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16 tháng 2 2018
a) \(\frac{x+5}{x-1}=\frac{x+1}{x-3}-\frac{8}{x^2-4x+3}\)
\(ĐKXĐ:\)\(x\ne1\)và \(x\ne3\)
\(\frac{\left(x+5\right)\left(x-3\right)}{\left(x-1\right)9x-3}=\frac{\left(x+1\right)\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}-\frac{8}{\left(x-3\right)\left(x-1\right)}\)
\(\Leftrightarrow\)\(x^2-3x+5x-15=x^2-x+x-1-8\)
\(\Leftrightarrow\)\(x^2-3x+5x-15-x^2+x-x+1+8=0\)
\(\Leftrightarrow\)\(2x-6=0\)
\(\Leftrightarrow\)\(2x=6\)
\(\Leftrightarrow\)\(x=3\)( loại )
Vậy \(S=\varnothing\)
b) \(\frac{y+1}{y-2}-\frac{5}{y+2}=\frac{12}{y^2-4}+1\)
\(ĐKXĐ:\)\(y\ne2\)và \(y\ne-2\)
\(\frac{\left(y+1\right)\left(y+2\right)}{\left(y-2\right)\left(y+2\right)}-\frac{5\left(y-2\right)}{\left(y-2\right)\left(y+2\right)}=\frac{12}{\left(y-2\right)\left(y+2\right)}+\frac{\left(y-2\right)\left(y+2\right)}{\left(y-2\right)\left(y+2\right)}\)
\(\Leftrightarrow\)\(y^2+2y+y+2-5y+10=12+y^2-4\)
\(\Leftrightarrow\)\(y^2+2y+y+2-5y+10-10-12-y^2+4=0\)
\(\Leftrightarrow\)\(-2y+4=0\)
\(\Leftrightarrow\)\(-2y=-4\)
\(\Leftrightarrow\)\(y=2\)( loại 0
Vậy \(S=\varnothing\)
20 tháng 3 2020
\(\frac{x+5}{x-1}=\frac{x+1}{x-3}-\frac{8}{x^2-4x+3}\left(x\ne1;x\ne3\right)\)
\(\Leftrightarrow\frac{x+5}{x-1}-\frac{x+1}{x-3}+\frac{8}{x^2-4x+3}=0\)
\(\Leftrightarrow\frac{\left(x+5\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}-\frac{\left(x+1\right)\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}+\frac{8}{\left(x-1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\frac{x^2+2x-15}{\left(x-1\right)\left(x-3\right)}-\frac{x^2-1}{\left(x-3\right)\left(x-1\right)}+\frac{8}{\left(x-1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\frac{x^2+2x-15-x^2+1+8}{\left(x-1\right)\left(x-3\right)}=0\)
\(\Rightarrow2x-4=0\)
<=> 2x=4
<=> x=2 (tmđk)
Vậy x=2
20 tháng 3 2020
b) \(\frac{x+1}{x-2}-\frac{5}{x+2}=\frac{12}{x^2-4}+1\left(x\ne\pm2\right)\)
\(\Leftrightarrow\frac{x+1}{x-2}-\frac{5}{x+2}-\frac{12}{\left(x-2\right)\left(x+2\right)}-1=0\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{12}{\left(x-2\right)\left(x+2\right)}-\frac{x^2-4}{x^2-4}=0\)
\(\Leftrightarrow\frac{x^2+3x+2-5x+10-12-x^2+4}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{-2x+2}{\left(x-2\right)\left(x+2\right)}=0\)
=> -2x+2=0
<=> -2x=-2
<=> x=1 (tmđk)
Vậy x=1
24 tháng 3 2020
a) 7x - 35 = 0
<=> 7x = 0 + 35
<=> 7x = 35
<=> x = 5
b) 4x - x - 18 = 0
<=> 3x - 18 = 0
<=> 3x = 0 + 18
<=> 3x = 18
<=> x = 5
c) x - 6 = 8 - x
<=> x - 6 + x = 8
<=> 2x - 6 = 8
<=> 2x = 8 + 6
<=> 2x = 14
<=> x = 7
d) 48 - 5x = 39 - 2x
<=> 48 - 5x + 2x = 39
<=> 48 - 3x = 39
<=> -3x = 39 - 48
<=> -3x = -9
<=> x = 3
22 tháng 2 2019
\(a.\Leftrightarrow\frac{5x^2+16}{\left(x+4\right)\left(x-4\right)}=\frac{\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)}{\left(x+4\right)\left(x-4\right)}DKXD:x\ne4;-4\)
\(\Rightarrow5x^2+16=2x^2-8x-x+4+3x^2+12x-x-4\)
\(\Leftrightarrow2x=16\)
\(\Leftrightarrow x=8\)
\(b.\Leftrightarrow\frac{\left(y+1\right)\left(y+2\right)-5\left(y-2\right)}{\left(y-2\right)\left(y+2\right)}=\frac{12+\left(y-2\right)\left(y+2\right)}{\left(y-2\right)\left(y+2\right)}.DKXD:y\ne2;-2\)
\(\Rightarrow y^2+2y+y+2-5y+10=12+y^2-4\)
\(\Leftrightarrow-2y=-4\)
\(\Leftrightarrow y=2\)
14 tháng 1 2017
Anh giải câu a thôi. Câu b hoàn toàn tương tự.
\(\left(x-1\right)\left(5x+3\right)-\left(x-1\right)\left(3x-8\right)=0\)
\(\left(x-1\right)\left(2x+11\right)=0\)
22 tháng 4 2019
a)
voi x=0 ta thay 0 o phai la no pt
voi x<>0 chia ca 2 ve cho x^2 ta dc
x^2-3x+6-3/x+1/x^2=0
(x^2+1/x^2)-3(x+1/x)+6=0 dat a=x+1/x ta co (x+1/x)^2=a^2=>x^2+1/x^2=a^2-2
=>a^2-3a+4=0=>pt vo no :(