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a)
Ta có:
cho nên x3 – 1 ≠ 0 khi x – 1 ≠ 0⇔ x ≠ 1
Vậy ĐKXĐ: x ≠ 1
Khử mẫu ta được:
b: Đặt \(x^2-6x-2=a\)
Theo đề, ta có: \(a+\dfrac{14}{a+9}=0\)
=>(a+2)(a+7)=0
\(\Leftrightarrow\left(x^2-6x\right)\left(x^2-6x+5\right)=0\)
=>x(x-6)(x-1)(x-5)=0
hay \(x\in\left\{0;1;6;5\right\}\)
c: \(\Leftrightarrow\dfrac{-8x^2}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(2x+1\right)}\)
\(\Leftrightarrow-32x^2=8x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)\)
\(\Leftrightarrow-32x^2=16x^2+8x-3\left(16x^2-8x+2x-1\right)\)
\(\Leftrightarrow-48x^2=8x-48x^2+18x+3\)
=>26x=-3
hay x=-3/26
a: =>10x-4=15-9x
=>19x=19
hay x=1
b: \(\Leftrightarrow3\left(10x+3\right)=36+4\left(8x+6\right)\)
=>30x+9=36+32x+24
=>30x-32x=60-9
=>-2x=51
hay x=-51/2
c: \(\Leftrightarrow2x+\dfrac{6}{5}=5-\dfrac{13}{5}-x\)
=>3x=6/5
hay x=2/5
d: \(\Leftrightarrow\dfrac{7x}{8}-\dfrac{5\left(x-9\right)}{1}=\dfrac{20x+1.5}{6}\)
\(\Leftrightarrow21x-120\left(x-9\right)=4\left(20x+1.5\right)\)
=>21x-120x+1080=80x+60
=>-179x=-1020
hay x=1020/179
e: \(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)
=>35x-5+60x=96-6x
=>95x+6x=96+5
=>x=1
f: \(\Leftrightarrow6\left(x+4\right)+30\left(-x+4\right)=10x-15\left(x-2\right)\)
=>6x+24-30x+120=10x-15x+30
=>-24x+96=-5x+30
=>-19x=-66
hay x=66/19
1: \(\Leftrightarrow x^2+6x+9-6x+3>x^2-4x\)
=>-4x<12
hay x>-3
2: \(\Leftrightarrow6+2x+2>2x-1-12\)
=>8>-13(đúng)
4: \(\dfrac{2x+1}{x-3}\le2\)
\(\Leftrightarrow\dfrac{2x+1-2x+6}{x-3}< =0\)
=>x-3<0
hay x<3
6: =>(x+4)(x-1)<=0
=>-4<=x<=1
\(1,\) thiếu đề
\(2,\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)
\(\Leftrightarrow\dfrac{5\left(5x+2\right)}{30}-\dfrac{10\left(8x-1\right)}{30}=\dfrac{6\left(4x+2\right)}{30}-\dfrac{150}{30}\)
\(\Leftrightarrow5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-150\)
\(\Leftrightarrow25x+10-80x+10=24x+12-150\)
\(\Leftrightarrow-55x+20=24x-138\)
\(\Leftrightarrow24x-138+55x-20=0\)
\(\Leftrightarrow79x-158=0\)
\(\Leftrightarrow x=2\)
\(3,ĐKXĐ:\left\{{}\begin{matrix}x\ne1\\x\ne-1\\x\ne3\end{matrix}\right.\\ \dfrac{x}{2x-6}+\dfrac{x}{2x-2}=\dfrac{-2x}{\left(x+1\right)\left(3-x\right)}\)
\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x-1\right)}+\dfrac{2x}{\left(x+1\right)\left(3-x\right)}=0\)
\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x-1\right)}-\dfrac{2x}{\left(x+1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow x\left(\dfrac{1}{2\left(x-3\right)}+\dfrac{1}{2\left(x-1\right)}-\dfrac{2}{\left(x+1\right)\left(x-3\right)}\right)=0\)
\(\Leftrightarrow x\left(\dfrac{\left(x-1\right)\left(x+1\right)}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}+\dfrac{\left(x-3\right)\left(x+1\right)}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}-\dfrac{4\left(x-1\right)}{2\left(x+1\right)\left(x-3\right)\left(x-1\right)}\right)=0\)
\(\Leftrightarrow x\left(\dfrac{x^2-1}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}+\dfrac{x^2-2x-3}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}-\dfrac{4x-4}{2\left(x+1\right)\left(x-3\right)\left(x-1\right)}\right)=0\)
\(\Leftrightarrow x.\dfrac{x^2-1+x^2-2x-3-4x+4}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x.\dfrac{2x^2-6x}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x.\dfrac{2x\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x.\dfrac{x}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x=0\)
a) \(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x+3}=1-\dfrac{4}{\left(x-1\right)\left(x+3\right)}\)
ĐKXĐ \(x-1\ne0\) hoặc \(x+3\ne0\)
\(\Rightarrow x\ne1\) và \(x\ne-3\)
\(\dfrac{\left(3x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\dfrac{\left(2x+5\right)\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}=\dfrac{\left(x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\dfrac{4}{\left(x-1\right)\left(x+3\right)}\)
\(\Leftrightarrow\left(3x-1\right)\left(x+3\right)-\left(2x+5\right)\left(x-1\right)=\left(x-1\right)\left(x+3\right)-4\)
\(\Leftrightarrow3x^2+9x-x-3-\left(2x^2-2x+5x-5\right)=x^2+3x-x-3-4\)
\(\Leftrightarrow3x^2+9x-x-3-2x^2+2x-5x+5=x^2+3x-x-3-4\)
\(\Leftrightarrow9x-x+2x-5x-3x+x=3-5-3-4\)
\(\Leftrightarrow3x=-9\)
\(\Leftrightarrow x=-3\) (không thỏa ĐK)
Vậy PTVN
b) \(\dfrac{13}{\left(x-3\right)\left(2x+7\right)}+\dfrac{1}{2x+7}=\dfrac{6}{\left(x-3\right)\left(x+3\right)}\)
ĐKXĐ: \(x-3\ne0\Rightarrow x\ne3\)
\(x+3\ne0\Rightarrow x\ne-3\)
\(2x+7\ne0\Rightarrow2x\ne-7\Rightarrow x\ne\dfrac{-7}{2}\)
\(\dfrac{13\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}+\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}=\dfrac{6\left(2x+7\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}\)
\(\Leftrightarrow13\left(x+3\right)+\left(x-3\right)\left(x+3\right)=6\left(2x+7\right)\)
\(\Leftrightarrow13x+39+x^2+3x-3x-9=12x+42\)
\(\Leftrightarrow x^2+x-12=0\)
\(\Leftrightarrow x^2-3x+4x-12=0\)
\(\Leftrightarrow x\left(x-3\right)+4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+4\right)=0\)
\(\left\{{}\begin{matrix}x-3=0\\x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3\left(KTĐK\right)\\x=-4\left(TĐK\right)\end{matrix}\right.\)
Vậy S={-4}
a) \(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x+3}=1-\dfrac{4}{\left(x-1\right)\left(x+3\right)}\) ( đk: x ≠ 1 ; x ≠ -3 )
\(\Leftrightarrow\left(3x-1\right)\left(x+3\right)-\left(2x+5\right)\left(x-1\right)=\left(x-1\right)\left(x+3\right)-4\)
\(\Leftrightarrow3x^2+9x-x-3-2x^2+2x-5x+5=x^2+3x-x-3-4\)
\(\Leftrightarrow3x=-9\)
\(\Rightarrow x=-3\left(KTM\right)\)
S = ∅
b) \(\dfrac{13}{\left(x-3\right)\left(2x+7\right)}+\dfrac{1}{2x+7}=\dfrac{6}{\left(x-3\right)\left(x+3\right)}\)
( đk: x ≠ ± 3 ; x ≠ \(\dfrac{-7}{2}\) )
\(\Leftrightarrow13\left(x+3\right)+\left(x-3\right)\left(x+3\right)=6\left(2x+7\right)\)
\(\Leftrightarrow13x+39+x^2-9=12x+42\)
\(\Leftrightarrow x^2-x-12=0\)
\(\Leftrightarrow x^2+3x-4x-12=0\)
\(\Leftrightarrow x\left(x+3\right)-4\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+3\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-4=0\\x-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4\left(TM\right)\\x=3\left(KTM\right)\end{matrix}\right.\)
S = \(\left\{4\right\}\)
4, \(\Leftrightarrow4x+4+9\left(2x+1\right)=4x+6\left(x+1\right)+7+12x\)
\(\Leftrightarrow22x+13=22x+13\)vậy pt có vô số nghiệm
5, \(\dfrac{2x}{3}+\dfrac{2x-1}{6}=4-\dfrac{x}{3}\Rightarrow4x+2x-1=24-2x\)
\(\Leftrightarrow8x=25\Leftrightarrow x=\dfrac{25}{8}\)
6, \(\dfrac{x-1}{2}+\dfrac{x-1}{4}=1-\dfrac{2\left(x-1\right)}{3}\Rightarrow6x-6+3x-3=12-8\left(x-1\right)\)
\(\Leftrightarrow9x-9=20-8x\Leftrightarrow17x=29\Leftrightarrow x=\dfrac{29}{17}\)
