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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: A=[(3x^2+3-x^2+2x-1-x^2-x-1)/(x-1)(x^2+x+1)]*(x-2)/2x^2-5x+5
=(x^2+x+1)/(x-1)(x^2+x+1)*(x-2)/2x^2-5x+5
=(x-2)/(2x^2-5x+5)(x-1)
Câu 1 :
a) Rút gọn P :
\(P=\dfrac{x+1}{3x-x^2}:\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{12x^2}{x^2-9}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left[\dfrac{\left(3+x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{\left(3-x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{12x^2}{\left(3-x\right)\left(3+x\right)}\right]\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left(\dfrac{9+6x+x^2-9+6x-x^2-12x^2}{\left(3-x\right)\left(3+x\right)}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\dfrac{12x-12x^2}{\left(3-x\right)\left(x+3\right)}\)
\(P=\dfrac{x+1}{x\left(3-x\right)}.\dfrac{\left(3-x\right)\left(x+3\right)}{12x\left(1-x\right)}\)
\(P=\dfrac{\left(x+1\right)\left(x+3\right)}{12x^2\left(1-x\right)}\)
\(ĐKXĐ:x\ne0;x\ne\pm3\)
\(A=\left(\frac{1}{3}+\frac{3}{x\left(x-3\right)}\right):\left(\frac{x^2}{3\left(9-x^2\right)}+\frac{1}{x+3}\right)\\ =\frac{x^2-3x+9}{3x\left(x-3\right)}\cdot\frac{-3\left(x+3\right)\left(x-3\right)}{x^2-3x+9}\\ =\frac{-x-3}{x}\)
b) Ta có :
\(A=\frac{-x-3}{x}< 1\\ \Leftrightarrow\frac{-x-3}{x}-1< 0\\ \Leftrightarrow\frac{-x}{x}-\frac{3}{x}-1< 0\\ \Leftrightarrow-1-1-\frac{3}{x}< 0\\ \Leftrightarrow-2-\frac{3}{x}< 0\\ \Leftrightarrow\frac{-3}{x}< 2\\ \Leftrightarrow2x< -3\\ \Rightarrow x>\frac{-3}{2}=-1,5\)
Vậy để A < 1 thì x > 1,5 / x ≠ 0 ; x ≠ 3 ; x ≠ -3
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) \(A = \frac{2x^2 - 16x+43}{x^2-8x+22}\) = \(\frac{2(x^2-8x+22)-1}{x^2-8x+22}\) = \(2 - \frac{1}{x^2-8x+22}\)
Ta có : \(x^2-8x+22 \) = \(x^2-8x+16+6 = ( x-4)^2 +6 \)
Vì \((x-4)^2 \ge 0 \) với \( \forall x\in R\) Nên \(( x-4)^2 +6 \ge 6 \)
\(\Rightarrow \) \(x^2-8x+22 \) \( \ge 6\)\(\Rightarrow \) \(\frac{1}{x^2-8x+22} \) \(\le \frac{1}{6}\) \(\Rightarrow \) - \(\frac{1}{x^2-8x+22} \) \(\ge - \frac{1}{6}\)
\(\Rightarrow \) A = \(2 - \frac{1}{x^2-8x+22}\) \( \ge 2-\frac{1}{6}\) = \(\frac{11}{6}\) Dấu "=" xảy ra khi và chỉ khi x=4
Vậy GTNN của A = \(\frac{11}{6}\) khi và chỉ khi x=4
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 :
| 3x+2 | 1 | -1 | 3 | -3 |
| x | -1/3 | -1 | 1/3 | -5/3 |
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
a: \(A=\dfrac{2x-5+x^2-4+x^2-9}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x^2+2x-18}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{2\left(x+3\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x+6}{x-3}\)
b: Để A/2=x+3/x-3 là số nguyên thì \(x-3+6⋮x-3\)
=>\(x-3\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
hay \(x\in\left\{4;51;6;0;9;-3\right\}\)
c: Để A=1/x-1 thì \(\dfrac{2x+6}{x-3}=\dfrac{1}{x-1}\)
=>2x^2-2x+6x-6=x-3
=>2x^2+5x-6-x+3=0
=>2x^2+4x-3=0
hay \(x=\dfrac{-2\pm\sqrt{10}}{2}\)
a) \(A=\left(\dfrac{1}{3}+\dfrac{3}{x^2-3x}\right):\left(\dfrac{x^2}{27-3x^2}+\dfrac{1}{x+3}\right)\)
\(\Rightarrow A=\dfrac{x^2-3x+9}{3\left(x^2-3x\right)}:\left(\dfrac{x^2}{3\left(9-x^2\right)}+\dfrac{1}{x+3}\right)\)
\(\Rightarrow A=\dfrac{x^2-3x+9}{3x.\left(x-3\right)}:\left(\dfrac{x^2}{3.\left(3-x\right).\left(3+x\right)}+\dfrac{1}{x+3}\right)\)
\(\Rightarrow A=\dfrac{x^2-3x+9}{3x.\left(x-3\right)}:\dfrac{x^2+3.\left(3-x\right)}{3.\left(3-x\right).\left(3+x\right)}\)
\(\Rightarrow A=\dfrac{x^2-3x+9}{3x.\left(x-3\right)}:\dfrac{x^2+9-3x}{3.\left(3-x\right).\left(3+x\right)}\)
\(\Rightarrow A=\dfrac{x^2-3x+9}{3x.\left(x-3\right)}.\dfrac{3.\left(3x-x\right).\left(3+x\right)}{x^2+9-3x}\)
\(\Rightarrow A=\dfrac{1}{x.\left(x-3\right)}.\left(-\left(x-3\right)\right).\left(3+x\right)\)
\(\Rightarrow A=\dfrac{1}{x}.\left(-1\right).\left(3+x\right)\)
\(\Rightarrow A=-\dfrac{1}{x}.\left(3+x\right)\)