Bài 1

Để A có giá trị nguyên thì \(\sqrt{x}-3\inƯ\left(2\right)\)

\(\sqrt{x}-3\in\left\lbrace-2;-1;1;2\right\rbrace\)

\(\sqrt{x}\in\left\lbrace1;2;4;5\right\rbrace\)

\(x\in\left\lbrace1;4;16;25\right\rbrace\)

ĐKXĐ: \(x\ne\pm3\)

a

Khi x = 1:

\(A=\dfrac{3.1+2}{1-3}=\dfrac{5}{-2}=-2,5\)

Khi x = 2:

\(A=\dfrac{3.2+2}{2-3}=-8\)

Khi x = \(\dfrac{5}{2}:\)

\(A=\dfrac{3.2,5+2}{2,5-3}=\dfrac{9,5}{-0,5}=-19\)

b

Để A nguyên => \(\dfrac{3x+2}{x-3}\) nguyên

\(\Leftrightarrow3x+2⋮\left(x-3\right)\\3\left(x-3\right)+11⋮\left(x-3\right) \)

Vì \(3\left(x-3\right)⋮\left(x-3\right)\) nên \(11⋮\left(x-3\right)\)

\(\Rightarrow\left(x-3\right)\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\\ \Rightarrow x\left\{4;2;-8;14\right\}\)

c

Để B nguyên => \(\dfrac{x^2+3x-7}{x+3}\) nguyên

\(\Rightarrow x\left(x+3\right)-7⋮\left(x+3\right)\)

\(\Rightarrow-7⋮\left(x+3\right)\\ \Rightarrow x+3\inƯ\left\{\pm1;\pm7\right\}\)

\(\Rightarrow x=\left\{-4;-11;-2;4\right\}\)

d

\(\left\{{}\begin{matrix}A.nguyên.\Leftrightarrow x=\left\{-8;2;4;14\right\}\\B.nguyên\Leftrightarrow x=\left\{-11;-4;-2;4\right\}\end{matrix}\right.\)

=> Để A, B cùng là số nguyên thì x = 4.

1.3+(-2)+x=5

-1+x=5

x=5-(-1)

x=6

Nhớ tick cho mình nha

\(\Leftrightarrow x^2-1\in\left\{-1;1;2\right\}\)

\(\Leftrightarrow x^2=0\)

hay x=0

Bài 3:

x - 14 = 3x + 18

3x - x = - 14 - 18

2x = - 32

x = - 32 : 2

x = - 16

Vậy x = - 16

Câu 3b:

2(x - 5) - 3(x -4) = - 6 + 15(-3)

2x - 10 - 3x + 12 = - 6 + 45

2x - 3x = - 6 + 45 + 10 - 12

- x = 39 + 10 - 12

- x = 49 - 12

- x = 37

x = 37 : (-1)

x = - 37

Vậy x = - 37