1: Ta có: \(\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}\)

\(\Leftrightarrow5x+20+12x-28=7x+2\)

\(\Leftrightarrow17x-7x=2+8=10\)

hay x=1

2: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)

\(\Leftrightarrow\dfrac{6x}{36}+\dfrac{4\left(1-3x\right)}{36}=\dfrac{3\left(-x+1\right)}{36}\)

\(\Leftrightarrow6x+4-12x=-3x+3\)

\(\Leftrightarrow-6x+3x=3-4\)

hay \(x=\dfrac{1}{3}\)

3: Ta có: \(\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}\)

\(\Leftrightarrow4x-12-x-2=6x-3\)

\(\Leftrightarrow3x-14-6x+3=0\)

\(\Leftrightarrow-3x=11\)

hay \(x=-\dfrac{11}{3}\)

4: Ta có: \(\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\)

\(\Leftrightarrow3x-6-8x-12=x+6\)

\(\Leftrightarrow-5x-x=6+18\)

hay x=-4

5: Ta có: \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)

\(\Leftrightarrow6x-3+2x-6=-1\)

\(\Leftrightarrow8x=8\)

hay x=1

k,\(\dfrac{x}{3}-\dfrac{2x+1}{2}=\dfrac{x}{6}-x\)

giúp mk câu k nhé đề bài như trên

b: \(\Leftrightarrow4x+8-9=4x-4\)

=>-1=-4(loại)

d: \(\Leftrightarrow3\left(x-2\right)+2\left(x+1\right)=8x\)

=>8x=3x-6+2x+2=5x-4

=>3x=-4

=>x=-4/3

f: \(\Leftrightarrow3\left(x+2\right)+4\left(2x-3\right)=2\left(x-12\right)\)

=>3x+6+8x-12=2x-24

=>11x-6=2x-24

=>9x=-18

=>x=-2

\(1,\left(dk:x\ne0,-1,4\right)\)

\(\Leftrightarrow\dfrac{9}{x+1}+\dfrac{2}{x-4}-\dfrac{11}{x}=0\)

\(\Leftrightarrow\dfrac{9x\left(x-4\right)+2x\left(x+1\right)-11\left(x+1\right)\left(x-4\right)}{x\left(x+1\right)\left(x-4\right)}=0\)

\(\Leftrightarrow9x^2-36x+2x^2+2x-11x^2+44x-11x+44=0\)

\(\Leftrightarrow-x=-44\)

\(\Leftrightarrow x=44\left(tm\right)\)

\(2,\left(đk:x\ne4\right)\)

\(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{2+x}{x-4}-\dfrac{3}{2\left(x-4\right)}+\dfrac{5}{6}=0\)

\(\Leftrightarrow\dfrac{14.2-6\left(2+x\right)-3.3+5\left(x-4\right)}{6\left(x-4\right)}=0\)

\(\Leftrightarrow28-12-6x-9+5x-20=0\)

\(\Leftrightarrow-x=13\)

\(\Leftrightarrow x=-13\left(tm\right)\)

a) Ta có: \(\dfrac{x+5}{3x-6}-\dfrac{1}{2}=\dfrac{2x-3}{2x-4}\)

\(\Leftrightarrow\dfrac{2\left(x+5\right)}{6\left(x-2\right)}-\dfrac{3\left(x-2\right)}{6\left(x-2\right)}=\dfrac{3\left(2x-3\right)}{6\left(x-2\right)}\)

Suy ra: \(2x+5-3x+6=6x-9\)

\(\Leftrightarrow-x+11-6x+9=0\)

\(\Leftrightarrow20-7x=0\)

\(\Leftrightarrow7x=20\)

hay \(x=\dfrac{20}{7}\)(thỏa ĐK)

Vậy: \(S=\left\{\dfrac{20}{7}\right\}\)

a/ \(\dfrac{3-x}{12}=\dfrac{2x+2}{8}\)

\(< =>\dfrac{2\left(3-x\right)}{24}=\dfrac{3\left(2x+2\right)}{24}\)

\(< =>6-2x-6x-6=0\)

\(< =>-8x=0\)

\(< =>x=0\)

Vậy tập nghiệm.....

b/ \(\dfrac{x+3}{x-4}+\dfrac{x-3}{x+4}=\dfrac{2\left(x^2+12\right)}{x^2-16}\)

Tìm ĐKXĐ của pt là: \(x\ne\pm4\)  (làm tắt, bạn làm rõ ra nhé)

\(\dfrac{x+3}{x-4}+\dfrac{x-3}{x+4}=\dfrac{2\left(x^2+12\right)}{x^2-16}\)

\(< =>\dfrac{\left(x+3\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}+\dfrac{\left(x-3\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{2\left(x^2+12\right)}{\left(x+4\right)\left(x-4\right)}\)

\(< =>x^2+3x+4x+12+x^2-3x-4x+12-2x^2-24=0\)

\(< =>0x=0\)

=> x có vô số nghiệm

Vậy ....

a) `(3-x)/12=(2x+2)/8`

`<=> (3-x)/12 =(x+1)/4`

`<=> 3-x=3(x+1)`

`<=>3-x=3x+3`

`<=> x=0`

Vậy `S={0}`.

b) ĐK: `x \ne \pm 4`

`(x+3)/(x-4)+(x-3)/(x+4)=(2(x^2+12))/(x^2-16)`

`<=> (x+3)(x+4)+(x-3)(x-4)=2(x^2+12)`

`<=> x^2+7x+12+x^2-7x+12=2x^2+24`

`<=> 0x=0`

Vậy PT có nghiệm với mọi x thỏa mãn điều kiện.

\(\dfrac{x+3}{x-4}+\dfrac{x-3}{x+4}=\dfrac{2\left(x^2+12\right)}{x^2-16}\)

⇔\(\dfrac{\left(x+3\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}+\dfrac{\left(x-3\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{2\left(x^2+12\right)}{\left(x-4\right)\left(x+4\right)}\)

⇔\(\dfrac{x^2+4x+3x+12}{\left(x-4\right)\left(x+4\right)}+\dfrac{x^2-4x-3x+12}{\left(x-4\right)\left(x+4\right)}=\dfrac{2x^2+24}{\left(x-4\right)\left(x+4\right)}\)

⇔\(\dfrac{x^2+7x+12}{\left(x-4\right)\left(x+4\right)}+\dfrac{x^2-7x+12}{\left(x-4\right)\left(x+4\right)}=\dfrac{2x^2+24}{\left(x-4\right)\left(x+4\right)}\)

⇒ \(x^2+7x+12+x^2-7x+12=2x^2+24\)

⇔ \(2x^2+24=2x^2+24\)

⇔ \(2x^2-2x^2=24-24\)

⇔ x=0

a) Ta có: \(\left(x+1\right)\left(2x-3\right)-3\left(x-2\right)=2\left(x-1\right)^2\)

\(\Leftrightarrow2x^2-3x+2x-3-3x+6=2\left(x^2-2x+1\right)\)

\(\Leftrightarrow2x^2-4x+3-2x^2+4x-2=0\)

\(\Leftrightarrow1=0\)(vô lý)

Vậy: \(S=\varnothing\)

Ai giúp vs

Câu 1:

Câu 2:

ĐKXĐ: \(\left[{}\begin{matrix}1-9x^2\ne0\\1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Rightarrow \left[{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)

\(\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\left(1\right)\)

\(\left(1\right):\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}-\dfrac{\left(1-3x\right)\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\dfrac{\left(1+3x\right)\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}=0\)

\(\Leftrightarrow 12-\left(1-3x-3x+9x^2\right)+\left(1+3x+3x+9x^2\right)=0\)

\(\Leftrightarrow 12-1+3x+3x-9x^2+1+3x+3x+9x^2=0\)

\(\Leftrightarrow12x+12=0\\ \Leftrightarrow12x=-12\\ \Leftrightarrow x=-1\left(TM\right)\)

Vậy \(S=\left\{-1\right\}\)

a) \(\dfrac{x+1}{2}+\dfrac{3x-2}{3}=\dfrac{x-7}{12}\)

\(\Leftrightarrow\dfrac{6\left(x+1\right)+4\left(3x-2\right)}{12}=\dfrac{x-7}{12}\)

\(\Leftrightarrow6\left(x+1\right)+4\left(3x-2\right)=x-7\)

\(\Leftrightarrow6x+6+12x-8=x-7\)

\(\Leftrightarrow6x+12x-x=-7-6+8\)

\(\Leftrightarrow17x=-5\)

\(\Leftrightarrow x=\dfrac{-5}{17}\)

Vậy .........................

b) \(\dfrac{2x}{x-3}-\dfrac{5}{x+3}=\dfrac{x^2+21}{x^2-9}\left(ĐKXĐ:x\ne\pm3\right)\)

\(\Leftrightarrow\dfrac{2x\left(x+3\right)-5\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+21}{\left(x-3\right)\left(x+3\right)}\)

\(\Rightarrow2x\left(x+3\right)-5\left(x-3\right)=x^2+21\)

\(\Leftrightarrow2x^2+6x-5x+15=x^2+21\)

\(\Leftrightarrow2x^2-x^2+x+15-21=0\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(n\right)\\x=-3\left(l\right)\end{matrix}\right.\)

Vậy \(S=\left\{2\right\}\)

d) \(\left(x-4\right)\left(7x-3\right)-x^2+16=0\)

\(\Leftrightarrow\left(x-4\right)\left(7x-3\right)-\left(x^2-16\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(7x-3\right)-\left(x-4\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(7x-3-x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(6x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\6x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{7}{6}\end{matrix}\right.\)

Vậy .........................

P/s: các câu còn lại tương tự, bn tự giải nha

\(a,\left(x-2\right)\left(x-3\right)-3\left(4x-2\right)=\left(x-4\right)^2\\ \Leftrightarrow x^2-5x+6-12x+6=x^2-8x+16\\ \Leftrightarrow-9x-4=0\\ \Leftrightarrow x=-\dfrac{4}{9}\)

\(b,\dfrac{2x^2+1}{8}-\dfrac{7x-2}{12}=\dfrac{x^2-1}{4}-\dfrac{x-3}{6}\\ \Leftrightarrow6x^2+3-14x+4=6x^2-6-4x+12\\ \Leftrightarrow10x=1\\ \Leftrightarrow x=\dfrac{1}{10}\)

\(c,x-\dfrac{2x-2}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\\ \Leftrightarrow30x-12x+12+5x+40=210+10x-10\\ \Leftrightarrow13x=148\\ \Leftrightarrow x=\dfrac{148}{13}\)

\(d,\left(2x+5\right)^2=\left(x+2\right)^2\\ \Leftrightarrow\left(2x+5\right)^2-\left(x+2\right)^2=0\\ \Leftrightarrow\left(2x+5-x-2\right)\left(2x+5+x+2\right)=0\\ \Leftrightarrow\left(x+3\right)\left(3x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{7}{3}\end{matrix}\right.\)

\(e,x^2-5x+6=0\\ \Leftrightarrow\left(x-2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

\(g,2x^3+6x^2=x^2+3x\\ \Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow x\left(2x-1\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=-3\end{matrix}\right.\)

\(h,\left(x+\dfrac{1}{x}\right)^2+2\left(x+\dfrac{1}{x}\right)-8=0\left(x\ne0\right)\)

Đặt \(x+\dfrac{1}{x}=t\), pt trở thành:

\(t^2+2t-8=0\\ \Leftrightarrow\left(t-2\right)\left(t+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}t=2\\t=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=2\\x+\dfrac{1}{x}=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1-2x=0\\x^2+1+4x=0\left(1\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\Delta\left(1\right)=16-4=12>0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\\left[{}\begin{matrix}x=-2+\sqrt{3}\\x=-2-\sqrt{3}\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2+\sqrt{3}\\x=-2-\sqrt{3}\end{matrix}\right.\)

Tick plzz

a: Ta có: \(\left(x-2\right)\left(x+3\right)-3\left(4x-2\right)=\left(x-4\right)^2\)

\(\Leftrightarrow x^2+x-6-12x+6-x^2+8x-16=0\)

\(\Leftrightarrow-3x=16\)

hay \(x=-\dfrac{16}{3}\)

b: Ta có: \(\dfrac{2x^2+1}{8}-\dfrac{7x-2}{12}=\dfrac{x^2-1}{4}-\dfrac{x-3}{6}\)

\(\Leftrightarrow6x^2+3-14x+4=6x^2-6-4x+12\)

\(\Leftrightarrow-14x+7+4x-6=0\)

\(\Leftrightarrow10x=1\)

hay \(x=\dfrac{1}{10}\)

c: Ta có: \(x-\dfrac{2x-5}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\)

\(\Leftrightarrow30x-12x+30+5x+40=210+10x-10\)

\(\Leftrightarrow23x+70=10x+200\)

\(\Leftrightarrow x=10\)