\(1,\dfrac{4x-3}{x-5}=\dfrac{29}{3}\left(ĐKXĐ:x\ne5\right)\)

\(\Rightarrow3\left(4x-3\right)=29\left(x-5\right)\)

\(\Leftrightarrow12x-9=29x-145\)

\(\Leftrightarrow12x-9-29x+145=0\)

\(\Leftrightarrow-17x+136=0\)

\(\Leftrightarrow-17x=-136\)

\(\Leftrightarrow x=8\left(tm\right)\)

Vậy \(S=\left\{8\right\}\)

\(2,\dfrac{2x-1}{5-3x}=2\left(ĐKXĐ:x\ne\dfrac{5}{3}\right)\)

\(\Rightarrow2x-1=2\left(5-3x\right)\)

\(\Leftrightarrow2x-1=10-6x\)

\(\Leftrightarrow2x-1-10+6x=0\)

\(\Leftrightarrow8x-11=0\)

\(\Leftrightarrow8x=11\)

\(\Leftrightarrow x=\dfrac{11}{8}\left(tm\right)\)

Vậy \(S=\left\{\dfrac{11}{8}\right\}\)

\(3,\dfrac{4x-5}{x-1}=2+\dfrac{x}{x-1}\left(ĐKXĐ:x\ne1\right)\)

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

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

\(\Leftrightarrow\dfrac{4x-5}{x-1}=\dfrac{3x-2}{x-1}\)

\(\Rightarrow4x-5=3x-2\)

\(\Leftrightarrow4x-5-3x+2=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\left(tm\right)\)

Vậy \(S=\left\{3\right\}\)

\(4,\dfrac{2x+5}{2x}-\dfrac{x}{x+5}=0\left(ĐKXĐ:x\ne\dfrac{1}{2};x\ne-5\right)\)

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

\(\Leftrightarrow\dfrac{2x^2+15x+25}{2x\left(x+5\right)}-\dfrac{2x^2}{2x\left(x+5\right)}=0\)

\(\Leftrightarrow\dfrac{15x+25}{2x\left(x+5\right)}=0\)

\(\Rightarrow15x+25=0\)

\(\Leftrightarrow15x=-25\)

\(\Leftrightarrow x=\dfrac{-5}{3}\left(tm\right)\)

Vậy \(S=\left\{\dfrac{-5}{3}\right\}\)

\(1,\dfrac{4x-3}{x-5}=\dfrac{29}{3}\)

\(\Leftrightarrow\dfrac{3\left(4x-3\right)-29\left(x-5\right)}{3\left(x-5\right)}=0\)

\(\Leftrightarrow12x-9-29x+145=0\)

\(\Leftrightarrow-17x=-136\)

\(\Leftrightarrow x=8\)

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

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

\(\Leftrightarrow2x-1-10+6x=0\)

\(\Leftrightarrow8x=11\)

\(\Leftrightarrow x=\dfrac{11}{8}\)

\(3,\dfrac{4x-5}{x-1}=2+\dfrac{x}{x-1}\)

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

\(\Leftrightarrow4x-5-2x+2+2x=0\)

\(\Leftrightarrow4x=3\)

\(\Leftrightarrow x=\dfrac{3}{4}\)

\(4,\dfrac{2x+5}{2x}-\dfrac{x}{x+5}=0\)

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

\(\Leftrightarrow2x^2+10x+5x+25-2x^2=0\)

\(\Leftrightarrow15x=-25\)

\(\Leftrightarrow x=-\dfrac{5}{3}\)

a: (3x-2)(4x+5)=0

=>3x-2=0 hoặc 4x+5=0

=>x=2/3 hoặc x=-5/4

b: (2,3x-6,9)(0,1x+2)=0

=>2,3x-6,9=0 hoặc 0,1x+2=0

=>x=3 hoặc x=-20

c: =>(x-3)(2x+5)=0

=>x-3=0 hoặc 2x+5=0

=>x=3 hoặc x=-5/2

a) (x + 6)(3x + 1) + x² - 36 = 0

<=> 3x² + x + 18x + 6 + x² - 36 = 0

<=> 4x² + 19x - 30 = 0

<=> 4x² + 24x - 5x - 30 = 0

<=> 4x(x + 6) - 5(x + 6) = 0

<=> (x + 6)(4x - 5) = 0

<=> x + 6 = 0 hoặc 4x - 5 = 0

<=> x = -6 hoặc x = 5/4

Bài 1 mình đã làm xong rồi, anh em nào giúp mình bài 2 với!

Ta có

4x-8=9x-3-2x+1

<=>-6=-3x(chuyển vế đổi dấu)

<=>x=2

b)

Ta có

Căn cả 2 vế ta đcx-5/ cawn3 =3 

<=>x=10.2

\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)

\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)

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

\(\Leftrightarrow x^2-9-x^2+3x=0\)

\(\Leftrightarrow3x-9=0\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\left(n\right)\)

Vậy \(S=\left\{3\right\}\)

\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)

\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)

\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)

\(\Leftrightarrow12x-9-12x+20+2x-7>0\)

\(\Leftrightarrow2x+4>0\)

\(\Leftrightarrow2x>-4\)

\(\Leftrightarrow x>-2\)

1) \(x^4-2x^2-144x+1295=0\)

\(\Rightarrow\)Cậu xem lại đề thử xem nhé !

2) \(x\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\)

\(\Leftrightarrow\left(x^2+2x\right)\left(x^2-1\right)-24=0\)

\(\Leftrightarrow x^4+2x^3-x^2-2x-24=0\)

\(\Leftrightarrow x^4+x^3+4x^2+x^3+x^2+4x-6x^2-6x-24=0\)

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

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

\(\Leftrightarrow\left(x^2+3x-2x-6\right)\left(x^2+x+4\right)=0\)

\(\Leftrightarrow\left[x\left(x+3\right)-2\left(x+3\right)\right]\left(x^2+x+4\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-2\right)\left(x^2+x+4\right)=0\)

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

hoặc \(x-2=0\)

hoặc \(x^2+x+4=0\)

\(\Leftrightarrow\)\(x=-3\left(tm\right)\)

hoặc   \(x=2\left(tm\right)\)

hoặc  \(\left(x+\frac{1}{2}\right)^2+\frac{15}{4}=0\left(ktm\right)\)

Vậy tập nghiệm của phương trình là : \(S=\left\{-3;2\right\}\)

3) \(x^4-2x^3+4x^2-3x-10=0\)

\(\Leftrightarrow x^4+x^3-3x^3-3x^2+7x^2+7x-10x-10=0\)

\(\Leftrightarrow x^3\left(x+1\right)-3x^2\left(x+1\right)+7x\left(x+1\right)-10\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^3-3x^2+7x-10\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^3-2x^2-x^2+2x+5x-10\right)=0\)

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

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

\(\Leftrightarrow\)\(x+1=0\)

hoặc \(x-2=0\)

hoặc \(x^2-x+5=0\)

\(\Leftrightarrow x=-1\left(tm\right)\)

hoặc \(x=2\left(tm\right)\)

hoặc \(\left(x-\frac{1}{2}\right)^2+\frac{19}{4}=0\left(ktm\right)\)

Vậy tập nghiệm của phương trình là :\(S=\left\{-1;2\right\}\)