c, Trừ hai vế cho 6 

Vế trái thì lấy từng số hạng trừ 1 là được

thế tức là phải như nào hả bạn

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

ttiiok

a,\(2x\left(x-3\right)=x-3.\)

\(\Leftrightarrow2x=1\)

\(\Leftrightarrow x=\frac{1}{2}\)

Vậy ..... 

b, \(\frac{x+2}{x-2}-\frac{5}{x}=\frac{8}{x^2-2x}\)

\(\Leftrightarrow\frac{\left(x+2\right)\cdot x}{\left(x-2\right)\cdot x}-\frac{5\left(x-2\right)}{x\left(x-2\right)}=\frac{8}{x^2-2x}\)

\(\Leftrightarrow\frac{x^2+2x-\left(5x-10\right)}{\left(x-2\right)x}=\frac{8}{x^2-2x}\)

\(\Leftrightarrow\frac{x^2+2x-5x+10}{x^2-2x}=\frac{8}{x^2-2x}\)

\(\Leftrightarrow x^2+2x-5x+10=8\)

\(\Leftrightarrow x^2-3x+10-8=0\)

\(\Leftrightarrow x^2-x-2x+2=0\)

\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}}\)

Vậy ....

\(\frac{2x+1}{4}-\frac{x-5}{3}< \frac{4x-1}{12}+12.\)

\(\Leftrightarrow\frac{\left(2x+1\right)\cdot3}{4\cdot3}-\frac{\left(x-5\right)\cdot4}{3\cdot4}< \frac{4x-1}{12}+12.\)

\(\Leftrightarrow\frac{6x+3}{12}-\frac{4x-20}{12}< \frac{4x-1}{12}+12\)

\(\Leftrightarrow\frac{6x+3-4x+20}{12}< \frac{4x-1}{12}+12\)

\(\Leftrightarrow\frac{2x+23}{12}< \frac{4x-1}{12}+12\)

\(\Leftrightarrow\frac{2x+23-4x+1}{12}< 12\)

\(\Leftrightarrow\frac{-2x+24}{12}< 12\)

\(\Leftrightarrow-2x+24< 144\)

\(\Leftrightarrow-2x< 120\)

\(\Leftrightarrow x< -60\)

\(\frac{5}{x^2+x-6}-\frac{2}{x^2+4x+3}=-\frac{3}{2x-1}\)

<=> \(\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{2}{\left(x+1\right)\left(x+3\right)}=-\frac{3}{2x-1}\)

<=> 5(x + 1)(2x - 1) - 2(x - 2)(2x - 1) = -3(x - 2)(x + 3)(x + 1)

<=> 6x² + 15x - 9 = -3x³ - 6x² + 15x + 18

<=> 6x² - 9 = -3x³ - 6x² + 18

<=> 6x² - 9 + 3x³ + 6x² - 18 = 0

<=> 12x² - 27 + 3x³ = 0

<=> 3(4x² - 9 + x³) = 0

<=> 3(x² + x - 3)(x + 3) = 0

<=> \(\orbr{\begin{cases}x=-3\\x=\frac{-1\pm\sqrt{13}}{2}\end{cases}}\)

DKXD \(x\ne\frac{1}{2};2;-1;3,;-3\)  

<=> \(\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{2}{\left(x+1\right)\left(x+3\right)}=\frac{-3}{2x-1}\)

<=> \(\frac{1}{x+3}\left(\frac{5}{x-2}-\frac{2}{x+1}\right)=\frac{-3}{2x-1}\)

<=> \(\frac{1}{x+3}\left(\frac{5x+5-2x+4}{\left(x-2\right)\left(x+1\right)}\right)=\frac{-3}{2x-1}\)

<=> \(\frac{1}{x+3}\left(\frac{3\left(x+3\right)}{\left(x-2\right)\left(x+1\right)}\right)=\frac{3}{1-2x}\)

<=> \(\frac{3}{\left(x-2\right)\left(x+1\right)}=\frac{3}{1-2x}\)

<=> \(x^2-x-2=1-2x\)

<=> \(x^2+x-3=0\)

<=> \(\orbr{\begin{cases}x=\frac{-1+\sqrt{13}}{2}\\x=\frac{-1-\sqrt{13}}{2}\end{cases}}\)

chuc ban hoc tot 

\(\frac{x+3}{x-4}-\frac{1}{x}=-\frac{5}{4x-x^2}\) (Điều kiện \(x\ne0\)và \(x\ne4\)

<=> \(\frac{x\left(x+3\right)-\left(x-4\right)}{x\left(x-4\right)}=\frac{5}{x\left(x-4\right)}\)

<=> x² + 3x -x+4=5

<=> x² + 2x -1=0

<=> (x+1)²-2=0

<=> \(\left(x+1-\sqrt{2}\right)\left(x+1+\sqrt{2}\right)=0\)

=> \(\hept{\begin{cases}x_1=-1+\sqrt{2}\\x_2=-1-\sqrt{2}\end{cases}}\)

Cách khác ạ =)

\(\frac{x+3}{x-4}-\frac{1}{x}=\frac{-5}{4x-x^2}\left(đkxđ:x\ne0;4\right)\)

\(< =>\frac{\left(x+3\right).x}{\left(x-4\right).x}-\frac{1\left(x-4\right)}{\left(x-4\right).x}=\frac{5}{x\left(x-4\right)}\)

\(< =>\left(x+3\right).x-\left(x-4\right)=5\)

\(< =>x^2+3x-x+4=5\)

\(< =>x^2+2x-1=0\)

Ta có : \(\Delta=2^2-4\left(-1\right)=0\)

Vì delta = 0 nên phương trình sẽ có nghiệm kép 

\(x_1=x_2=-\frac{b}{2a}=-\frac{2}{2}=-1\)

Vậy nghiệm của phương trình là -1

Đúng không nhỉ ?

a, 3-4x(25-2x)=8x^2+x-30

<=> 3-100x+8x^2=8x^2+x-30

<=>3-100x+8x^2-8x^2-x+30=0

<=>-101x+33=0

<=>-101x=-33

<=>x=\(\dfrac{33}{101}\)

Vậy S={\(\dfrac{33}{101}\) }

b,(2x+1)(3x-2)=(5x-8)(2x+1)

<=>(2x+1)(3x-2)-(5x-8)(2x+1)=0

<=>(2x+1)[(3x-2)-(5x-8)]=0

<=>(2x+1)(3x-2-5x+8)=0

<=>(2x+1)(-2x+6)=0

=> 2x+1=0 hoặc -2x+6=0

+) 2x+1=0

<=>2x=-1

<=>x=-1/2

+)-2x+6=0

<=>-2x=-6

<=>x=3

vậy S={-1/2;3}

c,d, do mình lười quá nên mình ghi luôn kết quả nhé : c, x= \(\dfrac{1}{2}\)

d, x=5