a, Ta có : \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\)

=> \(\frac{3\left(2x-1\right)}{15}-\frac{5\left(x-2\right)}{15}=\frac{x+7}{15}\)

=> \(3\left(2x-1\right)-5\left(x-2\right)=x+7\)

=> \(6x-3-5x+10-x-7=0\)

=> \(0=0\)

Vậy phương trình có vô số nghiệm .

b, Ta có : \(\frac{x+3}{2}-\frac{x-1}{3}=\frac{x+5}{6}+1\)

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

=> \(3\left(x+3\right)-2\left(x-1\right)=x+5+6\)

=> \(3x+9-2x+2-x-5-6=0\)

=> \(0=0\)

Vậy phương trình có vô số nghiệm .

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

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

=> \(4\left(x+5\right)+3\left(x+12\right)-5\left(x-2\right)=2x+66\)

=> \(4x+20+3x+36-5x+10-2x-66=0\)

=> \(0=0\)

Vậy phương trình có vô số nghiệm .

a)

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

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

\(\Leftrightarrow\frac{-10x-11}{12}=\frac{3x-60}{12}\)

\(\Leftrightarrow\frac{-10x-11-3x+60}{12}=0\)

\(\Leftrightarrow\frac{49-13x}{12}=0\)

\(\Rightarrow49-13x=0\)

\(\Rightarrow x=\frac{-49}{13}\)

b)

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

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

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

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

\(\Leftrightarrow\frac{-3x}{4}=0\)

\(\Rightarrow-3x=0\)

\(\Rightarrow x=0\)

e)

\(\frac{3\cdot\left(5x-2\right)}{4}-2=\frac{7x}{3}-5\cdot\left(x-7\right)\)

\(\Leftrightarrow\frac{45x-18-24-28x+60x-420}{12}=0\)

\(\Leftrightarrow\frac{77x-462}{12}=0\)

\(\Rightarrow77x-462=0\)

\(\Rightarrow x=\frac{462}{77}=6\)

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 !

1.

\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)

\(MC:12\)

Quy đồng :

\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)

\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)

\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)

\(\Leftrightarrow6x+9-3x=-4-9+16\)

\(\Leftrightarrow-7x=3\)

\(\Leftrightarrow x=\frac{-3}{7}\)

2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)

\(MC:20\)

Quy đồng :

\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)

\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)

\(\Leftrightarrow30x+15-20=15x-2\)

\(\Leftrightarrow15x=3\)

\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)

c,chia cả tử và mẫu cho x,sau đó đặt 3x+2/x=t

các câu còn lại hiện chưa giải đc vì chưa có giấy nháp,lúc nào rảnh mình chỉ cho cách làm

a) 3x - 2(5 + 2x) =45 - 2x

=> 3x - 10 - 4x = 45 - 2x

=> 3x - 4x + 2x = 45 + 10

=> x = 55

b) \(\frac{x-3}{5}=6-\frac{1-2x}{3}\)

=> \(\frac{x-3}{5}=\frac{2x+17}{3}\)

=> 5(2x + 17) = 3(x - 3)

=> 10x + 85 = 3x - 9

=> 7x = -94

=> x = -94/7

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

=> \(\frac{5x-3}{6}-\frac{7x-1}{4}=\frac{4x-33}{7}\)

=> \(\frac{10x-6}{12}-\frac{21x-3}{12}=\frac{4x-33}{7}\)

=> \(\frac{-11x-3}{12}=\frac{4x-33}{7}\)

=> (-11x - 3).7 = (4x - 33).12

= -77x - 21 = 48x - 396

=> x = 3

d) (x - 1)(5x + 3) = (3x - 8)(x - 1)

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

=> (x - 1)(2x + 11) = 0

=> \(\orbr{\begin{cases}x-1=0\\2x+11=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-5,5\end{cases}}\) 

e) (x - 1)(x² + 5x - 2) - (x³ - 1) = 0

=> (x - 1)(x² + 5x - 2) - (x - 1)(x² + x + 1) = 0

=> (x - 1)(4x - 3) = 0

=> \(\orbr{\begin{cases}x-1=0\\4x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=0,75\end{cases}}\)

f) \(\frac{x-17}{33}+\frac{x-21}{29}+\frac{x}{25}=4\) 

=> \(\left(\frac{x-17}{33}-1\right)+\left(\frac{x-21}{29}-1\right)+\left(\frac{x}{25}-2\right)=0\)

=> \(\frac{x-50}{33}+\frac{x-50}{29}+\frac{x-50}{25}=0\)

=> \(\left(x-50\right)\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)=0\)

=> x - 50 = 0 (Vì \(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\ne0\))

=> x = 50