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

A=(8xy-6x^2)/(12y^2-9xy)

A=2x(4y-3x)/3y(4y-3x)

A=2x/3y

B=(2x^3-18x)/(x^4-81)

B=2x(x^2-9)/(x^2-9)(x^2+9)

B=2x/(x^2+9)

C=(x^2-x-30)/(x^2-25)

C=(x^2+6x-5x-30)/(x^2-25)

C=(x(x+6)-5(x+6))/(x-5)(x+5)

C=(x+6)(x-5)/(x-5)(x+5)

C=(x+6)/(x+5)

Bài 2: \(a,\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\) 

 \(\frac{5-3x}{x^2-9}=\frac{5-3x}{\left(x-3\right)\left(x+3\right)}=\frac{\left(5-3x\right)2x}{2x\left(x-3\right)\left(x+3\right)}\)

\(b,\frac{x+1}{x-x^2}=\frac{x+1}{x\left(1-x\right)}=-\frac{x+1}{x\left(x+1\right)}=-\frac{2\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)^2}\) 

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

\(c,\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\) 

\(\frac{2x}{x^2+x+1}=\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(\frac{6}{x-1}=\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(d,\frac{7}{5x}=\frac{7.2\left(2y-x\right)\left(2y+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)

\(\frac{4}{x-2y}=-\frac{4}{2y-x}=-\frac{4.2.5x\left(2x+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)

\(\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2\left(4y^2-x^2\right)}=\frac{x-y}{2\left(2y-x\right)\left(2y+x\right)}=\frac{5x\left(x-y\right)}{2.5x.\left(2y-x\right)\left(2y+x\right)}\)

\(5X\left(X-2020\right)+X=2020\)

\(\Leftrightarrow5X^2-10100X+X=2020\)

\(\Leftrightarrow5X^2-10099X=2020\)

\(\Leftrightarrow5X^2-10099X-2020=0\)

\(\Leftrightarrow5X^2-10100X+x-2020=0\)

\(\Leftrightarrow5X\left(X-2020\right)+X-2020=0\)

\(\Leftrightarrow\left(X-2020\right)\left(5X+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2020\\x=-\frac{1}{5}\end{cases}}\)

\(4\left(x-5\right)^2-\left(2x+1\right)^2=0\)

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

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

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

\(\Leftrightarrow-11\left(4x-9\right)=0\)

\(\Leftrightarrow x=\frac{9}{4}\)

\(a,5x\left(x-2020\right)+x=2020\)

\(< =>5x\left(x-2020\right)+x-2020=0\)

\(< =>\left(5x+1\right)\left(x-2020\right)=0\)

\(< =>\orbr{\begin{cases}5x+1=0\\x-2020=0\end{cases}}\)

\(< =>\orbr{\begin{cases}5x=-1\\x=2020\end{cases}< =>\orbr{\begin{cases}x=-\frac{1}{5}\\x=2020\end{cases}}}\)

\(b,4\left(x-5\right)^2-\left(2x+1\right)^2=0\)

\(< =>4\left(x^2-20x+25\right)-\left(4x^2+4x+1\right)=0\)

\(< =>4x^2-80x+100-4x^2-4x-1=0\)

\(< =>-84x+99=0< =>84x=99< =>x=\frac{99}{84}\)

434

hình như bằng 434

434

a) ĐKXĐ: x≠0

Ta có: \(\frac{9}{x}+2=-6\)

⇔\(\frac{9}{x}+2+6=0\)

⇔\(\frac{9}{x}+8=0\)

⇔\(\frac{9}{x}+\frac{8x}{x}=0\)

⇔9+8x=0

⇔8x=-9

hay \(x=-\frac{9}{8}\)

Vậy: \(x=-\frac{9}{8}\)

b) ĐKXĐ: x≠0;x≠-1;x≠-3

Ta có: \(\frac{7}{x+1}+\frac{-18x}{x\left(x^2+4x+3\right)}=\frac{-4}{x+3}\)

⇔\(\frac{7}{x+1}+\frac{-18x}{x\left(x+1\right)\left(x+3\right)}-\frac{-4}{x+3}=0\)

⇔\(\frac{7x\left(x+3\right)}{\left(x+1\right)\cdot x\cdot\left(x+3\right)}+\frac{-18x}{\left(x+1\right)\cdot x\cdot\left(x+3\right)}-\frac{-4x\left(x+1\right)}{\left(x+3\right)\cdot x\cdot\left(x+1\right)}=0\)

⇔\(7x^2+21x-18x+4x\left(x+1\right)=0\)

\(\Leftrightarrow7x^2+21x-18x+4x^2+4x=0\)

⇔\(11x^2+7x=0\)

\(\Leftrightarrow x\left(11x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\11x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\11x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=\frac{-7}{11}\end{matrix}\right.\)

Vậy: \(x=\frac{-7}{11}\)

c) ĐKXĐ: x≠1; x≠-3

Ta có: \(\frac{3x-1}{x-1}-1=\frac{2x+5}{x+3}+\frac{4}{x^2-2x+3}\)

⇔\(\frac{3x-1}{x-1}-1-\frac{2x+5}{x+3}-\frac{4}{\left(x-1\right)\left(x+3\right)}=0\)

⇔\(\frac{\left(3x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{\left(x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{\left(2x+5\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}-\frac{4}{\left(x-1\right)\left(x+3\right)}=0\)

⇔\(\left(3x-1\right)\left(x+3\right)-\left(x-1\right)\left(x+3\right)-\left(2x+5\right)\left(x-1\right)-4=0\)

\(\Leftrightarrow3x^2+9x-x-3-\left(x^2+3x-x-3\right)-\left(2x^2-2x+5x-5\right)-4=0\)

\(\Leftrightarrow3x^2+8x-3-\left(x^2+2x-3\right)-\left(2x^2+3x-5\right)-4=0\)

\(\Leftrightarrow3x^2+8x-3-x^2-2x+3-2x^2-3x+5-4=0\)

\(\Leftrightarrow3x+1=0\)

\(\Leftrightarrow3x=-1\)

hay \(x=\frac{-1}{3}\)

Vậy: \(x=\frac{-1}{3}\)

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

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