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2. \(\frac{1}{x-1}-\frac{7}{x-2}=\frac{1}{\left(x-1\right)\left(2-x\right)}\) (ĐKXĐ:\(x\ne1,x\ne2\))
\(\Leftrightarrow\frac{1}{x-1}+\frac{7}{2-x}=\frac{1}{\left(x-1\right)\left(2-x\right)}\)
\(\Leftrightarrow\frac{2-x+7\left(x-1\right)}{\left(x-1\right)\left(2-x\right)}=\frac{1}{\left(x-1\right)\left(2-x\right)}\)
\(\Rightarrow2-x+7\left(x-1\right)=1\)
\(\Leftrightarrow2-x+7x-7=1\)
\(\Leftrightarrow-x+7x=1-2+7\)
\(\Leftrightarrow6x=6\)
\(\Leftrightarrow x=1\) (Không thỏa mãn ĐKXĐ)
Vậy phương trình trên vô nghiệm
ko phan tich duoc nha bn
chuc bn hoc gioi
happy new year
a, \(\frac{6x+1}{x^2+7x+10}+\frac{5}{x-2}=\frac{3}{x-5}\)
\(11x^3-31x^2-72x-240=3\left(x+2\right)\left(x+5\right)\left(x-2\right)\)
\(11x^3-31x^2-72x-240-3\left(x+2\right)\left(x+5\right)\left(x-2\right)=0\)
\(8x^3-46x^2-60x-180=0\)
=> vô nghiệm
b) \(\frac{2}{x^2-4}-\frac{x-1}{x\left(x-2\right)}+\frac{x-4}{x\left(x+2\right)}=0\left(x\ne0;x\ne\pm2\right)\)
\(\Leftrightarrow\frac{2x}{\left(x-2\right)\left(x+2\right)x}-\frac{\left(x+2\right)\left(x-1\right)}{x\left(x-2\right)\left(x+2\right)}+\frac{\left(x+4\right)\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{2x}{x\left(x-2\right)\left(x+2\right)}-\frac{x^2+x-2}{x\left(x-2\right)\left(x+2\right)}+\frac{x^2+2x-8}{x\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{2x-x^2-x+2+x^2+2x-8}{x\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{3x-6}{x\left(x-2\right)\left(x+2\right)}=0\)
=> 3x-6=0
<=> x=2 (ktm)
Vậy pt vô nghiệm
c, ĐKXĐ : \(\left\{{}\begin{matrix}x-1\ne0\\x-3\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne1\\x\ne3\end{matrix}\right.\)
- Ta có : \(\frac{6}{x-1}-\frac{4}{x-3}=\frac{8}{2x-6}\)
=> \(\frac{12\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)}-\frac{8\left(x-1\right)}{2\left(x-3\right)\left(x-1\right)}=\frac{8\left(x-1\right)}{2\left(x-3\right)\left(x-1\right)}\)
=> \(12\left(x-3\right)-8\left(x-1\right)=8\left(x-1\right)\)
=> \(12x-36-8x+8-8x+8=0\)
=> \(-4x-20=0\)
=> \(x=-5\) ( TM )
Vậy phương trình trên có tập nghiệm là \(S=\left\{-5\right\}\)
b, ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\2x-3\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne0\\x\ne\frac{3}{2}\end{matrix}\right.\)
Ta có : \(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)
=> \(\frac{x}{x\left(2x-3\right)}-\frac{3}{x\left(2x-3\right)}=\frac{5\left(2x-3\right)}{x\left(2x-3\right)}\)
=> \(x-3=5\left(2x-3\right)\)
=> \(x-3-10x+15=0\)
=> \(-9x=-12\)
=> \(x=\frac{4}{3}\) ( TM )
Vậy phương trình trên có nghiệm là \(S=\left\{\frac{4}{3}\right\}\)
\(a,\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\) \(Đkxđ:\left\{{}\begin{matrix}x\ne-1\\x\ne2\end{matrix}\right.\)
\(\Leftrightarrow\frac{2-x}{\left(x+1\right)\left(2-x\right)}+\frac{5x+5}{\left(2-x\right)\left(x+1\right)}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
\(\Leftrightarrow2-x+5x+5=15\)
\(\Leftrightarrow7+4x=15\)
\(\Leftrightarrow4x=8\)
\(\Leftrightarrow x=2\)
\(\Leftrightarrow Ptvn\)
\(b,\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\) \(Đkxđ:\left\{{}\begin{matrix}x\ne0\\x\ne\frac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\frac{x}{x\left(2x-3\right)}-\frac{3}{x\left(2x-3\right)}=\frac{10x-15}{x\left(2x-3\right)}\)
\(\Leftrightarrow x-3=10x-15\)
\(\Leftrightarrow x-3-10x+15=0\)
\(\Leftrightarrow-9x+12=0\)
\(\Leftrightarrow-9x=-12\)
\(\Leftrightarrow\frac{4}{3}\)
\(c,\frac{6}{x-1}-\frac{4}{x-3}=\frac{8}{2x-6}\) \(Đkxđ:\left\{{}\begin{matrix}x\ne1\\x\ne3\end{matrix}\right.\)
\(\Leftrightarrow\frac{6x-18}{\left(x-1\right)\left(x-3\right)}-\frac{4x-4}{\left(x-1\right)\left(x-3\right)}=\frac{4x-4}{\left(x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow6x-18-4x+4=4x-4\)
\(\Leftrightarrow2x-14=4x-4\)
\(\Leftrightarrow-2x=10\)
\(\Leftrightarrow x=-5\)
\(d,\frac{3}{\left(x-1\right)\left(x-2\right)}+\frac{2}{\left(x-3\right)\left(x-1\right)}=\frac{1}{\left(x-2\right)\left(x-3\right)}\) \(Đkxđ:\left\{{}\begin{matrix}x\ne1\\x\ne2\\x\ne3\end{matrix}\right.\)
\(\Leftrightarrow\frac{3x-9}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}+\frac{2x-4}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=\frac{x-1}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow3x-9+2x-4=x-1\)
\(\Leftrightarrow4x-12=0\)
\(\Leftrightarrow4x=12\)
\(\Leftrightarrow x=3\)
\(\Leftrightarrow Ptvn\)
Vậy .................................
a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ne\pm2\\x\ne0\end{matrix}\right.\)
Ta có : \(\frac{x-4}{x\left(x+2\right)}-\frac{1}{x\left(x-2\right)}=-\frac{2}{\left(x+2\right)\left(x-2\right)}\)
=> \(\frac{\left(x-4\right)\left(x-2\right)}{x\left(x+2\right)\left(x-2\right)}-\frac{x+2}{x\left(x-2\right)\left(x+2\right)}=-\frac{2x}{x\left(x+2\right)\left(x-2\right)}\)
=> \(\left(x-4\right)\left(x-2\right)-x-2=-2x\)
=> \(x^2-4x-2x+8-x-2=-2x\)
=> \(x^2-5x+6=0\)
=> \(\left(x-2\right)\left(x-3\right)=0\)
=> \(\left[{}\begin{matrix}x=2\\x=3\left(TM\right)\end{matrix}\right.\)
=> x = 3 .
Vậy phương trình trên có tập nghiệm là \(S=\left\{3\right\}\)
b, ĐKXĐ : \(x\ne0,-3,-6,-9,-12\)
Ta có : \(\frac{1}{x\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+9\right)}+\frac{1}{\left(x+9\right)\left(x+12\right)}=\frac{1}{16}\)
=> \(\frac{1}{x}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+9}+\frac{1}{x+9}-\frac{1}{x+12}=\frac{1}{16}\)
=> \(\frac{1}{x}-\frac{1}{x+12}=\frac{1}{16}\)
=> \(\frac{x+12}{x\left(x+12\right)}-\frac{x}{x\left(x+12\right)}=\frac{1}{16}\)
=> \(x\left(x+12\right)=192\)
=> \(x^2+12x-192=0\)
=> \(x^2+2x.6+36-228=0\)
=> \(\left(x+6\right)^2=288\)
=> \(\left[{}\begin{matrix}x=\sqrt{288}-6\\x=-\sqrt{288}-6\end{matrix}\right.\) ( TM )
Vậy phương trình có tập nghiệm là \(S=\left\{\pm\sqrt{288}-6\right\}\)
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 \(S=\left\{\frac{20}{3}\right\}\)
1.
\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
\(MC:12\)
Quy đồng :
\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)
\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)
\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)
\(\Leftrightarrow6x+9-3x=-4-9+16\)
\(\Leftrightarrow-7x=3\)
\(\Leftrightarrow x=\frac{-3}{7}\)
2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
\(MC:20\)
Quy đồng :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
\(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\)
<=> \(\frac{2x-1}{5}-\frac{x-2}{3}-\frac{x+7}{15}=0\)
<=> \(\frac{3\left(2x-1\right)}{5\cdot3}-\frac{5\left(x-2\right)}{3\cdot5}-\frac{x+7}{15}=0\)
<=> \(\frac{6x-3-5x+10-x-7}{15}=0\)
<=> \(\frac{-14}{15}=0\)
=> PT vô nghiệm
a) ĐKXĐ: \(x\ne-1;x\ne2\)
Ta có: \(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
⇔\(\frac{1}{x+1}-\frac{5}{x-2}+\frac{15}{\left(x+1\right)\left(x-2\right)}=0\)
⇔\(\frac{x-2}{\left(x+1\right)\left(x-2\right)}-\frac{5\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}+\frac{15}{\left(x+1\right)\left(x-2\right)}=0\)
⇔\(x-2-5x-5+15=0\)
⇔\(-4x+8=0\)
⇔\(-4x=-8\)
⇔\(x=\frac{-8}{-4}=2\)(loại)
Vậy: x không có giá trị
b) ĐKXĐ: \(x\ne0;x\ne\frac{3}{2}\)
Ta có: \(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)
⇔\(\frac{x}{\left(2x-3\right)\cdot x}-\frac{3}{x\left(2x-3\right)}-\frac{5\left(2x-3\right)}{x\left(2x-3\right)}=0\)
⇔\(x-3-10x+15=0\)
⇔\(-9x+12=0\)
⇔\(-9x=-12\)
⇔\(x=\frac{-12}{-9}=\frac{4}{3}\)
Vậy: \(x=\frac{4}{3}\)
c) ĐKXĐ:\(x\ne3;x\ne1\)
Ta có: \(\frac{6}{x-1}-\frac{4}{x-3}=\frac{8}{2x-6}\)
⇔\(\frac{6}{x-1}-\frac{4}{x-3}=\frac{8}{2\left(x-3\right)}\)
⇔\(\frac{6}{x-1}-\frac{4}{x-3}=\frac{4}{x-3}\)
⇔\(\frac{6}{x-1}-\frac{4}{x-3}-\frac{4}{x-3}=0\)
⇔\(\frac{6}{x-1}-\frac{8}{x-3}=0\)
⇔\(\frac{6\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}-\frac{8\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}=0\)
⇔\(6\left(x-3\right)-8\left(x-1\right)=0\)
⇔6x-18-8x+8=0
⇔-2x-10=0
⇔-2(x+5)=0
Vì 2≠0 nên x+5=0
hay x=-5
Vậy: x=-5
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
bài dưới chỗ quy đồng thôi nha
\(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{2\left(x+2\right)^2}{x^6-1}\)
<=>\(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{2\left(x+2\right)^2}{\left(x^3-1\right)\left(x^3+1\right)}\)
<=>\(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{2\left(x+2\right)^2}{\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)\left(x^2-x+1\right)}\)
<=>\(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{2\left(x+2\right)^2}{\left(x^2-1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)}\)
<=>\(\frac{\left(x^2-1\right)\left(x+1\right)\left(x^2-x+1\right)}{\left(x^2-1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)}-\frac{\left(x^2-1\right)\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2-1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)}\)=\(\frac{2\left(x+2\right)^2}{\left(x^2-1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)}\)
<=>\(\left(x^2-1\right)\left(x^3+1\right)-\left(x^2-1\right)\left(x^3-1\right)=2\left(x+2\right)^2\)
Quy đồng khử mẫu thôi mà :))