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Bài 2:
a: \(B=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{6}{3\left(x-2\right)}+\dfrac{1}{x-2}\right):\left(\dfrac{x^2-4+16-x^2}{x+2}\right)\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)
\(=\dfrac{x-x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{12}=\dfrac{-1}{6\left(x-2\right)}\)
b: Thay x=1/2 vào B, ta được:
\(B=\dfrac{-1}{6\cdot\left(\dfrac{1}{2}-2\right)}=\dfrac{-1}{6\cdot\dfrac{-3}{2}}=\dfrac{1}{9}\)
Thay x=-1/2 vào B, ta được:
\(B=\dfrac{-1}{6\cdot\left(-\dfrac{1}{2}-2\right)}=-\dfrac{1}{15}\)
c: Để B=2 thì \(\dfrac{-1}{6\left(x-2\right)}=2\)
=>6(x-2)=-1/2
=>x-2=-1/12
hay x=23/12
\(A=\frac{\left|x-1\right|+\left|x\right|-x}{3x^2+4x+1}=\frac{1-x-x-x}{3x^2+3x+x+1}=\frac{1-3x}{\left(x+1\right)\left(3x+1\right)}\)
\(B=\frac{\left|2x-1\right|+x}{3x^2-22x+7}=\frac{1-2x+x}{3x^2-21x-x+7}=\frac{1-x}{\left(x-7\right)\left(3x-1\right)}\)
a) \(\frac{5-2x}{3}+\frac{\left(x+1\right)\left(x-1\right)}{3x-1}=\frac{\left(x+2\right)\left(1-3x\right)}{9x-3}\)
<=> \(\frac{5-2x}{3}+\frac{\left(x+1\right)\left(x-1\right)}{3x-1}=-\frac{\left(x+3\right)\left(3x-1\right)}{3\left(3x-1\right)}\)
<=> \(\frac{5-2x}{3}+\frac{\left(x+1\right)\left(x-1\right)}{3x-1}=-\frac{x+2}{3}\)
<=> (5 - 2x)(3x - 1) + 3(x + 1)(x - 1) = -(x + 2)(3x - 1)
<=> 15x - 5 - 6x^2 + 2x + 3x^2 - 3x + 3x - 3 = -3x^2 - 6x + x + 2
<=> 17x - 8 = -5x + 2
<=> 17x - 8 + 5x = 2
<=> 22x - 8 = 2
<=> 22x = 2 + 8
<=> 22x = 10
<=> x = 10/22 = 5/11
b) \(\frac{2}{x-3}+\frac{x-5}{x-1}=1\)
<=> 2(x - 1) + (x - 5)(x - 3) = (x - 3)(x - 1)
<=> 2x - 2 + x^2 - 3x - 5x + 15 = x^2 - x - 3x + 3
<=> -6x + 13 = -4x + 3
<=> -6x + 13 + 4x = 3
<=> -2x + 13 = 3
<=> -2x = 3 - 13
<=> -2x = -10
<=> x = 5
a) \(\frac{5-2x}{3}+\frac{\left(x-1\right)\left(x+1\right)}{3x-1}=\frac{\left(x+2\right)\left(1-3x\right)}{9x-3}\left(x\ne\frac{1}{3}\right)\)
\(\Leftrightarrow\frac{5-2x}{3}+\frac{\left(x-1\right)\left(x+1\right)}{3x-1}-\frac{\left(x+2\right)\left(1-3x\right)}{3\left(3x-1\right)}=0\)
<=> \(\frac{\left(5-2x\right)\left(3x-1\right)}{3\left(3x-1\right)}+\frac{3\left(x^2-1\right)}{3\left(3x-1\right)}-\frac{x-3x^2+2-6x}{3\left(3x-1\right)}=0\)
<=> \(\frac{15x-5-6x^2+2x}{3\left(3x-1\right)}+\frac{3x^2-3}{3\left(3x-1\right)}-\frac{-3x^2-5x+2}{3\left(3x-1\right)}=0\)
<=> \(\frac{-6x^2+17x-5}{3\left(3x-1\right)}+\frac{3x^2-3}{3\left(3x-1\right)}-\frac{-3x^2-5x+2}{3\left(3x-1\right)}=0\)
<=> \(\frac{-6x^2+17x-5+3x^2-3+3x^2+5x-2}{3\left(3x-1\right)}=0\)
<=> \(\frac{22x-10}{3\left(3x-1\right)}=0\)
=> 22x-10=0
<=> \(x=\frac{5}{11}\)(tmđk)
a) Đề ( \(x\ne\pm1\))
>\(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}=\frac{4}{\left(x+1\right)\left(x-1\right)}\\ \Leftrightarrow\left(x+1\right)^2-\left(x-1\right)^2=4\\ \Leftrightarrow\left(x+1-x+1\right)\left(x+1+x-1\right)=4\\ \Leftrightarrow2.2x=4\Leftrightarrow x=1\left(kothỏa\right)\)
Vậy \(S=\varnothing\)
b) đề \(\left(x\ne-\frac{1}{2},\frac{1}{2}\right)\)
\(\frac{32x^2}{12\left(1-2x\right)\left(1+2x\right)}=\frac{-8x\left(1+2x\right)}{12\left(1-2x\right)\left(1+2x\right)}-\frac{3\left(1+8x\right)\left(1-2x\right)}{12\left(1-2x\right)\left(1+2x\right)}\\ \Leftrightarrow32x^2=-8x-16x^2-3-12x+48x^2\\ \Leftrightarrow20x+3=0\Leftrightarrow x=\frac{20}{3}\left(thỏadk\right)\)
Vậy \(S=\left\{\frac{20}{3}\right\}\)
a) Ta có: \(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow\frac{2\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}-\frac{x+1}{\left(x+1\right)\left(x-2\right)}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow\frac{x-5}{\left(x+1\right)\left(x-2\right)}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)\(\Rightarrow x-5=3x-11\Rightarrow x-3x=-11+5\Rightarrow-2x=-6\Rightarrow x=3\)
b)Ta có: \(\frac{15-6x}{3}>5\)
\(\Rightarrow15-6x>15\)
\(\Rightarrow6x< 0\)
\(\Rightarrow x< 0\).
Kb với mình nha!
a) x=3
b) x<0