\(\frac{1-6x}{x-2}-\frac{9x+4}{x+2}=\frac{x\left(3x-2+1\right)}{x^2-4}\)
Những câu hỏi liên quan
1. \(\frac{1-6x}{x-2}+\frac{9x+4}{x+2}=\frac{x\left(3x-2\right)+1}{x^2-4}\)
2. \(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}\)
3. \(\frac{1-6x}{x-2}+\frac{9x+4}{x+2}=\frac{x\left(3x-2\right)+1}{x^2-4}\)
Rút gọn
a) \(\left(\frac{4}{x^3-9x}+\frac{1}{x+3}\right):\left(\frac{x-3}{x^2+3x}-\frac{x}{3x+9}\right)\)
b) \(\left(\frac{2}{x-2}-\frac{2}{x+2}\right).\frac{x^2+4x+4}{8}\)
c) \(\left(\frac{3x}{1-3x}+\frac{2x}{3x+1}\right):\frac{6x^2+10x}{1-6x+9x^2}\)
\(\frac{1-6x}{x-2}-\frac{9x+4}{x+2}=\frac{x\left(3x-2\right)+1}{x^2-4}\)
ĐKXĐ: x∉{2;-2}
Ta có: \(\frac{1-6x}{x-2}-\frac{9x+4}{x+2}=\frac{x\left(3x-2\right)+1}{x^2-4}\)
\(\Leftrightarrow\frac{\left(1-6x\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{\left(9x+4\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{x\left(3x-2\right)+1}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow-6x^2-11x+2-\left(9x^2-14x-8\right)-\left(3x^2-2x+1\right)=0\)
\(\Leftrightarrow-6x^2-11x+2-9x^2+14x+8-3x^2+2x-1=0\)
\(\Leftrightarrow-18x^2+5x+9=0\)
Giải phương trình:
a, \(\frac{2}{\left(1-3x\right)\left(3x+11\right)}=\frac{1}{9x^2-6x+1}-\frac{3}{\left(3x+11\right)^2}\)
b,\(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^1-x+1}=\frac{3}{x\left(x^4+x^2+1\right)}\)
a) ĐKXĐ: \(x\notin\left\{\frac{1}{3};\frac{-11}{3}\right\}\)
Ta có: \(\frac{2}{\left(1-3x\right)\left(3x+11\right)}=\frac{1}{9x^2-6x+1}-\frac{3}{\left(3x+11\right)^2}\)
\(\Leftrightarrow\frac{2\left(1-3x\right)\left(3x+11\right)}{\left(1-3x\right)^2\cdot\left(3x+11\right)^2}=\frac{\left(3x+11\right)^2}{\left(1-3x\right)^2\cdot\left(3x+11\right)^2}-\frac{3\left(1-3x\right)^2}{\left(1-3x\right)^2\cdot\left(3x+11\right)^2}\)
\(\Leftrightarrow-18x^2-60x+22=9x^2+66x+121-3\left(1-6x+9x^2\right)\)
\(\Leftrightarrow-18x^2-60x+22-9x^2-66x-121+3\left(1-6x+9x^2\right)=0\)
\(\Leftrightarrow-27x^2-126x-99+3-18x+27x^2=0\)
\(\Leftrightarrow-144x-96=0\)
\(\Leftrightarrow-144x=96\)
hay \(x=\frac{-2}{3}\)(tm)
Vậy: \(x=\frac{-2}{3}\)
Đúng 0
Bình luận (0)
Giải các PT sau :
a,frac{1-6x}{x-2}+frac{9x+4}{x+2}frac{xleft(3x-2right)+1}{left(x+2right)left(x-2right)}
b,(2 - 3x) (x + 11) (3x - 2) (2 - 5x)
c,frac{3x-2}{6}-5frac{3-2left(x+7right)}{4}
d,frac{x}{x-3}+frac{x}{x
+1}frac{2x}{left(x+1right)left(x-3right)}
e,frac{x+1}{5}+frac{x+2}{4}frac{x+3}{3}+frac{x+4}{2}
Đọc tiếp
Giải các PT sau :
a,\(\frac{1-6x}{x-2}+\frac{9x+4}{x+2}=\frac{x\left(3x-2\right)+1}{\left(x+2\right)\left(x-2\right)}\)
b,(2 - 3x) (x + 11) = (3x - 2) (2 - 5x)
c,\(\frac{3x-2}{6}-5=\frac{3-2\left(x+7\right)}{4}\)
d,\(\frac{x}{x-3}+\frac{x}{x +1}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\)
e,\(\frac{x+1}{5}+\frac{x+2}{4}=\frac{x+3}{3}+\frac{x+4}{2}\)
bài tập. Giải các pt1, frac{5}{x-2}+frac{6}{3-4x}02,frac{x+1}{x-2}frac{1}{x^2-4}3,frac{x+2}{x}-frac{x^2+5x+4}{xleft(x+2right)}frac{x}{x+2}4,frac{1-6x}{x-2}+frac{9x+4}{x+2}frac{xleft(3x-2right)+1}{x^2-4}5,frac{3x+2}{3x-2}-frac{6}{2+3x}frac{9x^2}{9x^2-4}6,frac{x+1}{x}+frac{1}{x+1}frac{2x-1}{2x^2+2}7,frac{2}{x+1}-frac{3x+1}{left(x+1right)}frac{1}{left(x+1right)left(x-2right)}8,frac{4}{x^2+2x-3}frac{2x-5}{x+3}-frac{2x}{x-1}9,frac{3}{x^2+x-2}-frac{1}{x-1}frac{-7}{x+2}
Đọc tiếp
bài tập. Giải các pt
1, \(\frac{5}{x-2}+\frac{6}{3-4x}=0\)
2,\(\frac{x+1}{x-2}=\frac{1}{x^2-4}\)
3,\(\frac{x+2}{x}-\frac{x^2+5x+4}{x\left(x+2\right)}=\frac{x}{x+2}\)
4,\(\frac{1-6x}{x-2}+\frac{9x+4}{x+2}=\frac{x\left(3x-2\right)+1}{x^2-4}\)
5,\(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4}\)
6,\(\frac{x+1}{x}+\frac{1}{x+1}=\frac{2x-1}{2x^2+2}\)
7,\(\frac{2}{x+1}-\frac{3x+1}{\left(x+1\right)}=\frac{1}{\left(x+1\right)\left(x-2\right)}\)
8,\(\frac{4}{x^2+2x-3}=\frac{2x-5}{x+3}-\frac{2x}{x-1}\)
9,\(\frac{3}{x^2+x-2}-\frac{1}{x-1}=\frac{-7}{x+2}\)
ĐKXĐ : \(\hept{\begin{cases}x-2\ne0\\3-4x\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne2\\x\ne\frac{3}{4}\end{cases}}}\)
\(\frac{5}{x-2}+\frac{6}{3-4x}=0\)
\(\frac{5\left(3-4x\right)}{\left(x-2\right)\left(3-4x\right)}+\frac{6\left(x-2\right)}{\left(3-4x\right)\left(x-2\right)}=0\)
\(15-20x+6x-12=0\)
\(3-14x=0\Leftrightarrow14x=3\Leftrightarrow x=\frac{3}{14}\)theo ĐKXĐ : x thỏa mãn
bài 1:giải các pt sau:
a/frac{1-x}{x+1}+3frac{2x+3}{x+1}
b/frac{left(x+2right)^2}{2x-3}-1frac{x^2+10}{2x-3}
c/frac{1}{x+1}-frac{5}{x-2}frac{15}{left(x+1right)left(2-xright)}
d/frac{1-6x}{x-2}+frac{9x+4}{x+2}frac{xleft(3x-2right)+1}{x^2-4}
e/frac{12}{1-9x^2}frac{1-3x}{1+3x}-frac{1+3x}{1-3x}
ffrac{x+4}{x^2-3x+2}+frac{x+1}{x^2-4x+3}frac{2x+5}{x^2-4x+3}
Đọc tiếp
bài 1:giải các pt sau:
a/\(\frac{1-x}{x+1}\)+3=\(\frac{2x+3}{x+1}\)
b/\(\frac{\left(x+2\right)^2}{2x-3}-1=\frac{x^2+10}{2x-3}\)
c/\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
d/\(\frac{1-6x}{x-2}+\frac{9x+4}{x+2}=\frac{x\left(3x-2\right)+1}{x^2-4}\)
e/\(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)
f\(\frac{x+4}{x^2-3x+2}+\frac{x+1}{x^2-4x+3}=\frac{2x+5}{x^2-4x+3}\)
Giải bài toán sau:
\(\frac{1-6x}{x-2}-\frac{9x+4}{x+2}=\frac{x\left(3x-2\right)+1}{x^2-4}\)
Q= \(\left(x^3-1-\frac{7-x^3}{3+x^3}\right).\frac{4}{x^5+3x^2}:\left(\frac{6x^4-24}{x^9+6x^6+9x^3}.\frac{2x}{3x^3+6}\right)\)