Bài làm

a) \(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x-4}\)

\(\Leftrightarrow\frac{3x+2}{3x-2}-\frac{6}{3x+2}=\frac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

\(\Leftrightarrow\frac{(3x+2)\left(3x+2\right)}{(3x-2)\left(3x+2\right)}-\frac{6\left(3x-2\right)}{(3x+2)\left(3x-2\right)}=\frac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

\(\Rightarrow\left(3x+2\right)^2-\left(18x-12\right)=9x^2\)

\(\Leftrightarrow9x^2+12x+4-18x+12x-9x^2=0\)

\(\Leftrightarrow6x+4=0\)

\(\Leftrightarrow x=-\frac{4}{6}\)

\(\Leftrightarrow x=-\frac{2}{3}\)

Vậy x = -2/3 là nghiệm.

@Tao Ngu :))@ 9x-4 không tách thành (3x+4)(3x-4) được đâu bạn. Chỗ đó phải là: 9x²-4

Bài thiếu đkxđ của x \(\hept{\begin{cases}3x-2\ne0\\2+3x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}3x\ne2\\3x\ne-2\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne\frac{2}{3}\\x\ne\frac{-2}{3}\end{cases}\Leftrightarrow}x\ne\pm\frac{2}{3}}\)

b) Bạn kiểm tra lại đề bài

c) \(\frac{3}{1-4x}=\frac{2}{4x+1}-\frac{8}{16x^2-1}\left(x\ne\pm\frac{1}{4}\right)\)

\(\Leftrightarrow\frac{3}{1-4x}-\frac{2}{4x+1}+\frac{8}{16x^2-1}=0\)

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

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

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

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

=> -20x+13=0

<=> -20x=-13

<=> \(x=\frac{13}{20}\left(tmđk\right)\)

a) 7x - 35 = 0

<=> 7x = 0 + 35

<=> 7x = 35

<=> x = 5

b) 4x - x - 18 = 0

<=> 3x - 18 = 0

<=> 3x = 0 + 18

<=> 3x = 18

<=> x = 5

c) x - 6 = 8 - x

<=> x - 6 + x = 8

<=> 2x - 6 = 8

<=> 2x = 8 + 6

<=> 2x = 14

<=> x = 7

d) 48 - 5x = 39 - 2x

<=> 48 - 5x + 2x = 39

<=> 48 - 3x = 39

<=> -3x = 39 - 48

<=> -3x = -9

<=> x = 3

có bị viết nhầm thì thông cảm nha!

la`thu'hai nga`y 19 nhe

a) 4 ( x + 5 )( x + 6 )( x + 10 )( x + 12 ) = 3x²
Do x = 0 không là nghiệm pt nên chia 2 vế pt cho \(x^2\ne0\), ta được :

\(\frac{4}{x^2}\left(x^2+60+17x\right)\left(x^2+60+16x\right)=3\)

\(\Leftrightarrow4\left(x+\frac{60}{x}+17\right)\left(x+\frac{60}{x}+16\right)=3\)

Đến đây ta đặt  \(x+\frac{60}{x}+16=t\left(1\right)\)

Ta được :

\(4t\left(t+1\right)=3\Leftrightarrow4t^2+4t-3=0\Leftrightarrow\left(2t+3\right)\left(2t-1\right)=0\)

Từ đó ta lắp vào ( 1 ) tính được x 

Đây là những bài cơ bản mà bạn!

\(\frac{5x-2}{3}=\frac{5-3x}{2}\)

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

\(< =>\left(5x-2\right).2=\left(5-3x\right).3\)

\(< =>10x-4=15-9x\)

\(< =>10x+9x=15+4\)

\(< =>19x=19< =>x=1\)

\(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)

\(< =>\frac{\left(10x+3\right).3}{36}=\frac{36}{36}+\frac{\left(6+8x\right).4}{36}\)

\(< =>\left(10x+3\right).3=36+\left(6+8x\right).4\)

\(< =>30x+9=36+24+32x\)

\(< =>32x-30x=9-36-24\)

\(< =>2x=9-60=-51< =>x=-\frac{51}{2}\)

a) <=> \(6x^2-5x+3-2x+3x\left(3-2x\right)=0\)

<=> \(6x^2-5x+3-2x+9x-6x^2=0\)

<=> \(2x+3=0\)

<=> \(x=\frac{-3}{2}\)

b) <=> \(10\left(x-4\right)-2\left(3+2x\right)=20x+4\left(1-x\right)\)

<=> \(10x-40-6-4x=20x+4-4x\)

<=> \(6x-46-16x-4=0\)

<=> \(-10x-50=0\)

<=> \(-10\left(x+5\right)=0\)

<=> \(x+5=0\)

<=> \(x=-5\)

c) <=> \(8x+3\left(3x-5\right)=18\left(2x-1\right)-14\)

<=> \(8x+9x-15=36x-18-14\)

<=> \(8x+9x-36x=+15-18-14\)

<=> \(-19x=-14\)

<=> \(x=\frac{14}{19}\)

d) <=>\(2\left(6x+5\right)-10x-3=8x+2\left(2x+1\right)\)

<=> \(12x+10-10x-3=8x+4x+2\)

<=> \(2x-7=12x+2\)

<=> \(2x-12x=7+2\)

<=> \(-10x=9\)

<=> \(x=\frac{-9}{10}\)

e) <=> \(x^2-16-6x+4=\left(x-4\right)^2\)

<=> \(x^2-6x-12-\left(x-4^2\right)=0\)

<=> \(x^2-6x-12-\left(x^2-8x+16\right)=0\)

<=> \(x^2-6x-12-x^2+8x-16=0\)

<=> \(2x-28=0\)

<=> \(2\left(x-14\right)=0\)

<=> x-14=0

<=> x=14

Luffy , cậu sai câu c nhé , kia là -17 ạ => x=17/19

d: =>4x+6=15x-12

=>4x-15x=-12-6=-18

=>-11x=-18

hay x=18/11

e: =>\(45x+27=12+24x\)

=>21x=-15

hay x=-5/7

f: =>35x-5=96-6x

=>41x=101

hay x=101/41

g: =>3(x-3)=90-5(1-2x)

=>3x-9=90-5+10x

=>3x-9=10x+85

=>-7x=94

hay x=-94/7

\(\frac{3x-7}{5}=\frac{2x-1}{3}\)

\(\Leftrightarrow9x-21=10x-5\)

\(\Leftrightarrow-x=16\Leftrightarrow x=-16\)

\(\frac{4x-7}{12}-x=\frac{3x}{8}\)

\(\Leftrightarrow\frac{4x-7-12x}{12}=\frac{3x}{8}\)

\(\Leftrightarrow\frac{-7-8x}{12}=\frac{3x}{8}\)

\(\Leftrightarrow-56-64x=36x\)

\(\Leftrightarrow-56=100x\Leftrightarrow x=\frac{-14}{25}\)

\(\frac{x-2009}{1234}+\frac{x-2009}{5678}-\frac{x-2009}{197}=0\)

\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)=0\)

Vì \(\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)\ne0\)nên x - 2019 = 0

Vậy x = 2019

\(\frac{5x-8}{3}=\frac{1-3x}{2}\)

\(\Leftrightarrow10x-16=3-9x\)

\(\Leftrightarrow19x=19\Leftrightarrow x=1\)

\(\frac{x-5}{6}-\frac{x-9}{4}=\frac{5x-3}{8}+2\)

\(\Rightarrow\frac{4x-20-6x+54}{24}=\frac{5x-3+16}{8}\)

\(\Rightarrow\frac{-2x+34}{24}=\frac{5x+13}{8}\)

\(\Rightarrow-16x-272=120x+312\)

\(\Leftrightarrow-136x=584\Leftrightarrow x=\frac{-73}{17}\)

a) \(\frac{x+5}{4}\)-\(\frac{2x-5}{3}\)=\(\frac{6x-1}{3}\)+\(\frac{2x-3}{12}\)

⇔\(\frac{3\left(x+5\right)}{12}\)-\(\frac{4\left(2x-5\right)}{12}\)=\(\frac{4\left(6x-1\right)}{12}\)+\(\frac{2x-3}{12}\)

⇒ 3x+15-8x+20=24x-4+2x-3

⇔3x+15-8x+20-24x+4-2x+3=0

⇔-31x+42=0

⇔x=\(\frac{42}{31}\)

Vậy tập nghiệm của phương trình đã cho là:S={\(\frac{42}{31}\)}

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

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

⇒4x+6=15-12x

⇔16x=9

⇔ x=\(\frac{9}{16}\)

Vậy tập nghiệm của phương trình đã cho là:S={\(\frac{9}{16}\)}

Phạm Thái Hải hay lm 2 câu trog 4 hay 5 câu thôi nhỉ

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

\(2\left(3x-2\right)-60=3\left(-2x-11\right)\)

\(6x-4-60=-6x-33\)

\(6x-4-60+6x+33=0\)

\(12x-31=0\)

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