a: =>\(\dfrac{5x-15+4x-8}{\left(x-2\right)\left(x-3\right)}=\dfrac{1}{x}\)

=>\(\dfrac{9x-23}{\left(x-2\right)\left(x-3\right)}=\dfrac{1}{x}\)

=>9x^2-23x=x^2-5x+6

=>8x^2-18x-6=0

=>\(x=\dfrac{9\pm\sqrt{129}}{8}\)

b: =>\(\dfrac{12x+1}{11x-4}=\dfrac{20x+17-20x+8}{18}=\dfrac{25}{18}\)

=>216x+18=275x-100

=>-59x=-118

=>x=2

\(\frac{12x+1}{11x-4}+\frac{10x-4}{9}=\frac{20x+17}{18}\left(ĐKXĐ:x\ne\frac{4}{11}\right).\)

\(\Leftrightarrow\frac{18.\left(12x+1\right)}{18.\left(11x-4\right)}+\frac{2.\left(11x-4\right).\left(10x-4\right)}{18.\left(11x-4\right)}=\frac{\left(11x-4\right).\left(20x+17\right)}{18.\left(11x-4\right)}\)

\(\Rightarrow18.\left(12x+1\right)+2.\left(11x-4\right).\left(10x-4\right)=\left(11x-4\right).\left(20x+17\right)\)

\(\Leftrightarrow216x+18+\left(22x-8\right).\left(10x-4\right)=220x^2+187x-80x-68\)

\(\Leftrightarrow216x+18+220x^2-88x-80x+32=220x^2+107x-68\)

\(\Leftrightarrow48x+50+220x^2=220x^2+107x-68\)

\(\Leftrightarrow48x+50+220x^2-220x^2-107x+68=0\)

\(\Leftrightarrow118-59x=0\)

\(\Leftrightarrow59x=118-0\)

\(\Leftrightarrow59x=118\)

\(\Leftrightarrow x=118:59\)

\(\Leftrightarrow x=2\left(TM\right).\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2\right\}.\)

Chúc bạn học tốt!

f: =>\(\dfrac{14}{3\left(x-4\right)}-\dfrac{x+2}{x-4}=\dfrac{-3}{2\left(x-4\right)}-\dfrac{5}{6}\)

=>28-6(x+2)=-9-5(x-4)

=>28-6x-12=-9-5x+20

=>-6x+16=-5x+11

=>-x=-5

=>x=5

d: =>\(\dfrac{12x+1}{11x-4}=\dfrac{20x+17-20x+8}{18}=\dfrac{25}{18}\)

=>25(11x-4)=18(12x+1)

=>275x-100=216x+18

=>59x=118

=>x=2

f: =>\(\dfrac{14}{3\left(x-4\right)}-\dfrac{x+2}{x-4}=\dfrac{-3}{2\left(x-4\right)}-\dfrac{5}{6}\)

=>28-6(x+2)=-9-5(x-4)

=>28-6x-12=-9-5x+20

=>-6x+16=-5x+11

=>-x=-5

=>x=5

bạn tách một câu vài câu hỏi chứ đừng gộp như thế này ko ai trả lời đâu

a: =>\(4x-5=2x-2+x=3x-2\)

=>x=3

b: \(\Leftrightarrow7x-35=3x+6\)

=>4x=41

=>x=41/4

c: =>(2x+5)(x+5)-2x^2=0

=>2x^2+10x+5x+25-2x^2=0

=>15x=-25

=>x=-5/3

e: \(\Leftrightarrow\dfrac{11}{x}=\dfrac{9x-36+2x+2}{\left(x+1\right)\left(x-4\right)}\)

=>11(x^2-3x-4)=x(11x-34)

=>11x^2-33x-44=11x^2-34x

=>x=44

a) \(\dfrac{4x-1}{3x^2y}-\dfrac{7x-1}{3x^2y}\)

\(=\dfrac{\left(4x-1\right)-\left(7x-1\right)}{3x^2y}\)

\(=\dfrac{4x-1-7x+1}{3x^2y}\)

\(=\dfrac{-3x}{3x^2y}\)

\(=\dfrac{-1}{xy}\)

b) \(\dfrac{4x+5}{2x-1}-\dfrac{5-9x}{2x-1}\)

\(=\dfrac{\left(4x+5\right)-\left(5-9x\right)}{2x-1}\)

\(=\dfrac{4x+5-5+9x}{2x-1}\)

\(=\dfrac{13x}{2x-1}\)

c) \(\dfrac{11x}{2x-3}-\dfrac{x-18}{3-2x}\)

\(=\dfrac{11x}{2x-3}+\dfrac{x-18}{2x-3}\)

\(=\dfrac{11x+\left(x-18\right)}{2x-3}\)

\(=\dfrac{11x+x-18}{2x-3}\)

\(=\dfrac{12x-18}{2x-3}\)

\(=\dfrac{6\left(2x-3\right)}{2x-3}\)

\(=\dfrac{6}{1}\)

\(=6\)

d) \(\dfrac{2x-7}{10x-4}-\dfrac{3x+5}{4-10x}\)

\(=\dfrac{2x-7}{10x-4}+\dfrac{3x+5}{10x-4}\)

\(=\dfrac{\left(2x-7\right)+\left(3x+5\right)}{10x-4}\)

\(=\dfrac{2x-7+3x+5}{10x-4}\)

\(=\dfrac{5x-2}{10x-4}\)

\(=\dfrac{5x-2}{2\left(5x-2\right)}\)

\(=\dfrac{1}{2}\)

a)

b)

c)

d)

1. ( 3x + 2)² - 4

= (3x+2-2)(3x+2+2)

= 3x(3x+4)

2. 4x² - 25y²

= (2x-5y)(2x+5y)

3. 4x²- 49

=(2x-7)(2x+7)

4. 8z³ + 27

=(2z+3)(4x²-6z+9)

5. \(\dfrac{9}{25}x^4-\dfrac{1}{4}\)

= \((\dfrac{3}{5}x^2-\dfrac{1}{2})(\dfrac{3}{5}x^2+\dfrac{1}{2})\)

6. x³²  - 1

=(x¹⁶-1)(x¹⁶+1)

7. 4x² + 4x + 1

=(2x+1)²

8. x² - 20x + 100

=(x-10)²

9. y⁴ -14y² + 49

=(y²-7)²

10.  125x³ - 64y³

= (5x-4y)(25x²+20xy+16y²)

1) \(\left(3x+2\right)^2-4=\left(3x+2+2\right)\left(3x+2-2\right)=3x\left(3x+4\right)\)

2) \(4x^2-25y^2=\left(2x-5y\right)\left(2x+5y\right)\)

3) \(4x^2-49=\left(2x-7\right)\left(2x+7\right)\)

4) \(8z^3+27=\left(2z+3\right)\left(4z^2-6z+9\right)\)

5) \(\dfrac{9}{25}x^4-\dfrac{1}{4}=\left(\dfrac{3}{5}x^2-\dfrac{1}{2}\right)\left(\dfrac{3}{5}x^2+\dfrac{1}{2}\right)\)

6) \(x^{32}-1=\left(x^{16}-1\right)\left(x^{16}+1\right)\)

\(=\left(x^8-1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)

7) \(4x^2+4x+1=\left(2x+1\right)^2\)

8) \(x^2-20x+100=\left(x-10\right)^2\)

9) \(y^4-14y^2+49=\left(y^2-7\right)^2\)

lam kieu gi