1) điều kiện xác định : \(x\notin\left\{-1;-3;-5;-7\right\}\)

ta có : \(\dfrac{1}{x^2+4x+3}+\dfrac{1}{x^2+8x+15}+\dfrac{1}{x^2+12x+35}=\dfrac{1}{9}\)

\(\Leftrightarrow\dfrac{x^2+12x+35+x^2+8x+7+x^2+4x+3}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)}=\dfrac{1}{9}\)

\(\Leftrightarrow\dfrac{3x^2+24x+45}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)}=\dfrac{1}{9}\)

\(\Leftrightarrow9\left(3x^2+24x+45\right)=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)\)

\(\Leftrightarrow27\left(x^2+8x+15\right)=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)\)

\(\Leftrightarrow27\left(x+3\right)\left(x+5\right)=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)\)

\(\Leftrightarrow27=\left(x+1\right)\left(x+7\right)\) ( vì điều kiện xác định )

\(\Leftrightarrow27=x^2+8x+7\Leftrightarrow x^2+8x-20=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+10\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+10=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-10\end{matrix}\right.\left(tmđk\right)\)

vậy \(x=2\) hoặc \(x=-10\)

câu còn lại lm tương tự nha bn

\(\dfrac{1}{x^2-5x+6}+\dfrac{1}{x^2-7x+12}+\dfrac{1}{x^2-9x+20}+\dfrac{1}{x^2-11x+30}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}+\dfrac{1}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-4}+\dfrac{1}{x-4}-\dfrac{1}{x-5}+\dfrac{1}{x-5}-\dfrac{1}{x-6}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x-2}-\dfrac{1}{x-6}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{x-6-x+2}{\left(x-2\right)\left(x-6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{4}{\left(x-2\right)\left(x-6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow32=\left(x-2\right)\left(x-6\right)\)

\(\Leftrightarrow32=x^2-8x+12\)

\(\Leftrightarrow x^2+8x-20=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=10\end{matrix}\right.\)

bạn xem lại nhé

cái này là pt có chứa ẩn ở mẫu nên phải có điều kiện, đối chiếu điều kiện  và từ dòng có pt chứa ẩn ở mẫu sang dòng có pt đưa dc về dạng ax+b=0 thì dùng dấu ''=>'' nhé

b: Đặt \(x^2-6x-2=a\)

Theo đề, ta có: \(a+\dfrac{14}{a+9}=0\)

=>(a+2)(a+7)=0

\(\Leftrightarrow\left(x^2-6x\right)\left(x^2-6x+5\right)=0\)

=>x(x-6)(x-1)(x-5)=0

hay \(x\in\left\{0;1;6;5\right\}\)

c: \(\Leftrightarrow\dfrac{-8x^2}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(2x+1\right)}\)

\(\Leftrightarrow-32x^2=8x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)\)

\(\Leftrightarrow-32x^2=16x^2+8x-3\left(16x^2-8x+2x-1\right)\)

\(\Leftrightarrow-48x^2=8x-48x^2+18x+3\)

=>26x=-3

hay x=-3/26

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

a.

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

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

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

Đặt \(a=x^2+x-1\) , ta có pt:

\(\left(a+1\right)\left(a-1\right)-24=0\)

\(\Leftrightarrow a^2-1-24=0\)

\(\Leftrightarrow a^2-25=0\)

\(\Leftrightarrow\left(a-5\right)\left(a+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=5\\a=-5\end{matrix}\right.\)

*Với a = 5 ta được:

\(x^2+x-1=5\)

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

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

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

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

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

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

*Với a = -5 ta được:

\(x^2+x-1=-5\)

\(\Leftrightarrow x^2+x+4=0\)

\(\Leftrightarrow x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{15}{4}=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}=0\) ( loại)

Vậy pt có tập nghiệm là: \(s=\left\{-3;2\right\}\)

c)(ĐKXĐ: x khác 30;29)

\(\Leftrightarrow\dfrac{x-29}{30}-1+\dfrac{x-30}{29}-1=\dfrac{29}{x-30}-1+\dfrac{30}{x-29}-1\)

\(\Leftrightarrow\dfrac{x-59}{30}+\dfrac{x-59}{29}=\dfrac{x-59}{30-x}+\dfrac{x-59}{29-x}\)

\(\Leftrightarrow x=59\)(tm) or \(\dfrac{1}{30}+\dfrac{1}{29}-\dfrac{1}{30-x}-\dfrac{1}{29-x}=0\)

\(\Leftrightarrow\dfrac{-x}{30\left(30-x\right)}+\dfrac{-x}{29\left(29-x\right)}=0\)

\(\Leftrightarrow x=0\)(tm) or \(\dfrac{1}{30\left(30-x\right)}+\dfrac{1}{29\left(29-x\right)}=0\)

\(\Leftrightarrow1741-59x=0\)

\(\Leftrightarrow x=\dfrac{1741}{59}\left(tm\right)\)

Vậy S={0;\(\dfrac{1741}{59}\);59}

ĐKXĐ: x khác 2;3;4;5;6

\(\dfrac{1}{x^2-5x+6}+\dfrac{1}{x^2-7x+12}+\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2-11x+30}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}+\dfrac{1}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-4}+\dfrac{1}{x-4}-\dfrac{1}{x-5}+\dfrac{1}{x-5}-\dfrac{1}{x-6}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x-6}-\dfrac{1}{x-2}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{x+6-x+2}{\left(x-2\right)\left(x-6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{4}{\left(x-2\right)\left(x-6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow32=x^2-8x+12\)

\(\Leftrightarrow x^2+8x-20=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=10\end{matrix}\right.\)

a: \(=\dfrac{4x^3+8x^2-11x+3-\left(x^2-5\right)\left(2x-1\right)-2x^3-5x^2+x+1}{\left(2x-1\right)^3}\)

\(=\dfrac{2x^3+3x^2-10x+4-2x^3+x^2+10x-5}{\left(2x-1\right)^3}\)

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

b: \(=\dfrac{1+x+1-x}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{2+2x^2+2-2x^2}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{4+4x^4+4-4x^4}{1-x^8}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{8+8x^8+8-8x^8}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{32}{1+x^{32}}\)