5\right)-\left(5x^4-3x^3+2x^2+4x-7\right)\\ =\left(6x^4-5x^4\right)+3x^3+\left(-3x^2-2x^2\right)-4x+\left(-5+7\right)\\ =x^4+3x^3-5x^2-4x+2\)