a)\(\dfrac{3x+1}{x^2+1}\ge0\)
b) A=\(\dfrac{6x}{2x-1}\) . Tìm các giá trị nguyên của x để A nhận giá trị nguyên
c) B=\(\dfrac{x+1}{3x-x^2}:\left(\dfrac{3+x}{3-x}-\dfrac{3+x}{3-x}-\dfrac{12x^2}{x^2-9}\right)\). Rút gọn B
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Những câu hỏi liên quan
5 tháng 1 2021
ĐKXĐ: \(x\notin\left\{2;-2;-1\right\}\)
a) Ta có: \(A=\left(\dfrac{x}{x^2-4}-\dfrac{4}{2-x}+\dfrac{1}{x+2}\right):\dfrac{3x+3}{x^2+2x}\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}+\dfrac{4\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{1}{x+2}\right):\dfrac{3\left(x+1\right)}{x\left(x+2\right)}\)
\(=\left(\dfrac{x+4x+8}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}\right)\cdot\dfrac{x\left(x+2\right)}{3\left(x+1\right)}\)
\(=\dfrac{5x+8+x-2}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{x\left(x+2\right)}{3\left(x+1\right)}\)
\(=\dfrac{6x+6}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{x\left(x+2\right)}{3\left(x+1\right)}\)
\(=\dfrac{6\left(x+1\right)}{x-2}\cdot\dfrac{x}{3\left(x+1\right)}\)
\(=\dfrac{2x}{x-2}\)
b) Để A nguyên thì \(2x⋮x-2\)
\(\Leftrightarrow2x-4+4⋮x-2\)
mà \(2x-4⋮x-2\)
nên \(4⋮x-2\)
\(\Leftrightarrow x-2\inƯ\left(4\right)\)
\(\Leftrightarrow x-2\in\left\{1;-1;2;-2;4;-4\right\}\)
\(\Leftrightarrow x\in\left\{3;1;4;0;6;-2\right\}\)
Kết hợp ĐKXĐ, ta được:
\(x\in\left\{0;1;3;4;6\right\}\)
Vậy: Khi \(x\in\left\{0;1;3;4;6\right\}\) thì A nguyên
16 tháng 9 2021
a: Ta có: \(A=\left(\dfrac{3x+3}{x-9}-\dfrac{2\sqrt{x}}{\sqrt{x}+3}-\dfrac{\sqrt{x}}{\sqrt{x}-3}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\dfrac{3x+3-2x+6\sqrt{x}-x-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{2\sqrt{x}-2-\sqrt{x}+3}\)
\(=\dfrac{3}{\sqrt{x}+3}\)
31 tháng 8 2023
a: \(P=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\cdot\dfrac{\sqrt{x}-7+\sqrt{x}+1}{\sqrt{x}+1}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\cdot\dfrac{2\sqrt{x}-6}{\sqrt{x}+1}\)
\(=\dfrac{-3\sqrt{x}-3}{\sqrt{x}+1}\cdot\dfrac{2}{\sqrt{x}+3}=-\dfrac{6}{\sqrt{x}+3}\)
b: P>=-1/2
=>P+1/2>=0
=>\(\dfrac{-6}{\sqrt{x}+3}+\dfrac{1}{2}>=0\)
=>\(\dfrac{-12+\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}>=0\)
=>căn x-9>=0
=>x>=81
c: căn x+3>=3
=>6/căn x+3<=6/3=2
=>-6/căn x+3>=-2
Dấu = xảy ra khi x=0
5 tháng 12 2021
\(a,A=\dfrac{x^2-6x+9-x^2+9}{x\left(x-3\right)}\cdot\dfrac{x}{2\left(x-1\right)}\\ A=\dfrac{-6x+18}{2\left(x-3\right)\left(x-1\right)}=\dfrac{-6\left(x-3\right)}{2\left(x-3\right)\left(x-1\right)}=\dfrac{-3}{x-1}\\ b,A\in Z\Leftrightarrow x-1\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow x\in\left\{-2;0;2;4\right\}\)
14 tháng 12 2018
a,ĐK: \(\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)
b, \(A=\left(\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right)\)
\(=\frac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\frac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)
\(=\frac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}.\frac{3x\left(x+3\right)}{-x^2+3x-9}=\frac{-3}{x-3}\)
c, Với x = 4 thỏa mãn ĐKXĐ thì
\(A=\frac{-3}{4-3}=-3\)
d, \(A\in Z\Rightarrow-3⋮\left(x-3\right)\)
\(\Rightarrow x-3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\Rightarrow x\in\left\{0;2;4;6\right\}\)
Mà \(x\ne0\Rightarrow x\in\left\{2;4;6\right\}\)
\(a.\dfrac{3x+1}{x^2+1}\ge0\)
Do : \(x^2+1>0\forall x\)
\(\Rightarrow3x+1\ge0\Leftrightarrow x\ge-\dfrac{1}{3}\)
KL ........
\(b.A=\dfrac{6x}{2x-1}=\dfrac{3\left(2x-1\right)+3}{2x-1}=3+\dfrac{3}{2x-1}\left(x\ne\dfrac{1}{2}\right)\)
Để : \(A\in Z\Leftrightarrow\dfrac{3}{2x-1}\in Z\Leftrightarrow2x-1\in\left\{\pm1;\pm3\right\}\)
\(\oplus2x-1=1\Leftrightarrow x=1\left(TM\right)\)
\(\oplus2x-1=-1\Leftrightarrow x=0\left(TM\right)\)
\(\oplus2x-1=3\Leftrightarrow x=2\left(TM\right)\)
\(\oplus2x-1=-3\Leftrightarrow x=-1\left(TM\right)\)
KL...........
\(c.B=\dfrac{x+1}{3x-x^2}:\left(\dfrac{3+x}{3-x}-\dfrac{3+x}{3-x}-\dfrac{12x^2}{x^2-9}\right)=\dfrac{x+1}{x\left(3-x\right)}:\dfrac{\left(3-x\right)\left(3+x\right)}{12x^2}=\dfrac{\left(x+1\right)\left(3+x\right)}{12x^3}\left(x\ne0;x\ne\pm3\right)\)