Thực ra cũng EZ thôi :

\(\frac{6}{x^2-9}-1+\frac{4}{x^2-11}-1-\frac{7}{x^2-8}+1-\frac{3}{x^2-12}+1=0=>\)

\(\frac{15-x^2}{x^2-9}+\frac{15-x^2}{x^2-11}-\frac{15-x^2}{x^2-8}-\frac{15-x^2}{x^2-12}=0\)

=> \(\left(15-x^2\right)\left(\frac{1}{x^2-9}+\frac{1}{x^2-11}-\frac{1}{x^2-8}-\frac{1}{x^2-12}\right)=0\)

=>\(15-x^2=0=>x=\pm\sqrt{15}\)

Hình như còn nghiệm , any body help me ?

ĐKXĐ: \(\left\{{}\begin{matrix}x^2\ne9\\x^2\ne11\\x^2\ne8\\x^2\ne12\end{matrix}\right.\Leftrightarrow x\notin\left\{3;-3;\sqrt{11};-\sqrt{11};2\sqrt{2};-2\sqrt{2};2\sqrt{3};-2\sqrt{3}\right\}\)

Đặt \(x^2-11=a\)(Điều kiện: \(a\notin\left\{-2;0;-3;1\right\}\))

PT\(\Leftrightarrow\frac{6}{a+2}+\frac{4}{a}-\frac{7}{a+3}-\frac{3}{a-1}=0\)

\(\Leftrightarrow\frac{6}{a+2}-1+\frac{4}{a}-1+\frac{-7}{a+3}+1+\frac{-3}{a-1}+1=0\)

\(\Leftrightarrow\frac{6-a-2}{a+2}+\frac{4-a}{a}+\frac{-7+a+3}{a+3}+\frac{-3+a-1}{a-1}=0\)

\(\Leftrightarrow-\frac{a-4}{a+2}-\frac{a-4}{a}+\frac{a-4}{a+3}+\frac{a-4}{a-1}=0\)

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

\(\Leftrightarrow a-4=0\)

hay a=4

\(\Leftrightarrow x^2-11=4\)

\(\Leftrightarrow x^2=15\)

hay \(x=\pm\sqrt{15}\)

a) \(pt\Leftrightarrow\frac{6}{x^2+2}-1+\frac{7}{x^2+3}-1+\frac{12}{x^2+8}-1-\frac{3x^2+16}{x^2+10}+2=0\)

\(\Leftrightarrow\frac{4-x^2}{x^2+2}+\frac{4-x^2}{x^2+3}+\frac{4-x^2}{x^2+8}+\frac{4-x^2}{x^2+10}=0\)

\(\Leftrightarrow\left(4-x^2\right)\left(\frac{1}{x^2+2}+\frac{1}{x^2+3}+\frac{1}{x^2+8}+\frac{1}{x^2+10}\right)=0\)

\(\Leftrightarrow4-x^2=0\)(do \(\frac{1}{x^2+2}+\frac{1}{x^2+3}+\frac{1}{x^2+8}+\frac{1}{x^2+10}>0,\forall x\))

\(\Leftrightarrow x^2=4\Leftrightarrow x=\pm2\)

\(KL...\)

2x(8x - 1)²(4x - 1) = 9

<=> 512x⁴ - 256x³ + 40x² - 2x = 9

<=> 512x⁴ - 256x³ + 40x² - 2x - 9 = 0

<=> (2x - 1)(4x + 1)(64x⁴ - 16x + 9) = 0

vì 64x⁴ - 16x + 9 khác 0 nên:

<=> 2x - 1 = 0 hoặc 4x + 1 = 0

<=> x = 1/2 hoặc x = -1/4

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 !

\( a)5\left( {x - 3} \right) - 4 = 2\left( {x - 1} \right) + 7\\ \Leftrightarrow 5x - 15 - 4 = 2x - 2 + 7\\ \Leftrightarrow 5x - 19 = 2x + 5\\ \Leftrightarrow 5x - 2x = 5 + 19\\ \Leftrightarrow 3x = 24\\ \Leftrightarrow x = 8\\ b)\dfrac{{8x - 3}}{4} - \dfrac{{3x - 2}}{2} = \dfrac{{2x - 1}}{2} + \dfrac{{x + 3}}{4}\\ \Leftrightarrow 8x - 3 - \left( {3x - 2} \right).2 = \left( {2x - 1} \right).2 + x + 3\\ \Leftrightarrow 8x - 3 - 6x + 4 = 4x - 2 + x + 3\\ \Leftrightarrow 2x + 1 = 5x + 1\\ \Leftrightarrow 2x - 5x = 0\\ \Leftrightarrow - 3x = 0\\ \Leftrightarrow x = 0 \)

\( c)\dfrac{{2\left( {x + 5} \right)}}{3} + \dfrac{{x + 12}}{2} - \dfrac{{5\left( {x - 2} \right)}}{6} = \dfrac{x}{3} + 11\\ \Leftrightarrow 4\left( {x + 5} \right) + 3\left( {x + 12} \right) - \left[ {5\left( {x - 2} \right)} \right] = 2x + 66\\ \Leftrightarrow 4x + 20 + 3x + 36 - 5x + 10 = 2x + 66\\ \Leftrightarrow 2x + 66 = 2x + 66\\ \Leftrightarrow 0x = 0\left( {VSN} \right)\\ \Leftrightarrow x = 0 \)

\(d)\dfrac{x-10}{1994}+\dfrac{x-8}{1996}+\dfrac{x-6}{1998}+\dfrac{x-4}{2000}+\dfrac{x-2}{2002}=\dfrac{x-2002}{2}+\dfrac{x-2000}{4}+\dfrac{x-1998}{6}+\dfrac{x-1996}{8}+\dfrac{x-1994}{10}\\ \Leftrightarrow \dfrac{x-10}{1994}-1+\dfrac{x-8}{1996}-1+\dfrac{x-6}{1998}-1+\dfrac{x-4}{2000}-1+\dfrac{x-2}{2002}-1=\dfrac{x-2002}{2}-1+\dfrac{x-2000}{4}-1+\dfrac{x-1998}{6}-1+\dfrac{x-1996}{8}-1+\dfrac{x-1994}{10}-1\\ \Leftrightarrow \dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}+\dfrac{x-2004}{2000}\dfrac{x-2004}{2002}=\dfrac{x-2004}{2}+\dfrac{x-2004}{4}+\dfrac{x-2004}{6}+\dfrac{x-2004}{8}+\dfrac{x-2004}{10}\\ \Leftrightarrow \dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}+\dfrac{x-2004}{2000}\dfrac{x-2004}{2002}-\dfrac{x-2004}{2}-\dfrac{x-2004}{4}-\dfrac{x-2004}{6}-\dfrac{x-2004}{8}-\dfrac{x-2004}{10}=0\\ \Leftrightarrow \left(x-2004\right)\left(\dfrac{1}{1994}+\dfrac{1}{1996}+\dfrac{1}{1998}+\dfrac{1}{2000}+\dfrac{1}{2002}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{10}=0\right)\\ \Leftrightarrow x-2004=0\\ \Leftrightarrow x=2004\)

a, 5(x-3)-4=2(x-1)+7

<=>\(5x-15-4=2x-2+7\)

\(\Leftrightarrow5x-2x=15+4-2+7\)

\(\Leftrightarrow3x=24\)

\(\Leftrightarrow x=8\)

b, \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)

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

\(\Rightarrow8x-3-6x+4=4x-2+x+3\)

\(\Leftrightarrow8x-6x-4x-x=3+4-2+3\)

\(\Leftrightarrow-3x=8\)

\(\Leftrightarrow x=\frac{-8}{3}\)

c,\(\frac{2\left(x+5\right)}{3}+\frac{x+12}{2}-\frac{5\left(x-2\right)}{6}=\frac{x}{3}+11\)

<=>\(\frac{4\left(x+5\right)}{6}+\frac{3\left(x+12\right)}{6}-\frac{5\left(x-2\right)}{6}=\frac{2x}{6}+\frac{66}{6}\)

\(\Leftrightarrow\frac{4x+20}{6}+\frac{3x+36}{6}-\frac{5x-10}{6}=\frac{2x}{6}+\frac{66}{6}\)

\(\Rightarrow4x+20+3x+36-5x+10=2x+66\)

\(\Leftrightarrow4x+3x-5x-2x=66-20-36-10\)

\(\Leftrightarrow0=0\)

a) \(\frac{x}{3}-\frac{5x}{6}-\frac{15x}{12}=\frac{x}{4}-5\)

\(\Leftrightarrow\frac{4x-10x-15x}{12}=\frac{3x-60}{12}\)

\(\Leftrightarrow-21x=3x-60\)

\(\Leftrightarrow24x=60\)

\(\Leftrightarrow x=\frac{5}{2}\)

Vậy tập nghiệm của phương trình là \(S=\left\{\frac{5}{2}\right\}\)

b) \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)

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

\(\Leftrightarrow8x-3-6x+4=4x-2+x+3\)

\(\Leftrightarrow2x+1=5x+1\)

\(\Leftrightarrow2x=5x\)

\(\Leftrightarrow x=0\)

Vậy tập nghiệm của phương trình là \(S=\left\{0\right\}\)

c) \(\frac{x-1}{2}-\frac{x+1}{15}-\frac{2x-13}{6}=0\)

\(\Leftrightarrow\frac{15\left(x-1\right)-2\left(x+1\right)-5\left(2x-13\right)}{30}=0\)

\(\Leftrightarrow15x-15-2x-2-10x+65=0\)

\(\Leftrightarrow3x+48=0\)

\(\Leftrightarrow x=-16\)

Vậy tập nghiệm của phương trình là \(S=\left\{-16\right\}\)

d) \(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\)

\(\Leftrightarrow\frac{9\left(3-x\right)+16\left(5-x\right)}{24}=\frac{12\left(1-x\right)-48}{24}\)

\(\Leftrightarrow27-9x+80-16x=12-12x-48\)

\(\Leftrightarrow-25x+107=-12x-36\)

\(\Leftrightarrow-13x+143=0\)

\(\Leftrightarrow x=11\)

Vậy tập nghiệm của phương trình là \(S=\left\{11\right\}\)

\(\Leftrightarrow1-\frac{6}{x^2+2}+1-\frac{12}{x^2+8}+1-\frac{7}{x^2+3}=0\)

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

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

\(\Leftrightarrow x^2-4=0\Rightarrow x=\pm2\)

@Nguyễn Việt Lâm

Câu 1 :

a, Ta có : \(3\left(x-1\right)-2\left(x+3\right)=-15\)

=> \(3x-3-2x-6=-15\)

=> \(3x-3-2x-6+15=0\)

=> \(x=-6\)

Vậy phương trình có nghiệm là x = -6 .

b, Ta có : \(3\left(x-1\right)+2=3x-1\)

=> \(3x-3+2=3x-1\)

=> \(3x-3+2-3x+1=0\)

=> \(0=0\)

Vậy phương trình có vô số nghiệm .

c, Ta có : \(7\left(2-5x\right)-5=4\left(4-6x\right)\)

=> \(14-35x-5=16-24x\)

=> \(14-35x-5-16+24x=0\)

=> \(-35x+24x=7\)

=> \(x=\frac{-7}{11}\)

Vậy phương trình có nghiệm là \(x=\frac{-7}{11}\) .

Bài 2 :

a, Ta có : \(\frac{x}{30}+\frac{5x-1}{10}=\frac{x-8}{15}-\frac{2x+3}{6}\)

=> \(\frac{x}{30}+\frac{3\left(5x-1\right)}{30}=\frac{2\left(x-8\right)}{30}-\frac{5\left(2x+3\right)}{30}\)

=> \(x+3\left(5x-1\right)=2\left(x-8\right)-5\left(2x+3\right)\)

=> \(x+15x-3=2x-16-10x-15\)

=> \(x+15x-3-2x+16+10x+15=0\)

=> \(24x+28=0\)

=> \(x=\frac{-28}{24}=\frac{-7}{6}\)

Vậy phương trình có nghiệm là \(x=\frac{-7}{6}\) .

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

=> \(\frac{6\left(x+4\right)}{30}-\frac{30x}{30}+\frac{120}{30}=\frac{10x}{30}-\frac{15\left(x-2\right)}{30}\)

=> \(6\left(x+4\right)-30x+120=10x-15\left(x-2\right)\)

=> \(6x+24-30x+120=10x-15x+30\)

=> \(6x+24-30x+120-10x+15x-30=0\)

=> \(-19x+114=0\)

=> \(x=\frac{-114}{-19}=6\)

Vậy phương trình có nghiệm là x = 6 .

pt đầu \(\Leftrightarrow x+1+\frac{1}{x+1}+x+7+\frac{7}{x+7}=x+3+\frac{3}{x+3}+x+5+\frac{5}{x+5}\)

\(\Rightarrow\frac{1}{x+1}+\frac{7}{x+7}=\frac{3}{x+3}+\frac{5}{x+5}\\ \Rightarrow\frac{8x+14}{x^2+8x+7}=\frac{8x+30}{x^2+8x+15}\)

\(\Leftrightarrow\left(4x+7\right)\left(x^2+8x+15\right)=\left(4x+15\right)\left(x^2+8x+7\right)\)

Đặt a=4x+7

      b=x² +8x+7

như vậy ta được pt mới có dạng \(a\left(b+8\right)=b\left(a+8\right)\Leftrightarrow ab+8a=ab+8b\Rightarrow a=b\)

hay\(4x+7=x^2+8x+7\Rightarrow x^2+4x=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=-4\end{cases}}\)