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\(\frac{x^3+2x^2}{2x^2+10x}\)+\(\frac{2x^2-10x+10x-50}{2x^2-10x}\)+\(\frac{50-5x}{2x^2+10x}\)=\(\frac{x^3+4x^2-5x}{2x^2-10x}\)=\(\frac{x\left(x^2+4x-5\right)}{2x\left(x-5\right)}\)=\(\frac{x\left(x-1\right)\left(x-5\right)}{2x\left(x-5\right)}\)=\(\frac{x-1}{2}\)
\(A=\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)
\(=\frac{x\left(x+2\right)}{2\left(x+5\right)}+\frac{x-5}{x}+\frac{5\left(10-x\right)}{2x\left(x+5\right)}\)
\(=\frac{x^2\left(x+2\right)+2\left(x+5\right)\left(x-5\right)+5\left(10-x\right)}{2x\left(x+5\right)}\)
\(=\frac{x^3+2x+2x^2-50+50-5x}{2x\left(x+5\right)}\)
\(=\frac{x^3-3x+2x^2}{2x\left(x+5\right)}=\frac{x\left(x^2+2x-3\right)}{2x\left(x+5\right)}\)
\(=\frac{\left(x-1\right)\left(x+3\right)}{2\left(x+5\right)}\)
\(B=\frac{x+5}{2x}-\frac{x-6}{5-x}-\frac{2x^2-2x-50}{2x^2-10x}\)
\(B=\frac{x+5}{2x}-\left(\frac{x-6}{5-x}\right)-\left(\frac{2x^2-2x-50}{2x^2-10x}\right)\)
\(B=\frac{-2x^4+30x^3-150x^2+250x}{-4x^4+40x^3-100x^2}\)
\(B=\frac{-2x^3+30x^2-150x+250}{-4x^3+40x^2-100x}\)
\(B=\frac{-x^3+15x^2-75x+125}{-2x^3+20x^2-50x}\)
\(B=\frac{\left(-x+5\right)\left(x-5\right)\left(x-5\right)}{2x\left(-x+5\right)\left(x-5\right)}\)
\(B=\frac{x-5}{2x}\)
\(a,\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\) (x khác -3; khác 0)
\(=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x.\left(x+3\right)}=\frac{3x}{2x.\left(x+3\right)}-\frac{x-6}{2x.\left(x+3\right)}=\frac{3x-x+6}{2x.\left(x+3\right)}=\frac{2x+6}{x.\left(2x+6\right)}=\frac{1}{x}\)
\(b,\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5}\) (x khác 0 , khác 1/2 khác -1/2 )
\(=\left(\frac{\left(2x+1\right)^2}{\left(2x-1\right)\left(2x+1\right)}-\frac{\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}\right).\frac{10x-5}{4x}\)
\(=\left(\frac{4x^2+4x+1}{\left(2x-1\right)\left(2x+1\right)}-\frac{4x^2-4x+1}{\left(2x-1\right)\left(2x+1\right)}\right).\frac{10x-5}{4x}\)
\(=\frac{8x}{\left(2x-1\right)\left(2x+1\right)}.\frac{5.\left(2x-1\right)}{4x}=\frac{10}{2x+1}\)
ĐK của A \(x\ne4\),ĐK của B \(\hept{\begin{cases}x\ne0\\x\ne5\end{cases}}\)
a, \(x^2-3x=0\Rightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
Với \(x=0\Rightarrow A=\frac{-5}{-4}=\frac{5}{4}\)
Với \(x=3\Rightarrow A=\frac{3-5}{3-4}=2\)
b. \(B=\frac{x+5}{2x}+\frac{x-6}{x-5}-\frac{2x^2-2x-50}{2x\left(x-5\right)}=\frac{\left(x+5\right)\left(x-5\right)+2x\left(x-6\right)-2x^2+2x+50}{2x\left(x-5\right)}\)
\(=\frac{x^2-10x+25}{2x\left(x-5\right)}=\frac{\left(x-5\right)^2}{2x\left(x-5\right)}=\frac{x-5}{2x}\)
c. \(P=\frac{A}{B}=\frac{x-5}{x-4}.\frac{2x}{x-5}=\frac{2x}{x-4}=\frac{2x-8}{x-4}+\frac{8}{x-4}=2+\frac{8}{x-4}\)
P nguyên \(\Leftrightarrow x-4\inƯ\left(8\right)\Rightarrow x-4\in\left\{-8;-4;-2;-1;1;2;4;8\right\}\)
\(\Rightarrow x\in\left\{-4;0;2;3;5;6;8;12\right\}\)
So sánh điều kiện ta thấy \(x\in\left\{-4;2;3;6;8;12\right\}\)thì P nguyên
P=x3+2x2-2x2-10x+10x+50+50-5x / 2x(x+5)
=x3+100-5x / 2x2+10x
=x3+100-1 / 2x2+2
đây là câu a nha ban mih ko ghi lai cái đề
