\(\frac{2x+1}{x^2-5x+4}+\frac{5}{x-1}=\frac{2}{x-4}\)ĐKXĐ : \(x\ne1;4\)

\(\Leftrightarrow\frac{2x+1}{\left(x-1\right)\left(x-4\right)}+\frac{5\left(x-4\right)}{\left(x-1\right)\left(x-4\right)}=\frac{2\left(x-1\right)}{\left(x-1\right)\left(x-4\right)}\)

\(\Leftrightarrow2x+1+5x-20=2x-2\)

\(\Leftrightarrow2x+5x-2x=-1+20-2\)

\(\Leftrightarrow5x=17\)

\(\Leftrightarrow x=\frac{17}{5}\)

KL : Nghiệm của PT là S={ 17/5 }

\(\frac{7}{8x}-\frac{x-5}{4x^2-8x}=\frac{x-1}{2x\left(x-2\right)}+\frac{1}{8x-16}\) ĐKXĐ : \(x\ne0;2\)

\(\Leftrightarrow\frac{7}{8x}-\frac{x-5}{4x\left(x-2\right)}=\frac{x-1}{2x\left(x-2\right)}+\frac{1}{8\left(x-2\right)}\)

\(\Leftrightarrow\frac{7\left(x-2\right)}{8x\left(x-2\right)}-\frac{2\left(x-5\right)}{8x\left(x-2\right)}=\frac{4\left(x-1\right)}{8x\left(x-2\right)}+\frac{x}{8x\left(x-2\right)}\)

\(\Leftrightarrow7x-14-2x+10=4x-4+x\)

\(\Leftrightarrow7x-2x-4x-x=14-10-4\)

\(\Leftrightarrow0x=0\)

=> PT vô số nghiệm 

<=> \(\frac{7}{8x}+\frac{5-x}{4x\left(x-2\right)}=\frac{x-1}{2x\left(x-2\right)}+\frac{1}{8\left(x-2\right)}\)(DK: x khác 0 và 2)

<=>\(\frac{7x\left(x-2\right)}{8x\left(x-2\right)}+\frac{10-2x}{8x\left(x-2\right)}=\frac{4x-4}{8x\left(x-2\right)}=\frac{x}{8x\left(x-2\right)}\)

<=>\(7x^2-14x+10-2x=4x-4+x\)

<=>\(7x^2-14x-2x-4x-x=-4-10\)

<=>\(7x^2-21x+14=0\)

<=>\(7\left(x^2-3x+2\right)=0\)

<=>\(x^2-3x+2=0\)

<=>\(x^2-x-2x+2=0\)

<=>\(x\left(x-1\right)-2\left(x-1\right)=0\)

<=>\(\left(x-1\right)\left(x-2\right)=0\)

<=>\(\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\left(TMDK\right)\\x=2\left(KTMDK\right)\end{cases}}\)

Vậy: x=1

434

hình như bằng 434

434

a,\(\frac{3}{1-4x}=\frac{2}{4x+1}-\frac{3+6x}{16x^2-1}\)

ĐKXĐ: x≠1/4, x≠-1/4

⇔\(-\frac{3}{4x-1}=\frac{2}{4x+1}-\frac{3+6x}{16x^2-1}\)

⇔\(\frac{-3\left(4x+1\right)}{\left(4x-1\right)\left(4x+1\right)}=\frac{2\left(4x-1\right)}{\left(4x+1\right)\left(4x-1\right)}-\frac{3+6x}{16x^2-1}\)

⇒-12x-3=8x-2-3-6x

⇔8x-6x+12x=-3+2+3

⇔14x=2

⇔x=1/7(tmđk)

Vậy phương trình có nghiệm là x=1/7

b, \(\frac{5-x}{4x^2-8x}+\frac{7}{8x}=\frac{x-1}{2x\left(x-2\right)}+\frac{1}{8x-16}\) (2)

ĐKXĐ: x≠0, x≠2

(2)⇔\(\frac{2\left(5-x\right)}{2.4x\left(x-2\right)}+\frac{7\left(x-2\right)}{8x\left(x-2\right)}=\frac{4.\left(x-1\right)}{4.2x\left(x-2\right)}+\frac{x}{8.x\left(x-2\right)}\)

⇒10-2x+7x-14=4x-4+x

⇔-2x+7x-4x-x=-4-10+14

⇔0x=0

⇔ x∈R

Vậy phương trình có nghiệm là x∈R và x≠0, x≠2

c, \(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{3}{x\left(x^4+x^2+1\right)}\) (3)

ĐKXĐ: x≠0

(3)⇒x(x+1)(x²-x+1)-x(x-1)(x²+x+1)=3

⇔x⁴+x-x⁴+x=3

⇔2x=3

⇔x=3/2(tmđk)

Vậy phương trình có nghiệm là x=3/2