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\(Q=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)
\(\Leftrightarrow\) \(Q=\frac{\left(x-2\right)\left(x+2\right)}{\left(x+3\right)\left(2-x\right)}+\frac{5}{\left(x+3\right)\left(2-x\right)}+\frac{-1}{\left(x+3\right)\left(2-x\right)}\)
\(\Rightarrow\) \(Q=\left(x-2\right)\left(x+2\right)+5-1\)
\(\Leftrightarrow\) \(Q=x^2-4+5-1\)
\(\Leftrightarrow\) \(Q=x^2\)
Thay \(Q=\frac{-3}{4}\) ta được:
\(x^2=\frac{-3}{4}\)
Vì \(\frac{-3}{4}>0\forall x\)
\(\Rightarrow\) Pt vô nghiệm
Vậy không có giả trị nào của x thỏa mãn \(Q=\frac{-3}{4}\)
Chúc bn học tốt!!
\(ĐK:\hept{\begin{cases}x-1\ne0\\1+x\ne0\end{cases}\Rightarrow x\ne\pm1}\)
a) \(M=\left(\frac{1}{x-1}-\frac{x}{\left(1-x\right).\left(x^2+x+1\right)}\cdot\frac{x^2+x+1}{x+1}\right)\cdot\frac{x^2-1}{1}\)
\(M=\left(\frac{1}{x-1}-\frac{x}{1-x}\right)\cdot\frac{\left(x-1\right).\left(x+1\right)}{1}\)
\(M=\left(\frac{x+1}{\left(x-1\right).\left(x+1\right)}-\frac{-x}{\left(x-1\right).\left(x+1\right)}\right)\cdot\frac{\left(x-1\right).\left(x+1\right)}{1}\)
\(M=\frac{2x+1}{\left(x-1\right).\left(x+1\right)}\cdot\frac{\left(x-1\right).\left(x+1\right)}{1}=2x+1\)
b) \(M=2x+1=\frac{2.1}{2}+1=1+1=2\)
c) \(M=2x+1>0\Rightarrow2x>-1\Rightarrow x>-\frac{1}{2}\)và x khác +1,-1
a/ Ta có \(M=\frac{\frac{1}{x-1}-\frac{x}{1-x^3}.\frac{x^2+x+1}{x+1}}{\frac{1}{x^2-1}}\) với \(x\ne\pm1\)
\(M=\frac{\frac{1}{x-1}-\frac{x}{\left(1-x\right)\left(x^2+x+1\right)}.\frac{x^2+x+1}{x+1}}{\frac{1}{\left(x-1\right)\left(x+1\right)}}\)
\(M=\frac{\frac{1}{x-1}-\frac{x}{\left(1-x\right)\left(x+1\right)}}{\frac{1}{\left(x-1\right)\left(x+1\right)}}\)
\(M=\frac{\frac{1}{x-1}+\frac{x}{\left(x-1\right)\left(x+1\right)}}{\frac{1}{\left(x-1\right)\left(x+1\right)}}\)
\(M=\frac{x+1+x}{\left(x-1\right)\left(x+1\right)}.\left(x-1\right)\left(x+1\right)\)
\(M=2x+1\)
b/ Ta có \(x=\frac{1}{2}\)thoả mãn ĐKXĐ
Vậy với \(x=\frac{1}{2}\):
\(M=2x+1=2.\frac{1}{2}+1=2\)
c/ Khi M > 0
=> \(2x+1>0\)
=> \(x>-\frac{1}{2}\)
Vậy khi \(\hept{\begin{cases}x>-\frac{1}{2}\\x\ne\pm1\end{cases}}\)thì M > 0.
\(B=\left(\frac{1-x^3}{1-x}-x\right):\frac{1-x^2}{1-x-x^2+x^3}\) \(ĐKXĐ:x\ne\pm1\)
\(B=\left[\frac{\left(1-x\right)\left(x^2+x+1\right)}{\left(1-x\right)}-x\right]:\frac{\left(x+1\right)\left(1-x\right)}{\left(1-x\right)-x^2\left(1-x\right)}\)
\(B=\left(x^2+x+1-x\right):\frac{\left(x+1\right)\left(1-x\right)}{\left(1-x\right)\left(1-x^2\right)}\)
\(B=\left(x^2+1\right):\frac{x+1}{\left(x+1\right)\left(1-x\right)}\)
\(B=\frac{x^2+1}{1-x}\)
vậy \(B=\frac{x^2+1}{1-x}\)
b) \(x=-1\frac{2}{3}\)
\(x=\frac{-5}{3}\)
khi đó \(B=\frac{\left(\frac{-5}{3}\right)^2+1}{1+\frac{5}{3}}\)
\(B=\frac{\frac{25}{9}+1}{\frac{8}{3}}\)
\(B=\frac{34}{9}:\frac{8}{3}\)
\(B=\frac{17}{12}\)
vậy \(B=\frac{17}{12}\) khi \(x=-1\frac{2}{3}\)
c) \(B< 0\Leftrightarrow\frac{x^2+1}{1-x}< 0\)
\(\Leftrightarrow\hept{\begin{cases}x^2+1>0\\1-x< 0\end{cases}}\)hoặc \(\hept{\begin{cases}x^2+1< 0\\1-x>0\end{cases}}\)
đến đây bạn giải tiếp
a) \(A=\frac{4x}{x+2}+\frac{2}{x-2}+\frac{5x-6}{4-x^2}\)
\(A=\frac{4x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{5x-6}{\left(x-2\right)\left(x+2\right)}\)
\(A=\frac{4x^2-8x+2x+4-5x+6}{\left(x-2\right)\left(x+2\right)}\)
\(A=\frac{4x^2-11x+10}{\left(x-2\right)\left(x+2\right)}\)
\(a,A=\frac{4x}{x+2}+\frac{2}{x-2}+\frac{5x-6}{4-x^2}\)
\(=\frac{4x}{x+2}+\frac{2}{x-2}+\frac{6-5x}{x^2-4}\)
\(=\frac{4x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{6-5x}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{4x^2-8x+2x+4+6-5x}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{4x^2-11x+10}{\left(x-2\right)\left(x+2\right)}\)
a/ Ta có \(A=\frac{\frac{x}{x^2-4}+\frac{1}{x+2}-\frac{2}{x-2}}{1-\frac{x}{x+2}}\)với \(\hept{\begin{cases}x\ne\pm2\\x\ne0\end{cases}}\)
\(A=\frac{\frac{x}{x^2-4}+\frac{x-2-2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}}{\frac{x+2-x}{x+2}}\)
\(A=\frac{\frac{x}{x^2-4}+\frac{x-2-2x-4}{x^2-4}}{\frac{2}{x+2}}\)
\(A=\frac{\frac{x-x-6}{x^2-4}}{\frac{2}{x+2}}\)
\(A=\frac{-6}{x^2-4}.\frac{x+2}{2}\)
\(A=\frac{-3}{x-2}\)
b/ Ta có \(x=-4\)thoả mãn ĐKXĐ
Vậy với \(x=-4\):
\(A=\frac{-3}{x-2}=\frac{-3}{-4-2}=\frac{1}{2}\)
c/ Khi \(A\inℤ\)
=> \(\frac{-3}{x-2}\inℤ\)
=> \(-3⋮\left(x-2\right)\)
=> x - 2 là ước của -3
Ta có bảng sau:
| x-2 | -1 | -2 | -3 | -6 | 1 | 2 | 3 | 6 |
| x | 1 | 0 | -1 | -4 | 3 | 4 | 5 | 8 |
Mà ĐKXĐ \(\hept{\begin{cases}x\ne\pm2\\x\ne0\end{cases}}\)
=> \(x\in\left\{\pm1;\pm4;3;5;8\right\}\)
Vậy khi \(x\in\left\{\pm1;\pm4;3;5;8\right\}\)thì \(A\inℤ\).
\(a.\frac{x-6}{x-4}=\frac{x}{x-2}\\\Leftrightarrow \frac{\left(x-6\right)\left(x-2\right)}{\left(x-4\right)\left(x-2\right)}=\frac{x\left(x-4\right)}{\left(x-4\right)\left(x-2\right)}\\\Leftrightarrow \left(x-6\right)\left(x-2\right)=x\left(x-4\right)\\\Leftrightarrow \left(x-6\right)\left(x-2\right)-x\left(x-4\right)=0\\ \Leftrightarrow x^2-2x-6x+12-x^2+4x=0\\\Leftrightarrow -4x+12=0\\\Leftrightarrow -4x=-12\\ \Leftrightarrow x=3\)
\(b.1+\frac{2x-5}{x-2}-\frac{3x-5}{x-1}=0\\ \Leftrightarrow\frac{\left(x-2\right)\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}+\frac{\left(2x-5\right)\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}-\frac{\left(3x-5\right)\left(x-2\right)}{\left(x-2\right)\left(x-1\right)}=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)+\left(2x-5\right)\left(x-1\right)-\left(3x-5\right)\left(x-2\right)=0\\ \Leftrightarrow x^2-x-2x+3+2x^2-2x-5x+5-3x^2+6x+5x-10=0\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\\ \)