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a)\(2+\frac{3}{x-5}=1\)
\(\Rightarrow\frac{3}{x-5}=-1\)
\(\Rightarrow3=-x+5\)
\(\Leftrightarrow x+3=5\)
\(\Rightarrow x=2\)
a) \(\frac{4x-8}{2x^2+1}=0\)
\(\Rightarrow4x-8=0\left(2x^2+1\ne0\right)\)
\(\Leftrightarrow4x=8\)
\(\Leftrightarrow x=2\)
Vậy x=2
b)
\(\frac{x^2-x-6}{x-3}=0\)
\(\Leftrightarrow\frac{\left(x-3\right)\left(x+2\right)}{x-3}=0\)
\(\Rightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
Vậy x=-2
\(a,ĐKXĐ:x\ne\pm\frac{1}{2}\)
Ta có: \(\frac{2}{2x+1}-\frac{3}{2x-1}=\frac{4}{4x^2-1}\)
\(\Leftrightarrow2\left(2x-1\right)-3\left(2x+1\right)=4\)
\(\Leftrightarrow4x-2-6x-3=4\)
\(\Leftrightarrow-2x=9\)
\(\Leftrightarrow x=-\frac{9}{2}\)(Tm ĐKXĐ)
Vậy pt có nghiệm duy nhất \(x=-\frac{9}{2}\)
\(b,ĐKXĐ:x\ne\pm1;-3\)
Ta có: \(\frac{2x}{x+1}+\frac{18}{x^2+2x-3}=\frac{2x-5}{x+3}\)
\(\Leftrightarrow\frac{2x}{x+1}+\frac{18}{\left(x-1\right)\left(x+3\right)}=\frac{2x-5}{x+3}\)
\(\Leftrightarrow2x\left(x-1\right)\left(x+3\right)+18\left(x+1\right)=\left(2x-5\right)\left(x-1\right)\left(x+1\right)\)
\(\Leftrightarrow2x\left(x^2+2x-3\right)+18x+18=\left(2x-5\right)\left(x^2-1\right)\)
\(\Leftrightarrow2x^3+4x^2-6x+18x+18=2x^3-2x-5x^2+5\)
\(\Leftrightarrow9x^2+14x+13=0\)
\(\Leftrightarrow\left(9x^2+14x+\frac{49}{9}\right)+\frac{68}{9}=0\)
\(\Leftrightarrow\left(3x+\frac{7}{3}\right)^2+\frac{68}{9}=0\)
Pt vô nghiệm
\(c,ĐKXĐ:x\ne1\)
Ta có: \(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)
\(\Leftrightarrow x^2+x+1+2x^2-5=x-1\)
\(\Leftrightarrow3x^2=3\)
\(\Leftrightarrow x^2=1\)
\(\Leftrightarrow x=\pm1\)
Kết hợp vs ĐKXĐ được x = -1
Vậy pt có nghiệm duy nhất x = -1
làm lần lượt nha(bài nào k bt bỏ qua)
\(a,\frac{2}{2x+1}-\frac{3}{2x-1}=\frac{4}{4x^2-1}\)
\(\Rightarrow\frac{2\left(2x-1\right)-3\left(2x+1\right)}{4x^2-1}=\frac{4}{4x^2-1}\)
\(\Rightarrow-2x-5=4\)
\(\Rightarrow-2x=9\)
\(\Rightarrow x=\frac{9}{-2}\)
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) \(\frac{2x}{x-1}+\frac{4}{x^2+2x-3}=\frac{2x-5}{x+3}\)ĐKXĐ : \(x\ne1;-3\)
\(\Leftrightarrow\frac{2x\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}+\frac{4}{\left(x-1\right)\left(x+3\right)}=\frac{\left(2x-5\right)\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}\)
\(\Leftrightarrow\frac{2x^2+6x+4}{\left(x-1\right)\left(x+3\right)}=\frac{2x^2-7x+5}{\left(x-1\right)\left(x+3\right)}\)
\(\Rightarrow2x^2+6x+4=2x^2-7x+5\)
\(\Leftrightarrow2x^2+5x+4-2x^2+7x-5=0\)
\(\Leftrightarrow12x-1=0\)
\(\Leftrightarrow x=\frac{1}{12}\)( thỏa mãn ĐKXĐ )
b) c) tương tự
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