Question

Answer 1

\(1.A=\dfrac{x+5}{x+1}=\dfrac{x+1+4}{x+1}=1+\dfrac{4}{x+1}\)

Để A nguyên thì 4 ⋮ x + 1

=> x + 1 ∈ Ư(4) = {1; -1; 2; -2; 4; -4}

=> x ∈ {0; -2; 1; -3; 3; -5} 

\(2.B=\dfrac{2x+4}{x+3}=\dfrac{2\left(x+3\right)-2}{x+3}=2-\dfrac{2}{x+3}\)

Để B nguyên thì 2 ⋮ x + 3 

=> x + 3 ∈ Ư(2) = {1; -1; 2; -2}

=> x ∈ {-2; -4; -1; -5} 

\(3,C=\dfrac{3x+8}{x-1}=\dfrac{3\left(x-1\right)+11}{x-1}=3+\dfrac{11}{x-1}\)

Để C nguyên thì: 11 ⋮ x - 1

=> x - 1 ∈ Ư(11) = {1; -1; 11; -11}

=> x ∈ {2; 0; 12; -10} 

Answer 2

1: ĐKXĐ: \(x\ne-1\)

Để A 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 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 \(3x+8⋮x-1\)

=>\(3x-3+11⋮x-1\)

=>\(11⋮x-1\)

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

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

5: ĐKXĐ: \(x\ne-5\)

Để E nguyên thì \(5x+9⋮x+5\)

=>\(5x+25-16⋮x+5\)

=>\(-16⋮x+5\)

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

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

6: ĐKXĐ: \(x\ne-\dfrac{1}{2}\)

Để F nguyên thì \(4x+9⋮2x+1\)

=>\(4x+2+7⋮2x+1\)

=>\(7⋮2x+1\)

=>\(2x+1\in\left\{1;-1;7;-7\right\}\)

=>\(2x\in\left\{0;-2;6;-8\right\}\)

=>\(x\in\left\{0;-1;3;-4\right\}\)

7: ĐKXĐ: \(x\ne\dfrac{1}{2}\)

Để G nguyên thì \(6x+5⋮2x-1\)

=>\(6x-3+8⋮2x-1\)

=>\(8⋮2x-1\)

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

=>\(2x\in\left\{2;0;3;-1;5;-3;9;-7\right\}\)

=>\(x\in\left\{1;0;\dfrac{3}{2};-\dfrac{1}{2};\dfrac{5}{2};-\dfrac{3}{2};\dfrac{9}{2};-\dfrac{7}{2}\right\}\)

mà x nguyên

nên \(x\in\left\{0;1\right\}\)

9: ĐKXĐ: \(x\ne-2\)

Để I nguyên thì \(4x+4⋮2x+4\)

=>\(4x+8-4⋮2x+4\)

=>\(-4⋮2x+4\)

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

=>\(2x\in\left\{-3;-5;-2;-6;0;-8\right\}\)

=>\(x\in\left\{-\dfrac{3}{2};-\dfrac{5}{2};-1;-3;0;-4\right\}\)

mà x nguyên

nên \(x\in\left\{-1;-3;0;-4\right\}\)

10:

ĐKXĐ: \(x\ne-1\)

Để K là số nguyên thì \(4x+6⋮2x+2\)

=>\(4x+4+2⋮2x+2\)

=>\(2⋮2x+2\)

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

=>\(2x\in\left\{-1;-3;0;-4\right\}\)

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

mà x nguyên

nên \(x\in\left\{0;-2\right\}\)

Answer 3

\(5,E=\dfrac{5x+9}{x+5}=\dfrac{5\left(x+5\right)-16}{x+5}=5-\dfrac{16}{x+5}\)

Để E nguyên thì 16 ⋮ x + 5

=> x + 5 ∈ Ư(16) = {1; -1; 2; -2; 4; -4; 8; -8; 16; -16}

=> x ∈ {-4; -6; -3; -7; -1; -9; 3; -13; 11; -21} 

\(6,F=\dfrac{4x+9}{2x+1}=\dfrac{2\left(2x+1\right)+7}{2x+1}=2+\dfrac{7}{2x+1}\)

Để F nguyên thì 7 ⋮ 2x + 1

=> 2x + 1 ∈ Ư(7) = {1; -1; 7; -7}

=> 2x ∈ {0; -2; 6; -8}

=> x ∈ {0; -1; 3; -4} 

\(7.G=\dfrac{6x+5}{2x-1}=\dfrac{3\left(2x-1\right)+8}{2x-1}=3+\dfrac{8}{2x-1}\) 

Để G nguyên thì 8 ⋮ 2x - 1 

Mà 2x - 1 là số lẻ => 2x - 1 ∈ {1; -1}

=> 2x ∈ {2; 0}

=> x ∈ {0;1}