a) \(\frac{2x+1}{x-1}\)=\(\frac{5\left(x-1\right)}{x+1}\):dkxd x\(\ne\)\(\pm\)1
\(\Rightarrow\)(2x+1)(x+1)=5(x-1)2
\(\Leftrightarrow\)2x2+2x+x+1=5(x2-2x+1)
\(\Leftrightarrow\)2x2+2x+x+1=5x2-10x+5
\(\Leftrightarrow\)2x2+2x+x+1-5x2+10x-5=0
\(\Leftrightarrow\)-3x2+13x-4=0
\(\Leftrightarrow\)-3x2+12x+1x-4=0
\(\Leftrightarrow\)-4x(x-4)+(x-4)=0
\(\Leftrightarrow\)(x-4)(-4x+1)=0
\(\Leftrightarrow\)x-4=0 hoac -4x+1=0
\(\Leftrightarrow\)x=4(tmdkxd) \(\Leftrightarrow\)x=1/4(tmdkxd)
vay s={4;1/4}
b)\(\frac{x}{x-1}\)-\(\frac{2x}{x^{ }2^{ }-1}\)=0 dkxd x\(\ne\)\(\pm\)1
\(\Leftrightarrow\)\(\frac{x\left(X+1\right)-2x^{ }}{\left(x-1\right)\left(x+1\right)}\)=0
\(\Rightarrow\)x2+x-2x=0
\(\Leftrightarrow\)x2-x=0
\(\Leftrightarrow\)x(x-1)=0
\(\Leftrightarrow\)x=0 hoac x-1=0
\(\Leftrightarrow\)x=0(tmdkxd)\(\Leftrightarrow\)x=1(ktmdkxd)
vay s={0}
c.\(\frac{1}{x-2}\)+3=\(\frac{x-3}{2-x}\) dkxd x\(\ne\)2
\(\Leftrightarrow\)\(\frac{1}{x-2}\)+3=\(\frac{-\left(x-3\right)}{x-2}\)
\(\Leftrightarrow\)\(\frac{1+3\left(x-2\right)}{x-2}\)=\(\frac{-x+3}{x-2}\)
\(\Rightarrow\)1+3x-6=-x+3
\(\Leftrightarrow\)4x=8
\(\Leftrightarrow\)x=2(ktmdkxd)
vay s=\(\varnothing\)
chuc ban hoc tot
a.\(\frac{2x+1}{x-1}\) = \(\frac{5\left(x-1\right)}{x+1}\)
\(\leftrightarrow\) 2x+1 = 5x - 5
\(\leftrightarrow\) 2x - 5= -1-5
\(\leftrightarrow\) -3x = -6
\(\leftrightarrow\) x =2
Vậy S=\(\left\{2\right\}\)
b.\(\frac{x}{x-1}\) - \(\frac{2x}{x^2-1}\) =0
\(\leftrightarrow\) \(\frac{x}{x-1}\) - \(\frac{2x}{\left(x-1\left(x+1\right)\right)}\)= 0 (ĐK : x\(_{\ne}\) -1 và 1)
\(\leftrightarrow\)\(\frac{x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\) - \(\frac{2x}{\left(x-1\left(x+1\right)\right)}\) =0
\(\leftrightarrow\) x2 + x -2x = 0
\(\leftrightarrow\)(x2 + x) -2x =0
\(\leftrightarrow\)x(x+1) -2x =0
\(\leftrightarrow\) x =0 -> x=0
x+1 =0 -> x = -1(Loại)
-2x = 0 -> x= 2(TM)
Vậy x =\(\left\{0,2\right\}\)
(BẠN NHỚ COI LẠI CÁI CÂU TRẢ LỜI Ở CUỐI MỖI BÀI NHA ,MÌNH KO CHẮC CÂU TRẢ LỜI ĐÓ )
