a, \(\frac{x-3}{5}\) = 6 - \(\frac{1-2x}{3}\)

⇔ 3(x - 3) = 90 - 5(1 - 2x)

⇔ 3x - 9 = 90 - 5 + 10x

⇔ 3x - 10x = 90 - 5 + 9

⇔ -7x = 94

⇔ x = \(\frac{-94}{7}\)

S = { \(\frac{-94}{7}\) }

b, \(\frac{3x-2}{6}\) - 5 = \(\frac{3-2\left(x+7\right)}{4}\)

⇔ 2(3x - 2) - 60 = 9 - 6(x + 7)

⇔ 6x - 4 - 60 = 9 - 6x - 42

⇔ 6x + 6x = 9 - 42 + 60 + 4

⇔ 12x = 31

⇔ x = \(\frac{31}{12}\)

S = { \(\frac{31}{12}\) }

c, \(\frac{x+8}{6}\) - \(\frac{2x-5}{5}\) = \(\frac{x+1}{3}\) - x + 7

⇔ 5(x+ 8) - 6(2x - 5) = 10(x+1) - 30x+210

⇔ 5x+ 40 - 12x+ 30 = 10x+ 10 - 30x+210

⇔ 5x - 12x - 10x+ 30x = 10+ 210 - 30- 40

⇔ 13x = 150

⇔ x = \(\frac{150}{13}\)

S = { \(\frac{150}{13}\) }

d, \(\frac{7x}{8}\) - 5(x - 9) = \(\frac{2x+1,5}{6}\)

⇔ 21x - 120(x - 9) = 4(2x + 1,5)

⇔ 21x - 120x + 1080 = 8x + 6

⇔ 21x - 120x - 8x = 6 - 1080

⇔ -107x = -1074

⇔ x = \(\frac{1074}{107}\)

S = { \(\frac{1074}{107}\) }

e, \(\frac{5\left(x-1\right)+2}{6}\) - \(\frac{7x-1}{4}\) = \(\frac{2\left(2x+1\right)}{7}\) - 5

⇔ 140(x-1)+56 - 42(7x-1) = 48(2x+1)-840

⇔ 140x -140+56 -294x+42= 96x+48 -840

⇔ 140x -294x -96x = 48 -840 -42 -56+140

⇔ -250x = -750

⇔ x = 3

S = { 3 }

f, \(\frac{x+1}{3}\) + \(\frac{3\left(2x+1\right)}{4}\) = \(\frac{2x+3\left(x+1\right)}{6}\) + \(\frac{7+12x}{12}\)

⇔ 4(x+1)+9(2x+1) = 4x+6(x+1)+7+12x

⇔ 4x+4+18x+9 = 4x+6x+6+7+12x

⇔ 4x+18x - 4x - 6x - 12x = 6+7- 9 - 4

⇔ 0x = 0

S = R

Chúc bạn học tốt !

Bài 1:

ĐKXĐ: x≠1

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

\(\Leftrightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{4\left(x-1\right)}{\left(x^2+x-1\right)\left(x-1\right)}=0\)

\(\Leftrightarrow x^2+x+1+2x^2-5-4\left(x-1\right)=0\)

\(\Leftrightarrow x^2+x+1+2x^2-5-4x+4=0\)

\(\Leftrightarrow3x^2-3x=0\)

\(\Leftrightarrow3x\left(x-1\right)=0\)

Vì 3≠0

nên \(\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=0\)

Vậy: x=0

Bài 2:

ĐKXĐ: x≠2; x≠3; \(x\ne\frac{1}{2}\)

Ta có: \(\frac{x+4}{2x^2-5x+2}+\frac{x+1}{2x^2-7x+3}=\frac{2x+5}{2x^2-7x+3}\)

\(\Leftrightarrow\frac{x+4}{\left(x-2\right)\left(2x-1\right)}+\frac{x+1-\left(2x+5\right)}{\left(x-3\right)\left(2x-1\right)}=0\)

\(\Leftrightarrow\frac{x+4}{\left(x-2\right)\left(2x-1\right)}+\frac{x+1-2x-5}{\left(x-3\right)\left(2x-1\right)}=0\)

\(\Leftrightarrow\frac{\left(x+4\right)\left(x-3\right)}{\left(x-2\right)\left(2x-1\right)\left(x-3\right)}+\frac{\left(-x-4\right)\left(x-2\right)}{\left(x-3\right)\left(2x-1\right)\left(x-2\right)}=0\)
\(\Leftrightarrow x^2+x-12-x^2-2x+8=0\)

\(\Leftrightarrow-x-4=0\)

\(\Leftrightarrow-x=4\)

hay x=-4(tm)

Vậy: x=-4

Bài 3:

ĐKXĐ: x≠1; x≠-1

Ta có: \(\frac{x+1}{x-1}-\frac{x-1}{x+1}=3x\left(1-\frac{x-1}{x+1}\right)\)

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

\(\Leftrightarrow\frac{x+1}{x-1}-\frac{x-1}{x+1}-3x+\frac{3x\left(x-1\right)}{x+1}=0\)

\(\Leftrightarrow\frac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\frac{3x\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{3x\left(x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\)

\(\Leftrightarrow\left(x^2+2x+1\right)-\left(x^2-2x+1\right)-3x\left(x^2-1\right)+3x\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow x^2+2x+1-x^2+2x-1-3x^3+3x+3x^3-6x^2+3x=0\)

\(\Leftrightarrow-6x^2+10x=0\)

\(\Leftrightarrow2x\left(-3x+5\right)=0\)

Vì 2≠0

nên \(\left[{}\begin{matrix}x=0\\-3x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\-3x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{5}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{0;\frac{5}{3}\right\}\)

Bài 4:

ĐKXĐ: x≠1; x≠-3

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

\(\Leftrightarrow\frac{2x\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}+\frac{4}{\left(x-1\right)\left(x+3\right)}-\frac{\left(2x-5\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}=0\)

\(\Leftrightarrow2x^2+6x+4-\left(2x^2-7x+5\right)=0\)

\(\Leftrightarrow2x^2+6x+4-2x^2+7x-5=0\)

\(\Leftrightarrow13x-1=0\)

\(\Leftrightarrow13x=1\)

hay \(x=\frac{1}{13}\)(tm)

Vậy: \(x=\frac{1}{13}\)

Bài 5:

ĐKXĐ: x≠1; x≠-2

Ta có: \(\frac{1}{x-1}-\frac{7}{x+2}=\frac{3}{x^2+x-2}\)

\(\Leftrightarrow\frac{x+2}{\left(x-1\right)\left(x+2\right)}-\frac{7\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}-\frac{3}{\left(x+2\right)\left(x-1\right)}=0\)

\(\Leftrightarrow x+2-7\left(x-1\right)-3=0\)

\(\Leftrightarrow x+2-7x+7-3=0\)

\(\Leftrightarrow-6x+6=0\)

\(\Leftrightarrow-6\left(x-1\right)=0\)

Vì -6≠0

nên x-1=0

hay x=1(ktm)

Vậy: x∈∅

Bài 6:

ĐKXĐ: x≠4; x≠2

Ta có: \(\frac{x+3}{x-4}+\frac{x-1}{x-2}=\frac{2}{6x-8-x^2}\)

\(\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}-\frac{2}{6x-8-x^2}=0\)

\(\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}-\frac{2}{-\left(x^2-6x+8\right)}=0\)

\(\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}+\frac{2}{\left(x-4\right)\left(x-2\right)}=0\)

\(\Leftrightarrow\frac{\left(x+3\right)\left(x-2\right)}{\left(x-4\right)\left(x-2\right)}+\frac{\left(x-1\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}+\frac{2}{\left(x-4\right)\left(x-2\right)}=0\)

\(\Leftrightarrow x^2+x-6+x^2-5x+4+2=0\)

\(\Leftrightarrow2x^2-4x=0\)

\(\Leftrightarrow2x\left(x-2\right)=0\)

Vì 2≠0

nên \(\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=0\)

Vậy: x=0

Bài 7:

ĐKXĐ: x≠1; x≠-2; x≠-1

Ta có: \(\frac{1}{x-1}-\frac{7}{x+2}=\frac{3}{1-x^2}\)

\(\Leftrightarrow\frac{1}{x-1}-\frac{7}{x+2}+\frac{3}{x^2-1}=0\)

\(\Leftrightarrow\frac{\left(x+1\right)\left(x+2\right)}{\left(x-1\right)\left(x+1\right)\left(x+2\right)}-\frac{7\left(x-1\right)\left(x+1\right)}{\left(x+2\right)\left(x-1\right)\left(x+1\right)}+\frac{3\left(x+2\right)}{\left(x-1\right)\left(x+1\right)\left(x+2\right)}=0\)

\(\Leftrightarrow x^2+3x+2-7\left(x^2-1\right)+3x+6=0\)

\(\Leftrightarrow x^2+3x+2-7x^2+7x+3x+6=0\)

\(\Leftrightarrow-6x^2+13x+8=0\)
\(\Leftrightarrow-6x^2+16x-3x+8=0\)

\(\Leftrightarrow2x\left(-3x+8\right)+\left(-3x+8\right)=0\)

\(\Leftrightarrow\left(-3x+8\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-3x+8=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x=-8\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{8}{3}\\x=\frac{-1}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{8}{3};\frac{-1}{2}\right\}\)

\( 1)\dfrac{1}{{x - 1}} + \dfrac{{2{x^2} - 5}}{{{x^3} - 1}} = \dfrac{4}{{{x^2} + x + 1}}\\ DK:x \ne 1\\ \Leftrightarrow \dfrac{{{x^2} + x + 1 + 2{x^2} - 5}}{{{x^3} - 1}} = \dfrac{{4\left( {x - 1} \right)}}{{{x^3} - 1}}\\ \Leftrightarrow {x^2} + x + 1 + 2{x^2} - 5 = 4x - 4\\ \Leftrightarrow 3{x^2} - 3x = 0\\ \Leftrightarrow 3x\left( {x - 1} \right) = 0 \Leftrightarrow \left[ \begin{array}{l} x = 0\left( {tm} \right)\\ x = 1\left( {ktm} \right) \end{array} \right.\\ 2)\dfrac{{x + 4}}{{2{x^2} - 5x + 2}} + \dfrac{{x + 1}}{{2{x^2} - 7x + 3}} = \dfrac{{2x + 5}}{{2{x^2} - 7x + 3}}\\ + DK:x \ne \dfrac{1}{2};x \ne 2;x \ne 3\\ \Leftrightarrow \dfrac{{x + 4}}{{\left( {2x - 1} \right)\left( {x - 2} \right)}} + \dfrac{{x + 1}}{{\left( {x - 3} \right)\left( {2x - 1} \right)}} = \dfrac{{2x + 5}}{{\left( {x - 3} \right)\left( {2x - 1} \right)}}\\ \Leftrightarrow \left( {x + 4} \right)\left( {x - 3} \right) + \left( {x + 1} \right)\left( {x - 2} \right) = \left( {2x + 5} \right)\left( {x - 2} \right)\\ \Leftrightarrow {x^2} + x - 12 + {x^2} - x - 2 = 2{x^2} + x - 10\\ \Leftrightarrow x = - 4\left( {tm} \right)\\ 3)\dfrac{{x + 1}}{{x - 1}} - \dfrac{{x - 1}}{{x + 1}} = 3x\left( {1 - \dfrac{{x - 1}}{{x + 1}}} \right)\\ DK:x \ne \pm 1\\ \Leftrightarrow {\left( {x + 1} \right)^2} - {\left( {x - 1} \right)^2} = 3x\left( {x - 1} \right)\left( {x + 1 - x + 1} \right)\\ \Leftrightarrow {x^2} + 2x + 1 - {x^2} + 2x - 1 = 6x\left( {x - 1} \right)\\ \Leftrightarrow 4x = 6{x^2} - 6x\\ \Leftrightarrow 2x\left( {3x - 5} \right) = 0 \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = \dfrac{5}{3} \end{array} \right.\left( {tm} \right) \)

Còn lại tương tự mà làm nhé!

1) Ta có: x-4=2x+4

\(\Leftrightarrow x-4-2x-4=0\)

\(\Leftrightarrow-x-8=0\)

\(\Leftrightarrow-x=8\)

hay x=-8

Vậy: S={8}

2) Ta có: \(\frac{2x-1}{2}-\frac{x}{3}=x-\frac{x}{6}\)

\(\Leftrightarrow\frac{3\left(2x-1\right)}{6}-\frac{2x}{6}=\frac{6x}{6}-\frac{x}{6}\)

\(\Leftrightarrow3\left(2x-1\right)-2x-6x+x=0\)

\(\Leftrightarrow6x-3-2x-6x+x=0\)

\(\Leftrightarrow-x-3=0\)

\(\Leftrightarrow-x=3\)

hay x=-3

Vậy: S={-3}

3) ĐKXĐ: \(x\notin\left\{\frac{-1}{2};3\right\}\)

Ta có: \(\frac{x+3}{2x+1}-\frac{x}{x-3}=\frac{3x^2+x+9}{\left(2x+1\right)\left(x-3\right)}\)

\(\Leftrightarrow\frac{\left(x+3\right)\left(x-3\right)}{\left(2x+1\right)\left(x-3\right)}-\frac{x\left(2x+1\right)}{\left(x-3\right)\left(2x+1\right)}=\frac{3x^2+x+9}{\left(2x+1\right)\left(x-3\right)}\)

Suy ra: \(x^2-9-\left(2x^2+x\right)-3x^2-x-9=0\)

\(\Leftrightarrow-2x^2-x-18-2x^2-x=0\)

\(\Leftrightarrow-4x^2-2x-18=0\)

\(\Leftrightarrow-4\left(x^2+\frac{1}{2}x+\frac{4}{5}\right)=0\)

\(\Leftrightarrow x^2+\frac{1}{2}x+\frac{4}{5}=0\)

\(\Leftrightarrow x^2+2\cdot x\cdot\frac{1}{4}+\frac{1}{16}+\frac{59}{80}=0\)

\(\Leftrightarrow\left(x+\frac{1}{4}\right)^2+\frac{59}{80}=0\)(vô lý)

Vậy: S=\(\varnothing\)

4) Ta có: \(\frac{2x}{3}+\frac{2x-1}{6}=4-\frac{x}{3}\)

\(\Leftrightarrow\frac{4x}{6}+\frac{2x-1}{6}=\frac{24}{6}-\frac{2x}{6}\)

\(\Leftrightarrow4x+2x-1=24-2x\)

\(\Leftrightarrow6x-1-24+2x=0\)

\(\Leftrightarrow8x-25=0\)

\(\Leftrightarrow8x=25\)

hay \(x=\frac{25}{8}\)

Vậy: \(S=\left\{\frac{25}{8}\right\}\)

\(3x^2+7x-20=0\)

Ta có \(\Delta=7^2+4.3.20=289,\sqrt{\Delta}=17\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-7+17}{6}=\frac{5}{3}\\x=\frac{-7-17}{6}=-4\end{cases}}\)

a) \(2x-\frac{3x-1}{3}=2+\frac{x-3}{4}\)

<=> 24x - 4(3x - 1) = 24 + 3(x - 3)

<=> 24x - 12x - 4 = 24 + 3x - 9

<=> 12x + 4 = 24 + 3x - 9

<=> 12x + 4 = 3x + 15

<=> 12x = 3x + 15 - 4

<=> 12x = 3x + 11

<=> 12x - 3x = 11

<=> 9x = 11

<=> x = 11/9

Vậy: tập nghiệm phương trình: S = {11/9}

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

<=> 3(x - 5) + 3/2 = 2(x - 2) - 6x

<=> 3x - 15 + 3/2 = 2x - 4 - 6x

<=> 3x - 27/2 = -4x - 4

<=> 3x = -4x - 4 + 27/2

<=> 3x = -4x + 19/2

<=> 3x + 4x = 19/2

<=> 7x = 19/2

<=> x = 19/14

Vậy: tập nghiệm phương trình: S = {19/14}

c) \(\frac{5x-3}{6}-\frac{7x-1}{4}=\frac{4x+2}{8}-5\)

<=> \(\frac{5x-3}{6}-\frac{7x-1}{4}=\frac{2\left(2x+1\right)}{8}-5\)

<=> \(\frac{5x-3}{6}-\frac{7x-1}{4}=\frac{2x+1}{4}-5\)

<=> 2(5x - 3) - 3(7x - 1) = 3(2x + 1) - 60

<=> 10x - 6 - 21x + 3 = 6x + 3 - 60

<=> -11x - 3 = 6x - 57

<=> -3 = 6x - 57 + 11x

<=> -3 = 17x - 57

<=> -3 + 57 = 17x

<=> 54 = 17x

<=> x = 54/17

Vậy: tập nghiệm phương trình: S = {59/17}

d) 3x² + 7x - 20 = 0

<=> 3x² + 12x - 5x - 20 = 0

<=> 3x(x + 4) - 5(x + 4) = 0

<=> (x + 4)(3x - 5) = 0

<=> x + 4 = 0 hoặc 3x - 5 = 0

<=> x = -4 hoặc x = 5/3

Vậy: tập nghiệm phương trình: S = {-4; 5/3}

e) x³ - 3x + 2 = 0

<=> (x² + x - 2)(x - 1) = 0

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

<=> x - 1 = 0 hoặc x + 2 = 0

<=> x = 1 hoặc x = -2

Vậy: tập nghiệm phương trình: S = {1; -2}

\(\frac{1}{x^2-2x+2}-1+\frac{2}{x^2-2x+3}-1+2-\frac{6}{x^2-2x+4}=0\)

\(\Leftrightarrow\frac{-x^2+2x-1}{x^2-2x+2}+\frac{-x^2+2x-1}{x^2-2x+3}+\frac{2\left(x^2-2x+1\right)}{x^2-2x+4}=0\)

\(\Leftrightarrow\left(x^2-2x+1\right)\left(\frac{2}{x^2-2x+4}-\frac{1}{x^2-2x+2}-\frac{1}{x^2-2x+3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x+1=0\Rightarrow x=1\\\frac{2}{x^2-2x+4}-\frac{1}{x^2-2x+2}-\frac{1}{x^2-2x+3}=0\left(1\right)\end{matrix}\right.\)

Xét (1), đặt \(a=x^2-2x+3\) pt trở thành:

\(\frac{2}{a+1}-\frac{1}{a-1}-\frac{1}{a}=0\Leftrightarrow\frac{2\left(a-1\right)-\left(a+1\right)}{\left(a^2-1\right)}-\frac{1}{a}=0\)

\(\Leftrightarrow\frac{a-3}{a^2-1}=\frac{1}{a}\Leftrightarrow a^2-3a=a^2-1\Leftrightarrow3a=1\Rightarrow a=\frac{1}{3}\)

\(\Rightarrow x^2-2x+3=\frac{1}{3}\Leftrightarrow x^2-2x+1+\frac{5}{3}=0\)

\(\Leftrightarrow\left(x-1\right)^2+\frac{5}{3}=0\) (vô nghiệm)

Vậy \(x=1\)