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a)2x-5/x+5=3=>2x-5=3(x+5)=3x+15
=>2x=3x+20=>x=-20
b)(x^2-6)/x=x+3/2
=>(x^2-6)/x - x=3/2
=>-6/x[quy đồng]=3/2
=>x=-4
c)Để (x^2+2x)−(3x+6)/x−3=0
thì (x^2+2x)−(3x+6)=0
=x(x+2)-3(x+2)=(x-3)(x+2)=0
=>x=3 hoặc x=-2
Mà ở mẫu có x-3 nếu x=3 thì mẫu =0=>loại
Vậy x=2
d)5/3x+2=2x−1
=>5=(3x+2)(2x-1)
Tìm ước của 5 rùi thay vào 3x+2 và 2x-1 rùi tìm x,cái đó dễ nên bn tự lm nhé
e)
(2x−1/x−1)+1=1/x−1
=>1/x-1-2x-1/x-1=1
=>-2x/x-1=1
=>-2x=x-1
=>x=1/3
g)(x+3/x+1)+(x−2/x)=2
=>quy đồng rùi tính và tìm x nhé bn,mk mỏi tay rùi
nhớ tick cho mk nha,mk siêng lắm ms ghi cho bn nhiều thế này nè,nhớ tick nha,thanks
a) \(\frac{2x-5}{x+5}=3\)
\(\Leftrightarrow2x-5=3\left(x+5\right)\)
\(\Leftrightarrow2x-5=3x+15\)
\(\Leftrightarrow2x-3x=15+5\)
\(\Leftrightarrow-x=20\\ \)
\(\Leftrightarrow x=-20\)
b) \(\frac{x^2-6}{x}=x+\frac{3}{2}\)
\(\Leftrightarrow\frac{x^2-6}{x}=\frac{2x+3}{2}\)
\(\Leftrightarrow2\left(x^2-6\right)=x\left(2x+3\right)\)
\(\Leftrightarrow2x^2-12=2x^2+3x\)
\(\Leftrightarrow3x=-12\)
\(\Leftrightarrow x=-4\)
c) \(\frac{\left(x^2+2x\right)-\left(3x+6\right)}{x-3}=0\)
\(\Leftrightarrow\frac{x\left(x+2\right)-3\left(x+2\right)}{x-3}=0\)
\(\Leftrightarrow\frac{\left(x+2\right)\left(x-3\right)}{x-3}=0\)
\(\Leftrightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
d) \(\frac{5}{3x+2}=2x-1\)
\(\Leftrightarrow5=\left(2x-1\right)\left(3x+2\right)\)
\(\Leftrightarrow5=6x^2+x-2\)
\(\Leftrightarrow6x^2+x-7=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}1\\\frac{-7}{6}\end{array}\right.\)
e) \(\frac{2x-1}{x-1}+1=\frac{1}{x-1}\)
\(\Leftrightarrow2x-1+x-1=1\)
\(\Leftrightarrow3x=3\)
\(\Leftrightarrow x=1\)
g) \(\frac{x+3}{x+1}+\frac{x-2}{x}=2\)
\(\Leftrightarrow\frac{x\left(x+3\right)}{x\left(x+1\right)}+\frac{\left(x-2\right)\left(x+1\right)}{x\left(x+1\right)}=\frac{2x\left(x+1\right)}{x\left(x+1\right)}\)
\(\Leftrightarrow x\left(x+3\right)+\left(x-2\right)\left(x+1\right)=2x\left(x+1\right)\)
\(\Leftrightarrow x^2+3x+x^2-x-2=2x^2+2x\)
\(\Leftrightarrow2x-2x-2=0\)
\(\Leftrightarrow-2=0\) \(\Rightarrow\)Phương trình vô nghiệm
a) ĐKXĐ: x≠-5
Ta có: \(\frac{2x-5}{x+5}=3\)
\(\Leftrightarrow\frac{2x-5}{x+5}-3=0\)
\(\Leftrightarrow\frac{2x-5}{x+5}-\frac{3\left(x+5\right)}{x+5}=0\)
\(\Leftrightarrow2x-5-3\left(x+5\right)=0\)
\(\Leftrightarrow2x-5-3x-15=0\)
\(\Leftrightarrow-x-20=0\)
\(\Leftrightarrow-x=20\)
\(\Leftrightarrow x=-20\)(tmđk)
Vậy: x=-20
b) ĐKXĐ: x≠1;x≠-1
Ta có: \(\frac{2}{x-1}=\frac{6}{x+1}\)
\(\Leftrightarrow\frac{2}{x-1}-\frac{6}{x+1}=0\)
\(\Leftrightarrow\frac{2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{6\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\)
\(\Leftrightarrow2\left(x+1\right)-6\left(x-1\right)=0\)
\(\Leftrightarrow2x+2-6x+6=0\)
\(\Leftrightarrow-4x+8=0\)
\(\Leftrightarrow-4x=-8\)
\(\Leftrightarrow x=2\)(tmđk)
Vậy: x=2
c) ĐKXĐ: x≠1;x≠-1
Ta có: \(\frac{2x+1}{x-1}=\frac{5\left(x-1\right)}{x+1}\)
\(\Leftrightarrow\frac{2x+1}{x-1}-\frac{5\left(x-1\right)}{x+1}=0\)
\(\Leftrightarrow\frac{\left(2x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{5\left(x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x+1\right)-5\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow2x^2+2x+x+1-5x^2+10x-5=0\)
\(\Leftrightarrow-3x^2+13x-4=0\)
\(\Leftrightarrow-3x^2+x+12x-4=0\)
\(\Leftrightarrow x\left(-3x+1\right)+4\left(3x-1\right)=0\)
\(\Leftrightarrow x\left(1-3x\right)-4\left(1-3x\right)=0\)
\(\Leftrightarrow\left(1-3x\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}1-3x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=1\\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=4\end{matrix}\right.\)(thỏa mãn điều kiện)
Vậy: \(x\in\left\{\frac{1}{3};4\right\}\)
d) ĐKXĐ: x≠1;x≠-1
Ta có: \(\frac{x}{x-1}-\frac{2x}{x^2-1}=0\)
\(\Leftrightarrow\frac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{2x}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x\left(x+1\right)-2x=0\)
\(\Leftrightarrow x^2+x-2x=0\)
\(\Leftrightarrow x^2-x=0\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=1\left(ktm\right)\end{matrix}\right.\)
Vậy: x=0
e) ĐKXĐ: x≠2
Ta có: \(\frac{1}{x-2}+3=\frac{x-3}{2-x}\)
⇔\(\frac{1}{x-2}+3-\frac{x-3}{2-x}=0\)
⇔\(\frac{1}{x-2}+3+\frac{x-3}{x-2}=0\)
⇔\(\frac{1}{x-2}+\frac{3\left(x-2\right)}{x-2}+\frac{x-3}{x-2}=0\)
\(\Leftrightarrow1+3\left(x-2\right)+x-3=0\)
\(\Leftrightarrow1+3x-6+x-3=0\)
\(\Leftrightarrow4x-8=0\)
\(\Leftrightarrow4x=8\)
\(\Leftrightarrow x=2\)(không thỏa mãn)
Vậy: x∈∅
f) ĐKXĐ: \(x\ne\pm2\)
Ta có: \(\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)
⇔\(\frac{x+1}{x-2}+\frac{x-1}{x+2}-\frac{2\left(x^2+2\right)}{x^2-4}=0\)
⇔\(\frac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{2x^2+4}{\left(x+2\right)\left(x-2\right)}=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)+\left(x-1\right)\left(x-2\right)-2x^2-4=0\)
\(\Leftrightarrow x^2+2x+x+2+x^2-2x-x+2-2x^2-4=0\)
\(\Leftrightarrow0=0\)
Vậy: x∈R
g) ĐKXĐ: \(x\ne\pm2\)
Ta có: \(\frac{x+2}{x-2}+\frac{1}{x+2}=\frac{x\left(x-5\right)}{x^2-4}\)
⇔\(\frac{x+2}{x-2}+\frac{1}{x+2}-\frac{x\left(x-5\right)}{\left(x-2\right)\left(x+2\right)}=0\)
⇔\(\frac{\left(x+2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x+2\right)\left(x-2\right)}-\frac{x^2-5x}{\left(x-2\right)\left(x+2\right)}=0\)
⇔\(\left(x+2\right)^2+x-2-x^2+5x=0\)
\(\Leftrightarrow x^2+4x+4+x-2-x^2+5x=0\)
\(\Leftrightarrow10x-2=0\)
\(\Leftrightarrow10x=2\)
\(\Leftrightarrow x=\frac{2}{10}=\frac{1}{5}\)(thỏa mãn)
Vậy: \(x=\frac{1}{5}\)
\(b.\frac{12}{x^2-4}-\frac{x+1}{x-2}+\frac{x+7}{x+2}=0\left(dkxd:x\ne\pm2\right)\\ \Leftrightarrow\frac{12}{x^2-4}-\frac{\left(x+1\right)\left(x+2\right)}{x^2-4}+\frac{\left(x+7\right)\left(x-2\right)}{x^2-4}=0\\\Leftrightarrow 12-x^2-3x-2+x^2+5x-14=0\\ \Leftrightarrow2x-4=0\\\Leftrightarrow 2\left(x-2\right)=0\\\Leftrightarrow x-2=0\\\Leftrightarrow x=2\left(ktmdk\right)\)
Vô nghiệm
\(a.\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{16}{x^2-1}\left(dkxd:x\ne\pm1\right)\\\Leftrightarrow \frac{\left(x+1\right)^2}{x^2-1}-\frac{\left(x-1\right)^2}{x^2-1}=\frac{16}{x^2-1}\\\Leftrightarrow \left(x+1\right)^2-\left(x-1\right)^2=16\\\Leftrightarrow \left(x+1-x+1\right)\left(x+1+x-1\right)-16=0\\\Leftrightarrow 4x-16=0\\\Leftrightarrow 4\left(x-4\right)=0\\\Leftrightarrow x-4=0\\ \Leftrightarrow x=4\left(tmdk\right)\)
a) ĐKXĐ : \(x\ne-2;x\ne5\)
\(\frac{7}{x+2}=\frac{3}{x-5}\)
<=> 3(x + 2) = 7(x - 5)
<=> 3x + 6 = 7x - 35
<=> 4x = 41
<=>x = 41/4 (tm)
Vậy x = 41/4 là ngiệm phương trình
b) ĐKXĐ \(x\ne\pm3\)
\(\frac{2x-1}{x+3}=\frac{2x}{x-3}\)
<=> \(\frac{\left(2x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\frac{2x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)
<=> (2x - 1)(x - 3) = 2x(x + 3)
<=> 2x2 - 7x + 3 = 2x2 + 6x
<=> 13x = 3
<=> x = 3/13 (tm)
Vậy x = 3/13 là nghiệm phương trình
c) ĐKXĐ : \(x\ne-7;x\ne1,5\)
Khi đó \(\frac{3x-2}{x+7}=\frac{6x+1}{2x-3}\)
<=> \(\frac{\left(3x-2\right)\left(2x-3\right)}{\left(x+7\right)\left(2x-3\right)}=\frac{\left(6x+1\right)\left(x+7\right)}{\left(x+7\right)\left(2x-3\right)}\)
<=> (3x - 2)(2x - 3) = (6x + 1)(x + 7)
<=> 6x2 - 13x + 6 = 6x2 + 43x + 7
<=> 56x = -1
<=> x = -1/56 (tm)
Vậy x = -1/56 là nghiệm phương trình
d) ĐKXĐ : \(x\ne\pm1\)
Khi đó \(\frac{2x+1}{x-1}=\frac{5\left(x-1\right)}{x+1}\)
<=> \(\frac{\left(2x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{5\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\)
<=> (2x + 1)(x + 1) = 5(x - 1)2
<=> 2x2 + 3x + 1 = 5x2 - 10x + 5
<=> 3x2 - 13x + 4 = 0
<=> 3x2 - 12x - x + 4 = 0
<=> 3x(x - 4) - (x - 4) = 0
<=> (3x - 1)(x - 4) = 0
<=> \(\orbr{\begin{cases}3x-1=0\\x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\left(tm\right)\\x=4\left(tm\right)\end{cases}}\)
Vậy x \(\in\left\{\frac{1}{3};4\right\}\)là nghiệm phương trình
e) ĐKXĐ : \(x\ne1\)
Khi đó \(\frac{4x-5}{x-1}=2+\frac{x}{x-1}\)
<=> \(\frac{3x-5}{x-1}=2\)
<=> 3x - 5 = 2(x - 1)
<=> 3x - 5 = 2x - 2
<=> x = 3 (tm)
Vậy x = 3 là nghiệm phương trình
f) ĐKXĐ : \(x\ne-1\)
\(\frac{1-x}{x+1}+3=\frac{2x+3}{x+1}\)
<=> \(\frac{3x+2}{x+1}=3\)
<=> 3x + 2 = 3(x + 1)
<=> 3x + 2 = 3x + 3
<=> 0x = 1
<=> \(x\in\varnothing\)
Vậy tập nghiệm phương trình S = \(\varnothing\)
g) ĐKXĐ : \(x\ne2\)
Khi đó \(\frac{1}{x-2}+3=\frac{x-3}{2-x}\)
<=>\(\frac{x-2}{x-2}=3\)
<=> (x - 2) = 3(x - 2)
<=> x - 2 = 3x - 6
<=> -2x = -4
<=> x = 2 (loại)
Vậy tập nghiệm phương trình S = \(\varnothing\)
h) ĐKXĐ : \(x\ne7\)
Khi đó \(\frac{1}{7-x}=\frac{x-8}{x-7}-8\)
<=> \(\frac{x-7}{x-7}=8\)
<=> x - 7 = 8(x - 7)
<=> x - 7 = 8x - 56
<=> 7x = 49
<=> x = 7 (loại)
Vậy tập nghiệm phương trình S = \(\varnothing\)
i) ĐKXĐ : \(x\ne0;x\ne6\)
Ta có : \(\frac{x+6}{x}=\frac{1}{2}+\frac{15}{2\left(x-6\right)}\)
<=> \(\frac{x+6}{x}-\frac{15}{2\left(x-6\right)}=\frac{1}{2}\)
<=> \(\frac{2\left(x+6\right)\left(x-6\right)}{2x\left(x-6\right)}-\frac{15x}{2x\left(x-6\right)}=\frac{1}{2}\)
<=> \(\frac{2x^2-72-15x}{2x\left(x-6\right)}=\frac{1}{2}\)
<=> 4x2 - 144 - 30x = 2x(x - 6)
<=> 2x2 - 18x - 144 = 0
<=> x2 - 9x - 72 = 0
<=> x2 - 9x + 81/4 - 72- 81/4 = 0
<=> \(\left(x-\frac{9}{2}\right)^2-\frac{369}{4}=0\)
<=> \(\left(x-\frac{9}{2}+\sqrt{\frac{369}{4}}\right)\left(x-\frac{9}{2}-\sqrt{\frac{369}{4}}\right)=0\)
<=> \(\orbr{\begin{cases}x=\frac{9}{2}-\sqrt{\frac{369}{4}}\\x=\frac{9}{2}+\sqrt{\frac{369}{4}}\end{cases}}\)(tm)
Vậy x \(\in\left\{\frac{9}{2}-\sqrt{\frac{369}{4}};\frac{9}{2}+\sqrt{\frac{369}{4}}\right\}\)
a, \(\frac{x-1}{x+2}+1=\frac{1}{x-2}\)
ĐKXĐ: x + 2 \(\ne\) 0 và x - 2 \(\ne\) 0
\(\Rightarrow\) x \(\ne\) \(\pm\) 2
b, \(\frac{x-1}{1-2x}=1\)
ĐKXĐ: 1 - 2x \(\ne\) 0
\(\Leftrightarrow\) x \(\ne\) \(\frac{1}{2}\)
Bài 2:
a, \(\frac{x+2}{x}=\frac{2x+3}{x-2}\) (ĐKXĐ: x \(\ne\) 0; x \(\ne\) 2)
\(\Leftrightarrow\) \(\frac{\left(x+2\right)\left(x-2\right)}{x\left(x-2\right)}=\frac{x\left(2x+3\right)}{x\left(x-2\right)}\)
\(\Rightarrow\) (x + 2)(x - 2) = x(2x + 3)
\(\Leftrightarrow\) x2 - 4 = 2x2 + 3x
\(\Leftrightarrow\) x2 - 2x2 - 3x = 4
\(\Leftrightarrow\) -x2 - 3x = 4
\(\Leftrightarrow\) -x2 - 3x - 4 = 0
\(\Leftrightarrow\) -(x2 + 3x + 4) = 0
\(\Leftrightarrow\) x2 + 3x + 4 = 0
\(\Leftrightarrow\) x2 + 3x + \(\frac{9}{4}\) + \(\frac{7}{4}\) = 0
\(\Leftrightarrow\) (x + \(\frac{3}{2}\))2 + \(\frac{7}{4}\) = 0
Vì (x + \(\frac{3}{2}\))2 + \(\frac{7}{4}\) > 0 với mọi x
\(\Rightarrow\) Pt vô nghiệm
Vậy S = \(\varnothing\)
b, \(\frac{2x+5}{2x}-\frac{x}{x+5}=0\) (ĐKXĐ: x \(\ne\) 0; x \(\ne\) -5)
\(\Leftrightarrow\) \(\frac{\left(2x+5\right)\left(x+5\right)}{2x\left(x+5\right)}-\frac{2x^2}{2x\left(x+5\right)}=0\)
\(\Rightarrow\) (2x + 5)(x + 5) - 2x2 = 0
\(\Leftrightarrow\) 2x2 + 10x + 5x + 25 - 2x2 = 0
\(\Leftrightarrow\) 15x + 25 = 0
\(\Leftrightarrow\) x = \(\frac{-5}{3}\) (TMĐKXĐ)
Vậy S = {\(\frac{-5}{3}\)}
c, \(\frac{x+1}{3-x}=2\)
\(\Leftrightarrow\) \(\frac{x+1}{3-x}=\frac{2\left(3-x\right)}{3-x}\) (ĐKXĐ: x \(\ne\) 3)
\(\Rightarrow\) x + 1 = 2(3 - x)
\(\Leftrightarrow\) x + 1 - 2(3 - x) = 0
\(\Leftrightarrow\) x + 1 - 6 + 2x = 0
\(\Leftrightarrow\) 3x - 5 = 0
\(\Leftrightarrow\) x = \(\frac{5}{3}\) (TMĐKXĐ)
Vậy S = {\(\frac{5}{3}\)}
d, \(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{16}{x^2-1}\) (ĐKXĐ: x \(\ne\) \(\pm\) 1)
\(\Leftrightarrow\) \(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\frac{16}{\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow\) (x + 1)2 - (x - 1)2 = 16
\(\Leftrightarrow\) (x + 1 - x + 1)(x + 1 + x - 1) = 16
\(\Leftrightarrow\) 4x = 16
\(\Leftrightarrow\) x = 4 (TMĐKXĐ)
Vậy S = {4}
Chúc bn học tốt!!
kcj