\(\begin{array}{l} n) \Leftrightarrow \dfrac{{x + 1}}{7} + 1 + \dfrac{{x + 2}}{6} + 1 = \dfrac{{x + 3}}{5} + 1 + \dfrac{{x + 4}}{4} + 1\\ \Leftrightarrow \dfrac{{x + 8}}{7} + \dfrac{{x + 8}}{6} - \dfrac{{x + 8}}{5} - \dfrac{{x + 8}}{4} = 0\\ \Leftrightarrow \left( {x + 8} \right)\underbrace {\left( {\dfrac{1}{7} + \dfrac{1}{8} - \dfrac{1}{5} - \dfrac{1}{6}} \right)}_{ < 0} = 0\\ \Leftrightarrow x + 8 = 0\\ \Leftrightarrow x = - 8 \end{array}\)

k/

\(8-\dfrac{x-2}{3}=\dfrac{x}{4}\)

\(\Leftrightarrow\dfrac{96}{12}-\dfrac{4\left(x-2\right)}{12}=\dfrac{3x}{12}\)

\(\Leftrightarrow96-4x+8=3x\)

\(\Leftrightarrow96-4x+8-3x=0\)

\(\Leftrightarrow104-7x=0\)

\(\Leftrightarrow7x=104\)

\(\Leftrightarrow x=104:7\)

\(\Leftrightarrow x=\dfrac{104}{7}\)

Vậy tập nghiệm của phương trình là \(S=\left\{\dfrac{104}{7}\right\}\)

m/ 

\(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)

\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)

\(\Leftrightarrow9x+6-3x-1-12x-10=0\)

\(\Leftrightarrow-6x-5=0\)

\(\Leftrightarrow-6x=5\)

\(\Leftrightarrow x=-\dfrac{5}{6}\)

Vậy tập nghiệm của phương trình là \(S=\left\{-\dfrac{5}{6}\right\}\)

k) Ta có: \(8-\dfrac{x-2}{2}=\dfrac{x}{4}\)

\(\Leftrightarrow\dfrac{32}{4}-\dfrac{2\left(x-2\right)}{4}=\dfrac{x}{4}\)

\(\Leftrightarrow32-2x+4-x=0\)

\(\Leftrightarrow28-x=0\)

hay x=28

Vậy: S={28}

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

\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)

\(\Leftrightarrow9x+6-3x-1=12x+10\)

\(\Leftrightarrow6x+5-12x-10=0\)

\(\Leftrightarrow-6x=5\)

hay \(x=-\dfrac{5}{6}\)

Vậy: \(S=\left\{-\dfrac{5}{6}\right\}\)

n) Ta có: \(\dfrac{x+1}{7}+\dfrac{x+2}{6}=\dfrac{x+3}{5}+\dfrac{x+4}{4}\)

\(\Leftrightarrow\dfrac{x+1}{7}+1+\dfrac{x+2}{6}+1=\dfrac{x+3}{5}+1+\dfrac{x+4}{4}+1\)

\(\Leftrightarrow\dfrac{x+8}{7}+\dfrac{x+8}{6}=\dfrac{x+8}{5}+\dfrac{x+8}{4}\)

\(\Leftrightarrow\dfrac{x+8}{7}+\dfrac{x+8}{6}-\dfrac{x+8}{5}-\dfrac{x+8}{4}=0\)

\(\Leftrightarrow\left(x+8\right)\left(\dfrac{1}{7}+\dfrac{1}{6}-\dfrac{1}{5}-\dfrac{1}{4}\right)=0\)

mà \(\dfrac{1}{7}+\dfrac{1}{6}-\dfrac{1}{5}-\dfrac{1}{4}\ne0\)

nên x+8=0

hay x=-8

Vậy: S={-8}

a: =>3,6-1,7x=2,3-1,4-4=0,9-4=-3,1

=>1,7x=6,7

hay x=67/17

b: \(\Leftrightarrow30\left(5x+4\right)-15\left(3x+5\right)=24\left(4x+9\right)-40\left(x-9\right)\)

=>150x+120-45x-75=96x+216-40x+360

=>105x+45=56x+576

=>49x=531

hay x=531/49

a) ĐKXD: x ≠ 2

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

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

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

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

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

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

\(\Leftrightarrow x+3x=6+2\)

\(\Leftrightarrow4x=8\)

\(\Leftrightarrow x=2\) (loại vì không thỏa mãn điều kiện)

Vậy S = ∅

b) ĐKXĐ: x ≠ 7

 \(\dfrac{8-x}{x-7}-8=\dfrac{1}{x-7}\)

\(\Leftrightarrow\dfrac{8-x}{x-7}-\dfrac{1}{x-7}=8\)

\(\Leftrightarrow\dfrac{7-x}{x-7}=8\)

\(\Leftrightarrow-1=8\left(vô-lý\right)\)

Vậy S = ∅ 

P/s: Ko chắc ạ! 

c) ĐKXĐ: x ≠ 1

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

Quy đồng và khử mẫu ta được:

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

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

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

\(\Leftrightarrow x=1\) (loại vì ko t/m đk)

Vậy S = ∅

e) ĐK : \(\left\{{}\begin{matrix}1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x\ne-1\\3x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)

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

\(\Leftrightarrow12\left(1+3x\right)\left(1-3x\right)=\left(1-3x\right)\left(1+3x\right)\left(1-3x-1-3x\right)\left(1-3x+1+3x\right)\)

\(\Leftrightarrow12=\left(-6x\right).2\Leftrightarrow6=-6x\)

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

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

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

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

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

\(\Leftrightarrow-x=-44\)

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

\(2,\left(đk:x\ne4\right)\)

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

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

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

\(\Leftrightarrow-x=13\)

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

b: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)

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

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

=>(7x+10)(x-3)=0

hay \(x\in\left\{-\dfrac{10}{7};3\right\}\)

d: \(\Leftrightarrow\dfrac{13}{2x^2+7x-6x-21}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow26x+91+x^2-9-12x-14=0\)

\(\Leftrightarrow x^2+14x+68=0\)

hay \(x\in\varnothing\)

Để giải phương trình, ta sẽ thực hiện các bước sau: Bước 1: Giải các phép tính trong phương trình. 32x^(-1) + 2.9x^(-1) = 405(13)^(-1) + 5.(13)^2 + 1 = 1493(31)^(-1) + 5.(31)^2 + 1 = 9314(35)^(-1) Bước 2: Rút gọn các số hạng. 32x^(-1) + 2.9x^(-1) = 405/13 + 5.(13)^2 + 1 = 1493/31 + 5.(31)^2 + 1 = 9314/35 Bước 3: Đưa các số hạng về cùng mẫu số. 32x^(-1) + 2.9x^(-1) = (405/13).(31/31) + 5.(13)^2 + 1 = (1493/31).(13/13) + 5.(31)^2 + 1 = 9314/35 Bước 4: Tính toán các số hạng. 32x^(-1) + 2.9x^(-1) = 405.(31)/13.(31) + 5.(13)^2 + 1 = 1493.(13)/31.(13) + 5.(31)^2 + 1 = 9314/35 Bước 5: Tính tổng các số hạng. 32x^(-1) + 2.9x^(-1) = 405.(31)/403 + 5.(13)^2 + 1 = 1493.(13)/403 + 5.(31)^2 + 1 = 9314/35 Bước 6: Đưa phương trình về dạng chuẩn. 32x^(-1) + 2.9x^(-1) - 9314/35 = 0 Bước 7: Giải phương trình. Để giải phương trình này, ta cần biến đổi nó về dạng tương đương. Nhân cả hai vế của phương trình với 35 để loại bỏ mẫu số. 35.(32x^(-1) + 2.9x^(-1) - 9314/35) = 0 1120x^(-1) + 101.5x^(-1) - 9314 = 0 Bước 8: Tìm giá trị của x. Để tìm giá trị của x, ta cần giải phương trình này. Tuy nhiên, phương trình này không thể giải được vì x có mũ là -1.

\(a,3-x=x+1,8\)

\(\Rightarrow-x-x=1,8-3\)

\(\Rightarrow-2x=-1,2\)

\(\Rightarrow x=0,6\)

\(b,2x-5=7x+35\)

\(\Rightarrow2x-7x=35+5\)

\(\Rightarrow-5x=40\)

\(\Rightarrow x=-8\)

\(c,2\left(x+10\right)=3\left(x-6\right)\)

\(\Rightarrow2x+20=3x-18\)

\(\Rightarrow2x-3x=-18-20\)

\(\Rightarrow-x=-38\)

\(\Rightarrow x=38\)

\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)

\(\Rightarrow8x-3+1=1+6x+x\)

\(\Rightarrow8x-3=7x\)

\(\Rightarrow8x-7x=3\)

\(\Rightarrow x=3\)

\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)

\(\Rightarrow-3x+x=\dfrac{4}{3}-\dfrac{2}{9}\)

\(\Rightarrow-2x=\dfrac{10}{9}\)

\(\Rightarrow x=-\dfrac{5}{9}\)

\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)

\(\Rightarrow\dfrac{1}{2}x-\dfrac{3}{4}x=-\dfrac{1}{2}-\dfrac{5}{6}\)

\(\Rightarrow-\dfrac{1}{4}x=-\dfrac{4}{3}\)

\(\Rightarrow x=\dfrac{16}{3}\)

\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)

\(\Rightarrow x-4=5-x\)

\(\Rightarrow x+x=5+4\)

\(\Rightarrow2x=9\)

\(\Rightarrow x=\dfrac{9}{2}\)

\(k,7x^2-11=6x^2-2\)

\(\Rightarrow7x^2-6x^2=-2+11\)

\(\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

\(m,5\left(x+3\cdot2^3\right)=10^2\)

\(\Rightarrow5\left(x+24\right)=100\)

\(\Rightarrow x+24=20\)

\(\Rightarrow x=-4\)

\(n,\dfrac{4}{9}-\left(\dfrac{1}{6^2}\right)=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)

\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{4}{9}-\dfrac{1}{36}\)

\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{5}{12}\)

\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2=0\)

\(\Rightarrow x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}\)

#\(Urushi\text{☕}\)

a: 3-x=x+1,8

=>-2x=-1,2

=>x=0,6

b: 2x-5=7x+35

=>-5x=40

=>x=-8

c: 2(x+10)=3(x-6)

=>3x-18=2x+20

=>x=38

d; 8(x-3/8)+1=6(1/6+x)+x

=>8x-3+1=1+6x+x

=>8x-2=7x+1

=>x=3

e: =>-3x+x=4/3-2/9

=>-2x=12/9-2/9=10/9

=>x=-5/9

g: =>3/4x-1/2x=5/6+1/2

=>1/4x=5/6+3/6=8/6=4/3

=>x=4/3*4=16/3

h: =>x-4=-x+5

=>2x=9

=>x=9/2

a: =>2x-1=-2

=>2x=-1

hay x=-1/2

b: \(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\-\dfrac{2}{5}x-7=0\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{2}{3};-\dfrac{35}{2}\right\}\)

c: x/8=9/4

nên x/8=18/8

hay x=18

d: \(\Leftrightarrow\left(x-3\right)^2=36\)

=>x-3=6 hoặc x-3=-6

=>x=9 hoặc x=-3

e: =>-1,7x=6,12

hay x=-3,6

h: =>x-3,4=27,6

hay x=31

a) \(\dfrac{1}{3}\div\left(2x-1\right)=\dfrac{-1}{6}\)

\(\left(2x-1\right).\dfrac{1}{3}\div\left(2x-1\right)=\left(2x-1\right)\left(-\dfrac{1}{6}\right)\)

\(\dfrac{1}{3}=\left(2x-1\right)\left(-\dfrac{1}{6}\right)\)

\(\dfrac{1}{3}=-1\left(2x-1\right)\div6\)

\(\dfrac{1}{3}=-2x+1\div6\)

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

b) \(\left(3x+2\right)\left(\dfrac{-2}{5}x-7\right)=0\)

\(TH1:3x+2=0\)

\(3x=0-2\)

\(3x=-2\)

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

\(TH2:\left(-\dfrac{2}{5}x-7\right)=0\)

\(\left(\dfrac{-2}{5}x-7\right)=0\)

\(\left(\dfrac{-2x}{5}+\dfrac{5\left(-7\right)}{5}\right)=0\)

\(\left(\dfrac{-2x-35}{5}\right)=0\)

\(-2x-35=0\)

\(-2x=0+35\)

\(x=-\dfrac{35}{2}\)

c) \(\dfrac{x}{8}=\dfrac{9}{4}\)

\(\Leftrightarrow x=\dfrac{9.8}{4}=\dfrac{72}{4}=18\)

\(x=18\)

d) \(\dfrac{x-3}{2}=\dfrac{18}{x-3}\)

\(x-3=18+2\)

\(x=20-3\)

\(x=17\)

e) \(4,5x-6,2x=6,12\)

\(\dfrac{9x}{2}-6,2.x=6,12\)

\(\dfrac{9x}{2}+\dfrac{-31x}{5}=6,12\)

\(\dfrac{5.9x}{10}+\dfrac{2\left(-31\right)x}{10}=6.12\)

\(\dfrac{45x-62x}{10}=6.12\)

\(=-17x\div10=6.12\)

\(-17x=10.6.12\)

\(x=-3,6\)

h) \(11,4-\left(x-3,4\right)=-16,2\)

\(x-3,4=-16,2+11,4\)

\(x-3,4=-4,8\)

\(x=-1,4\)

a) \(\dfrac{x+5}{3\left(x-1\right)}+1=\dfrac{3x+7}{5\left(x-1\right)}\) ( đk: \(x\ne1\))

\(\Leftrightarrow\dfrac{5\left(x+5\right)}{15\left(x-1\right)}+\dfrac{15\left(x-1\right)}{15\left(x-1\right)}=\dfrac{3\left(3x+7\right)}{15\left(x-1\right)}\)

\(\Rightarrow5\left(x+5\right)+15\left(x-1\right)=3\left(3x+7\right)\)

\(\Leftrightarrow5x+25+15x-15=9x+21\)

\(\Leftrightarrow5x+15x-9x=21-25+15\)

\(\Leftrightarrow11x=11\Leftrightarrow x=1\) (loại)

Vậy tập nghiệm: \(S=\varnothing\)

b) \(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x+3}-\dfrac{8}{x^2+2x-3}=1\)  (đk: \(x\ne1,x\ne-3\))

\(\Leftrightarrow\dfrac{\left(3x-1\right)\left(x+3\right)}{x^2+2x-3}-\dfrac{\left(2x+5\right)\left(x-1\right)}{x^2+2x-3}-\dfrac{8}{x^2+2x-3}=\dfrac{x^2+2x-3}{x^2+2x-3}\)

\(\Rightarrow\left(3x-1\right)\left(x+3\right)-\left(2x+5\right)\left(x-1\right)-8=x^2+2x-3\)

\(\Leftrightarrow3x^2+9x-x-3-2x^2+2x-5x+5-8=x^2+2x-3\)

\(\Leftrightarrow3x=3\Leftrightarrow x=1\) (loại)

Vậy tập nghiệm: \(S=\varnothing\)