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

ĐK: ...

c) \(\frac{x+5}{x^2-5x}-\frac{x-5}{2x^2+10x}=\frac{x+25}{2x^2-50}\)

\(\Leftrightarrow\frac{2\left(x+5\right)^2}{2x\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)^2}{2x\left(x+5\right)\left(x-5\right)}=\frac{x\left(x+25\right)}{2x\left(x-5\right)\left(x+5\right)}\)

\(\Leftrightarrow2x^2+20x+50-x^2+10x-25=x^2+25x\)

\(\Leftrightarrow5x+25=0\)

\(\Leftrightarrow x=-5\)( ko t/m )

d) tương tự, ngại tính lắm

e) \(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}\)

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

\(\Leftrightarrow4x^2-3x-1=0\)

\(\Leftrightarrow\left(x-1\right)\left(4x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(l\right)\\x=\frac{-1}{4}\left(c\right)\end{matrix}\right.\)

\(\frac{7}{8x}+\frac{5-x}{4x^2-8x}=\frac{x-1}{2x\left(x-2\right)}+\frac{1}{8x-16}\\ =>\frac{7}{8x}+\frac{5-x}{4x\left(x-2\right)}=\frac{x-1}{2x\left(x-2\right)}+\frac{1}{8\left(x-2\right)}\\ =>\frac{7\left(x-2\right)}{8x\left(x-2\right)}+\frac{10-2x}{8x\left(x-2\right)}=\frac{4x-4}{8x\left(x-2\right)}+\frac{x}{8x\left(x-2\right)}\\ =>7x-14+10-2x=4x-4+x\\ =>7x-2x-4x-x=-4+14-10\\ \)

=>0x=0

=> ptvsn

chúc bn hk tốt

#Mai.T.Loan

\(\frac{7}{8x}\) thật à =((( bài mình chỉ \(\frac{7}{8}\) thôi nên ko áp dụng được =((((

Giải:

a) $\frac{}{}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4}$⇔⇔ 9x² + 12x + 4 - 18x + 12 = 9x² ⇔ 9x² + 12x + 4 - 18x + 12 - 9x² = 0

⇔ 16 + 6x = 0 ⇔ 2(8 + 3x) = 0 ⇔ 8 + 3x = 0 ⇔ x = \(\frac{-8}{3}\)

Vậy nghiệm của phương trình là x = \(\frac{-8}{3}\) .

b) \(\frac{3}{5x-1}+\frac{3}{3-5x}=\frac{4}{\left(1-5x\right)\left(5x-3\right)}\text{⇔ }\frac{-3}{1-5x}+\frac{-3}{5x-3}=\frac{4}{\left(1-5x\right)\left(5x-3\right)}\)

⇔ \(\frac{9-15x}{\left(1-5x\right)\left(5x-3\right)}+\frac{15x-3}{\left(1-5x\right)\left(5x-3\right)}=\frac{4}{\left(1-5x\right)\left(5x-3\right)}\) ⇔ 9 - 15x + 15x - 3 = 4

⇔ 8 = 4 ( vô lí)

Vậy phương trình trên vô nghiệm.

Mình chỉ làm 2 câu a, b thôi nhé! Các bài tập này cách làm giống nhau, bạn tự hoàn thành những bài còn lại nhé!

\(\frac{2x+1}{x^2-5x+4}+\frac{5}{x-1}=\frac{2}{x-4}\)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 

a) \(\frac{5-x}{4x^2-8x}\) + \(\frac{7}{8x}\) = \(\frac{x-1}{2x\left(x-2\right)}\) +\(\frac{1}{8x-16}\)                               ĐKXĐ : x #0, x#2, x#-2

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

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

=> 10 - 2x + 7x - 14 = 4x - 4 + x

<=>-2x + 7x - 4x + x  = -4 - 10 + 14

<=>x=-14