Câu 1:

\(\left(x-2\right)\left(x^2+2x+4\right)+25x=x\left(x+5\right)\left(x-5\right)+8\)

\(\Leftrightarrow x^3-8+25x=x\left(x^2-25\right)+8\)

\(\Leftrightarrow x^3-8+25x=x^3-25x+8\)

\(\Leftrightarrow x^3-8+25x-x^3+25x-8=0\)

\(\Leftrightarrow50x-16=0\)

\(\Leftrightarrow50x=16\)

\(\Leftrightarrow x=\dfrac{8}{25}\)

Câu 2 :

\(\dfrac{x+5}{4}+\dfrac{3+2x}{3}=\dfrac{6x-1}{3}-\dfrac{1-2x}{12}\)

<=> \(\dfrac{3\left(x+5\right)}{12}+\dfrac{4\left(3+2x\right)}{12}=\dfrac{4\left(6x-1\right)}{12}-\dfrac{1-2x}{12}\)

<=>\(\dfrac{3x+15+12+8x}{12}=\dfrac{24x-4-1+2x}{12}\)

<=> 3x + 15 + 12 + 8x = 24x - 4 - 1 +2x

<=> 11x+27 = 26x -5

<=> ( 26x - 5 ) - ( 11x + 27 ) = 0

<=> 15x - 32 = 0

<=> 15x = 32

<=> x = \(\dfrac{32}{15}\)

Câu 3:

x - 4/3 - 3x - 1/12 = 3x + 1/4 + 9x - 2/8

<=> 4x - 16 - 3x + 1/12 = 6x + 2 + 9x - 2/8

<=> x - 15/12 = 15x/8

<=> 8x - 120 = 180x

<=> 120 = -172x <=> x = -172/120 = -43/30

a: \(x\cdot\dfrac{3}{4}+x=\dfrac{7}{8}\)

\(\Leftrightarrow x\cdot\dfrac{7}{4}=\dfrac{7}{8}\)

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

\(a,=\dfrac{x^2+4x+3-2x^2+2x+x^2-4x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ b,=\dfrac{1-2x+3+2y+2x-4}{6x^3y}=\dfrac{2y}{6x^3y}=\dfrac{1}{x^2}\\ c,=\dfrac{75y^2+18xy+10x^2}{30x^2y^3}\\ d,=\dfrac{5x+8-x}{4x\left(x+2\right)}=\dfrac{4\left(x+2\right)}{4x\left(x+2\right)}=\dfrac{1}{x}\\ c,=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)

\(\dfrac{x-1}{x-2}+\dfrac{x+3}{x-4}=\dfrac{2}{-x^2+6x-8}\left(đk:x\ne2,x\ne4\right)\Leftrightarrow\dfrac{\left(x-1\right)\left(x-4\right)+\left(x+3\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}=\dfrac{-2}{x^2-6x+8}\Leftrightarrow\dfrac{2x^2-4x-2}{x^2-6x+8}=\dfrac{-2}{x^2-6x+8}\Leftrightarrow2x^2-4x-2=-2\Leftrightarrow2x^2-4x=0\Leftrightarrow2x\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)\(\Leftrightarrow x=0\)( do x≠2)

2)Biện luận PT

`m(mx-1)=x+1`

`<=>m^2x-m=x+1`

`<=>x(m^2-1)=m+1`

PT vô nghiệm `<=>{(m^2-1=0),(m+1\ne0):}<=>m=1`

PT vô số nghiệm `<=>{(m^2-1=0),(m+1=0):}<=>m=-1`

PT có nghiệm duy nhất `m^2-1\ne0<=>m^2\ne1<=>m\ne+-1=>x=(m+1)/(m^2-1)=1/(m-1)`

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

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}m-1=0\\m^2x-m=x+1\end{matrix}\right.\\\left\{{}\begin{matrix}x\left(m-1\right)-1=0\\m^2x-m=x+1\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}m=1\\x-1=x+1\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\m=2\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\m=2\end{matrix}\right.\)(do x-1≠x+1)

\(a,\Leftrightarrow\left[{}\begin{matrix}-\dfrac{4}{3}x+\dfrac{1}{2}=\dfrac{1}{2}\\-\dfrac{4}{3}x+\dfrac{1}{2}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{4}\end{matrix}\right.\\ c,\Leftrightarrow\left(\dfrac{1}{2}\right)^x\left(1+\dfrac{1}{4}\right)=\dfrac{5}{4}\\ \Leftrightarrow\left(\dfrac{1}{2}\right)^x=1\Leftrightarrow x=0\)

b: Ta có: \(3^x+3^{x+2}=20\)

\(\Leftrightarrow3^x\cdot10=20\)

\(\Leftrightarrow3^x=2\left(loại\right)\)

\(\dfrac{3}{2}x-0,2=\dfrac{3}{5}\)

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

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

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

   \(x=\dfrac{4}{5}:\dfrac{3}{2}\)

   \(x=\dfrac{4}{5}\cdot\dfrac{2}{3}\)

   \(x=\dfrac{8}{15}\)

\(\dfrac{1}{3}+x=\dfrac{3}{4}\)

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

        \(x=\dfrac{9}{12}-\dfrac{4}{12}\)

        \(x=\dfrac{5}{12}\)

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

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

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

\(\dfrac{3}{2}x=\dfrac{13}{20}\)

   \(x=\dfrac{13}{20}:\dfrac{3}{2}\)

   \(x=\dfrac{13}{20}\cdot\dfrac{2}{3}\)

   \(x=\dfrac{13}{30}\)

\(\dfrac{11}{8}-\dfrac{3}{8}\cdot x=\dfrac{1}{8}\)

          \(\dfrac{3}{8}\cdot x=\dfrac{11}{8}-\dfrac{1}{8}\)

          \(\dfrac{3}{8}\cdot x=\dfrac{5}{4}\)

                \(x=\dfrac{5}{4}:\dfrac{3}{8}\)   

                \(x=\dfrac{5}{4}\cdot\dfrac{8}{3}\)

                \(x=\dfrac{10}{3}\)

1) điều kiện xác định : \(x\notin\left\{-1;-2;-3;-4\right\}\)

ta có : \(\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}=\dfrac{1}{6}\)

\(\Leftrightarrow\dfrac{x^2+7x+12+x^2+5x+4+x^2+3x+2}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)

\(\Leftrightarrow\dfrac{3x^2+15x+18}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)

\(\Leftrightarrow6\left(3x^2+15x+18\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)

\(\Leftrightarrow18\left(x^2+5x+6\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)

\(\Leftrightarrow18\left(x+2\right)\left(x+3\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)

\(\Leftrightarrow18=\left(x+1\right)\left(x+4\right)\) ( vì điều kiện xác định )

\(\Leftrightarrow18=x^2+5x+4\Leftrightarrow x^2+5x-14=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+7=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-7\end{matrix}\right.\left(tmđk\right)\)

vậy \(x=2\) hoặc \(x=-7\) mấy câu kia lm tương tự nha bn

ĐK: ` x\ne 2; x \ne 4`.

`2/(-x^2+6x-8)-(x-1)/(x-2)=(x+3)/(x-4)`

`<=> -2-(x-1)(x-4)=(x+3)(x-2)`

`<=> −x^2+5x−6=x^2+x−6`

`<=> 2x^2-4x=0`

`<=> 2x(x-2)=0`

`<=>` \(\left[{}\begin{matrix}x=0\\x=2\left(L\right)\end{matrix}\right.\)

Vậy `S={0}`.

ĐKXĐ: \(x\neq 2;x\neq 4\)

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

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

\(\Leftrightarrow-2-\left(x^2-5x+4\right)=x^2+x-6\Leftrightarrow2x^2-4x=0\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=2\left(l\right)\end{matrix}\right.\)

Vậy x = 0

1) \(\dfrac{1}{27}+a^3=\left(\dfrac{1}{3}+a\right)\left(\dfrac{1}{9}-\dfrac{a}{3}+a^2\right)\)

2) \(=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)

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

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

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

6) \(=\left(x-4\right)\left(x^2+4x+16\right)\)

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

8) \(=\left(2x^2-3y\right)\left(4x^4+6x^2y+9y^2\right)\)

9) \(=\left(\dfrac{1}{4}x^2-5y\right)\left(\dfrac{1}{16}x^4+\dfrac{5}{4}x^2y+25y^2\right)\)

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

11) \(=\left(x+2\right)^3\)

12) \(=\left(x+3\right)^3\)