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a) \(A=\left(\dfrac{1}{3}+\dfrac{3}{x^2-3x}\right):\left(\dfrac{x^2}{27-3x^2}+\dfrac{1}{x+3}\right)\)
\(\Rightarrow A=\dfrac{x^2-3x+9}{3\left(x^2-3x\right)}:\left(\dfrac{x^2}{3\left(9-x^2\right)}+\dfrac{1}{x+3}\right)\)
\(\Rightarrow A=\dfrac{x^2-3x+9}{3x.\left(x-3\right)}:\left(\dfrac{x^2}{3.\left(3-x\right).\left(3+x\right)}+\dfrac{1}{x+3}\right)\)
\(\Rightarrow A=\dfrac{x^2-3x+9}{3x.\left(x-3\right)}:\dfrac{x^2+3.\left(3-x\right)}{3.\left(3-x\right).\left(3+x\right)}\)
\(\Rightarrow A=\dfrac{x^2-3x+9}{3x.\left(x-3\right)}:\dfrac{x^2+9-3x}{3.\left(3-x\right).\left(3+x\right)}\)
\(\Rightarrow A=\dfrac{x^2-3x+9}{3x.\left(x-3\right)}.\dfrac{3.\left(3x-x\right).\left(3+x\right)}{x^2+9-3x}\)
\(\Rightarrow A=\dfrac{1}{x.\left(x-3\right)}.\left(-\left(x-3\right)\right).\left(3+x\right)\)
\(\Rightarrow A=\dfrac{1}{x}.\left(-1\right).\left(3+x\right)\)
\(\Rightarrow A=-\dfrac{1}{x}.\left(3+x\right)\)
a: Để A nguyên thì \(x-3\in\left\{1;-1;2;-2\right\}\)
=>\(x\in\left\{4;2;5;1\right\}\)
b: Để B nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{3;1;5;-1\right\}\)
c: Để C nguyên thì \(3x^2+2x-3x-2+3⋮3x+2\)
=>\(3x+2\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{-\dfrac{1}{3};-1;\dfrac{1}{3};-\dfrac{5}{3}\right\}\)
a: \(A=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{\left(x-2\right)}+\dfrac{1}{x+2}\right):\dfrac{x^2-4+10-x^2}{x+2}\)
\(=\dfrac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{6}\)
\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{6}=\dfrac{-1}{x-2}\)
b: |x|=1/2 khi x=1/2 hoặc x=-1/2
Khi x=1/2 thì \(A=\dfrac{-1}{\dfrac{1}{2}-2}=-1:\dfrac{-3}{2}=\dfrac{2}{3}\)
Khi x=-1/2 thì \(A=\dfrac{-1}{-\dfrac{1}{2}-2}=-1:\dfrac{-5}{2}=\dfrac{2}{5}\)
c: Để A=2 thì x-2=-1/2
hay x=3/2
d:Để A<0 thì x-2>0
hay x>2
a: \(M=\left(\dfrac{-\left(x+2\right)}{x-2}-\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{x+2}\right):\dfrac{x^2\left(x-3\right)}{x^3\left(2-x\right)}\)
\(=\dfrac{-x^2-4x-4-4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-x\left(x-2\right)}{x-3}\)
\(=\dfrac{-4x^2-8x}{x+2}\cdot\dfrac{-x}{x-3}=\dfrac{4x^2+8x}{x+2}\cdot\dfrac{x}{x-3}\)
\(=\dfrac{4x^2}{x-3}\)
b: Để M là số nguyên thì \(4x^2⋮x-3\)
\(\Leftrightarrow4x^2-36+36⋮x-3\)
\(\Leftrightarrow x-3\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;9;-9;12;-12;18;-18;36;-36\right\}\)
hay \(x\in\left\{4;5;1;0;7;-1;9;-3;12;-6;15;-9;21;-12;39;-33\right\}\)
a/ Để A ∈ Z
⇒ \(3x^2-9x+2\) ⋮ \(x-3\)
⇒ \(3x\left(x-3\right)+2\) ⋮ \(x-3\)
Vì \(3x\left(x-3\right)\) ⋮ \(x-3\)
⇒ \(2\) ⋮ \(x-3\)
⇒ \(x-3\inƯ_{\left(2\right)}\)
⇒ \(x-3\in\left\{1;2;-1;-2\right\}\)
⇒ \(x\in\left\{4;5;2;1\right\}\)
Vậy ...
b.
Ta có:
\(A=\dfrac{3n+9}{n-4}=\dfrac{3\left(n-4\right)+21}{n-4}=3+\dfrac{21}{n-4}\)
Để A thuộc Z
=> \(\dfrac{21}{n-4}\in Z\) ( n khác 4)
=> \(21⋮\left(n-4\right)\)
\(\Rightarrow n-4\inƯ\left(21\right)=\left\{21;-21;7;-7;3;-3\right\}\)
\(\Rightarrow n\in\left\{25;-17;11;-3;-1;1\right\}\) ( nhận)
Để A thuộc Z thì \(3x^2-x+1⋮3x+2\)
\(3x^2+2x-3x-2+3⋮3x+2\)
\(x\left(3x+2\right)-\left(3x+2\right)+3⋮3x+2\)
\(\left(3x+2\right)\left(x-1\right)+3⋮3x+2\)
Mà \(\left(3x+2\right)\left(x-1\right)⋮3x+2\)
\(\Rightarrow3⋮3x+2\)
\(\Rightarrow3x+2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
Ta có bảng :
Mà x thuộc Z => x = -1
Vậy x = -1
\(A=\dfrac{3x^2+2x-3x+1}{3x+2}=\dfrac{3x^2+2x-3x-2+3}{3x+2}\)
\(A=\dfrac{x\left(3x+2\right)-\left(3x+2\right)+3}{3x+2}=x-1+\dfrac{3}{3x+2}\in Z\)
\(\Rightarrow3x+2\inƯ\left(3\right)\)
Xét ước thôi