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a) Rút gọn :
\(ĐKXĐ:x\ne\pm5\)
Ta có : \(P=\left(\frac{x}{\left(x-5\right)\left(x+5\right)}-\frac{x-5}{x\left(x+5\right)}\right):\frac{2x-5}{x\left(x+5\right)}-\frac{2x}{5-x}\)
\(=\left(\frac{x^2-\left(x-5\right)\left(x-5\right)}{x\left(x-5\right)\left(x+5\right)}\right):\frac{\left(2x-5\right)\left(x-5\right)+2x^2\left(x+5\right)}{x\left(x+5\right)\left(x-5\right)}\)
\(=\frac{10x-25}{x\left(x-5\right)\left(x+5\right)}\cdot\frac{x\left(x+5\right)\left(x-5\right)}{ }\)
Tui đang định làm tiếp đó, nhưng khẳng định đề này hơi sai sai ở vế bị chia. Bạn xem lại đc k ?
P/s : lười làm nên đăng hình ảnh zậy , viết mỏi tay lắm ( em lùng ảnh cũ , ko phải bây h mới làm , có kí tên nên ko pải hàng fake )
a) ĐKXĐ : \(x\ne\pm5,x\ne0,x\ne\frac{5}{2}\)
Rút gọn :
Ta có : \(P=\left(\frac{x}{\left(x-5\right)\left(x+5\right)}-\frac{x-5}{x\left(x+5\right)}\right):\frac{5\left(2x-5\right)}{x\left(x+5\right)}+\frac{x}{5-x}\)
\(=\frac{x^2-\left(x-5\right)\left(x-5\right)}{x\left(x-5\right)\left(x+5\right)}:\frac{5\left(2x-5\right)}{x\left(x+5\right)}+\frac{x}{5-x}\)
\(=\frac{5\left(2x-5\right)}{x\left(x-5\right)\left(x+5\right)}\cdot\frac{x\left(x+5\right)}{5\left(2x-5\right)}+\frac{x}{5-x}\)
\(=\frac{1}{x-5}-\frac{x}{x-5}=\frac{1-x}{x-5}\)
Vậy : \(P=\frac{1-x}{x-5}\) với \(x\ne\pm5,x\ne0,x\ne\frac{5}{2}\)
b) Để \(P=2013\Leftrightarrow\frac{1-x}{x-5}=2013\)
\(\Leftrightarrow\frac{1-x}{x-5}-2013=0\)
\(\Leftrightarrow\frac{1-x-2013\left(x-5\right)}{x-5}=0\)
\(\Rightarrow10066-2014x=0\)
\(\Leftrightarrow2014x=10066\)
\(\Leftrightarrow x=\frac{10066}{2014}\approx4,999\)( thỏa mãn )
c) Để P là số nguyên \(\Leftrightarrow1-x⋮x-5\)
\(\Leftrightarrow-\left(x-5\right)-4⋮x-5\)
\(\Leftrightarrow4⋮x-5\)
\(\Leftrightarrow x-5\inƯ\left(4\right)\)
\(\Leftrightarrow x-5\in\left\{-1,1,-2,2,-4,4\right\}\)
\(\Leftrightarrow x\in\left\{4,6,3,7,1,9\right\}\) ( thỏa mãn ĐKXĐ và \(x\inℤ\) )
Vậy \(x\in\left\{4,6,3,7,1,9\right\}\) để P là số nguyên .
\(a.ĐKXĐ:\hept{\begin{cases}1-3x\ne0\\3x+1\ne0\\x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{3}\\...\\x\ge0\end{cases}}}\)
\(b,M=\left(\frac{3x}{1-3x}+\frac{2x}{3x+1}\right):\frac{6x^2+10}{1-6x+9x^2}\)
\(=\left(\frac{3x\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\frac{2x\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}\right).\frac{\left(1-3x\right)^2}{6x^2+10}\)
\(=\left(\frac{3x+9x^2+2x-6x^2}{\left(1-3x\right)\left(1+3x\right)}\right).\frac{\left(1-3x\right)^2}{6x^2+10}\)
\(=\frac{5x+3x^2}{1+3x}.\frac{1-3x}{2\left(3x^2+5\right)}\)
==>Sai đề không mem
\(A=\left(\frac{x}{25+5x}+\frac{5x+50}{x^2+5x}-\frac{10-2x}{x}\right)\div\frac{3x+15}{7}\)
ĐK : \(\hept{\begin{cases}x\ne0\\x\ne-5\end{cases}}\)
\(=\left(\frac{x}{5\left(x+5\right)}+\frac{5\left(x+10\right)}{x\left(x+5\right)}-\frac{2\left(5-x\right)}{x}\right)\div\frac{3\left(x+5\right)}{7}\)
\(=\left(\frac{x^2}{5x\left(x+5\right)}+\frac{5\cdot5\cdot\left(x+10\right)}{5x\left(x+5\right)}-\frac{2\left(5-x\right)\cdot5\left(x+5\right)}{5x\left(x+5\right)}\right)\div\frac{3\left(x+5\right)}{7}\)
\(=\left(\frac{x^2}{5x\left(x+5\right)}+\frac{25x+250}{5x\left(x+5\right)}-\frac{10\left(25-x^2\right)}{5x\left(x+5\right)}\right)\div\frac{3\left(x+5\right)}{7}\)
\(=\left(\frac{x^2+25x+250-250+10x^2}{5x\left(x+5\right)}\right)\div\frac{3\left(x+5\right)}{7}\)
\(=\frac{11x^2+25x}{5x\left(x+5\right)}\times\frac{7}{3\left(x+5\right)}\)
\(=\frac{77x^2+175x}{15x\left(x+5\right)^2}\)
\(=\frac{77x^2+175x}{15x\left(x^2+10x+25\right)}=\frac{77x^2+175x}{15x^3+150x^2+375x}\)
\(=\frac{77x+175}{15x^2+150x+375}\)
\(A=\frac{\left(x^2+2x\right)x+\left(x-5\right)2\left(x+5\right)+20-5x}{2x\left(x+5\right)}\)
\(A=\frac{x^3+2x^2+2x^2-50+20-5x}{2x\left(x+5\right)}\)
\(A=\frac{x^3+4x^2-5x-30}{2x\left(x+5\right)}\)
a) \(P=\frac{x}{2x-2}+\frac{x^2+1}{2-2x^2}\)
\(P=\frac{x}{2\left(x-1\right)}+\frac{x^2+1}{2\left(1-x^2\right)}\)
\(P=\frac{x}{2\left(x-1\right)}-\frac{x^2+1}{2\left(x^2-1\right)}\)
\(P=\frac{x}{2\left(x-1\right)}-\frac{x^2+1}{2\left(x-1\right)\left(x+1\right)}\)
\(P=\frac{x\left(x+1\right)-x^2-1}{2\left(x-1\right)\left(x+1\right)}\)
\(P=\frac{x^2+x-x^2-1}{2\left(x-1\right)\left(x+1\right)}\)
\(P=\frac{x-1}{2\left(x-1\right)\left(x+1\right)}=\frac{1}{2\left(x+1\right)}\)
b) \(Q=\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)
\(Q=\frac{x\left(x+2\right)}{2\left(x+5\right)}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)
\(Q=\frac{x^2\left(x+2\right)+2\left(x+5\right)\left(x-5\right)+50-5x}{2x\left(x+5\right)}\)
\(Q=\frac{x^3+2x^2+2\left(x^2-25\right)+50-5x}{2x\left(x+5\right)}\)
\(Q=\frac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}\)
\(Q=\frac{x^3+4x^2-5x}{2x\left(x+5\right)}\)
\(Q=\frac{x\left(x^2+4x-5\right)}{2x\left(x+5\right)}=\frac{x^2+4x-5}{2\left(x+5\right)}\)
=\(\left(\frac{x}{\left(x-5\right).\left(x+5\right)}-\frac{\left(x-5\right)}{x.\left(x+5\right)}\right).\frac{x^2+5x}{2x-5}\)
=\(\left(\frac{x^2}{x.\left(x-5\right).\left(x+5\right)}-\frac{\left(x-5\right)^2}{x.\left(x-5\right).\left(x+5\right)}\right).\frac{x\left(x+5\right)}{2x-5}\)
=\(\frac{x^2-\left(x-5\right)^2}{x.\left(x-5\right).\left(x+5\right)}.\frac{x.\left(x+5\right)}{2x-5}\)
=\(\frac{\left(x-x+5\right).\left(x+x-5\right)}{x.\left(x-5\right)\left(x+5\right)}.\frac{x.\left(x+5\right)}{2x+5}\)
=\(\frac{5.\left(2x-5\right).x\left(x+5\right)}{x.\left(x-5\right).\left(x+5\right).\left(2x-5\right)}\)
=\(\frac{5}{x+5}\)