giải phương trình
a)\(\frac{7x+10}{x+1}\left(x^2-x-2\right)=\frac{7x+10}{x+1}\left(2x^2-3x-5\right)\)
b)\(\frac{1}{x^2-5x+6}+\frac{1}{x^2-7x+12}+\frac{1}{x^2-9x+20}+\frac{1}{x^2-11x+30}=\frac{1}{8}\)
c)\(x^2+\frac{1}{x^2}+\frac{9x}{2}-\frac{9}{2x}+7=0\)
Bài 1: Rút gọn biểu thức :A=\(\left(\frac{x^2-2x}{2x^2+8}-\frac{2x^2}{8-4x+2x^2-x^3}\right).\left(1-\frac{1}{x}-\frac{2}{x^2}\right)\)
Chứng minh đẳng thức
\(\text{[}\frac{2}{3x}-\frac{2}{x+1}\left(\frac{x+1}{3x}-x-1\right)\text{]}:\frac{x-1}{x}=\frac{2x}{x-1}\)
Cho biểu thức :M=\(\left(\frac{x^2-2x}{2x^x+8}-\frac{2x^2}{8-4x+2x^2-x^3}\right)\left(1-\frac{1}{x}-\frac{2}{x^2}\right)\)
a)Rút gọn M
b)Tìm x nguyên để M có giá trị là số nguyên dương
c) Tìm x để M ≥ -3
với những giá trị nào thì x của 2 phân thức kia bằng nhau
a\(\frac{x-2}{x^2-5x+6}\)và \(\frac{1}{x-3}\)
b) \(A=\frac{\left(2x+1\right)\left(x-2\right)}{3\left(2x+1\right)}\) và \(B=\frac{x-2}{3}vớix\in N\)
Rút gọn
P = \(\left(\frac{4x^2-3x+17}{x^3-1}+\frac{2x-1}{x^2+x+1}+\frac{6x}{x-x^2}\right):\frac{x+2}{x^3-1}\)
\(\frac{1}{2}x^2y^2.\left(2x+y\right).\left(2x-y\right)\)
\(\frac{1}{2}x^2y^2.\left(2x+y\right).\left(2x-y\right)\)
3) \(\frac{x-2}{x-5}-\frac{5}{x^2-5x}=\frac{1}{x}\)
\(\Leftrightarrow\frac{x-2}{x-5}-\frac{5}{x.\left(x-5\right)}=\frac{1}{x}\)
\(\Leftrightarrow\frac{x.\left(x-2\right)}{x.\left(x-5\right)}-\frac{5}{x.\left(x-5\right)}=\frac{1.\left(x-5\right)}{x.\left(x-5\right)}\)
Mc: \(x.\left(x-5\right)\)
\(\Leftrightarrow\) \(x^2\) - 2\(x\) - 5 = \(x\) - 5
\(\Leftrightarrow\) \(x^2\) - 2\(x\) - \(x\) - 5 + 5 = 0
\(\Leftrightarrow\) \(x^2\) - 3\(x\) = 0
\(\Leftrightarrow\) \(x\) . (\(x\) - 3) = 0
\(\Leftrightarrow\) \(x\) = 0 hoặc \(x\) - 3 = 0
\(\Leftrightarrow\) \(x\) = 0 hoặc \(x\) = 3
Vậy \(x\) = 0 hoặc \(x\) = 3
\(x-5\ne0\Rightarrow x\ne5\)
\(x^2-5\ne0\Rightarrow x\ne5\) và \(x\ne0\) \(\Rightarrow\left\{{}\begin{matrix}x\ne0\\x\ne5\end{matrix}\right.\)
\(x\ne0\)
Vậy S = {3}
4) \(\frac{x-4}{x+7}-\frac{1}{x}=\frac{-7}{x^2+7x}\)
\(\Leftrightarrow\frac{x-4}{x+7}-\frac{1}{x}=\frac{-7}{x.\left(x+7\right)}\)
\(\Leftrightarrow\frac{x.\left(x-4\right)}{x.\left(x+7\right)}-\frac{1.\left(x+7\right)}{x.\left(x+7\right)}=\frac{-7}{x.\left(x+7\right)}\)
Mc: \(x.\left(x+7\right)\)
\(\Leftrightarrow x^2-4x-x-7=-7\)
\(\Leftrightarrow x^2-4x-x=-7+7\)
\(\Leftrightarrow\) \(x^2-5x=0\)
\(\Leftrightarrow x.\left(x-5\right)=0\)
\(\Leftrightarrow x=0\) hoặc \(x-5=0\)
\(\Leftrightarrow x=0\) hoặc \(x=5\)
Vậy \(x=0\) hoặc \(x=5\)
\(x+7\ne0\Rightarrow x\ne-7\)
\(x^2+7\ne0\Rightarrow x\ne-7\) và \(x\ne0\) \(\Rightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-7\end{matrix}\right.\)
\(x\ne0\)
Vậy S = {5}
5) \(\frac{x+2}{x-2}+\frac{x-2}{x+2}=\frac{8x}{x^2-4}\)
\(\left\{{}\begin{matrix}x-2\ne0\\x+2\ne0\\x^2-4\ne0\end{matrix}\right.\Rightarrow TXĐ\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)
Mc : \(\left(x-2\right).\left(x+2\right)\)
\(\Leftrightarrow\frac{\left(x+2\right).\left(x+2\right)}{\left(x-2\right).\left(x+2\right)}+\frac{\left(x-2\right).\left(x-2\right)}{\left(x+2\right).\left(x-2\right)}=\frac{8x}{\left(x-2\right).\left(x+2\right)}\)
\(\Leftrightarrow x^2+2x+2x+4+x^2-2x-2x+4=8x\)
\(\Leftrightarrow x^2+x^2+2x+2x-2x-2x-8x+4+4=0\)
\(\Leftrightarrow2x^2-8x+8=0\)
\(\Leftrightarrow\) \(2x^2-4x-4x+8=0\)
\(\Leftrightarrow\) \(2x.\left(x-2\right)-4.\left(x-2\right)=0\)
\(\Leftrightarrow\left(2x-4\right).\left(x-2\right)=0\)
\(\Leftrightarrow2x-4=0\) hoặc \(x-2=0\)
\(\Leftrightarrow x=2\) hoặc \(x=2\)
\(\Leftrightarrow x=2\) (Loại) hoặc x = 2 (Loại)
Vậy S = \(\left\{\varnothing\right\}\)
6) \(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{4}{x^2-1}\)
\(\Leftrightarrow\frac{\left(x+1\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}-\frac{\left(x-1\right).\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}=\frac{4}{\left(x-1\right).\left(x+1\right)}\)
MC: \(\left(x-1\right).\left(x+1\right)\)
\(\Leftrightarrow x^2+x+x+1-x^2+x+x-1=4\)
\(\Leftrightarrow x^2-x^2+x+x+x+x+1-1-4=0\)
\(\Leftrightarrow4x-4=0\)
\(\Leftrightarrow4.\left(x-1\right)=0\)
\(\Leftrightarrow\) 4 = 0 hoặc \(x-1=0\)
\(\Leftrightarrow\) 4 = 0 hoặc \(x=1\)
\(\Leftrightarrow\) 4 = 0 (Loại) hoặc \(x=1\) (Loại)
Vậy S = \(\left\{\varnothing\right\}\)
7) \(\frac{x+1}{x-1}+\frac{-4x}{x^2-1}=\frac{x-1}{x+1}\)
\(\Leftrightarrow\frac{\left(x+1\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}+\frac{-4x}{\left(x-1\right).\left(x+1\right)}=\frac{\left(x-1\right).\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}\)
\(Mc:\left(x-1\right).\left(x+1\right)\)
\(\Leftrightarrow\) \(x^2+x+x+1-4x=x^2-x-x+1\)
\(\Leftrightarrow x^2-x^2+x+x-4x+x+x=-1+1\)
\(\Leftrightarrow0=0\) (Nhận)
Vậy S = \(\left\{x\in R;x\ne\pm1\right\}\)
