Cho biểu thức: \(M=\left(\dfrac{1+2x}{4+2x}-\dfrac{x}{3x-6}+\dfrac{2x^2}{12-3x^2}\right).\dfrac{24+12x}{6+13x}\). Rút gọn M
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TN
19 tháng 8 2017
\(A=\left(\dfrac{1+2x}{4+2x}-\dfrac{x}{3x-6}+\dfrac{2x^2}{12-3x^2}\right).\dfrac{24-12x}{6+13x}\)\(=\left(\dfrac{1+2x}{2\left(x+2\right)}-\dfrac{x}{3\left(x-2\right)}-\dfrac{2x^2}{3\left(x-2\right)\left(x+2\right)}\right).\dfrac{-12\left(x-2\right)}{13x+6}\)\(=\left(\dfrac{3\left(1+2x\right)\left(x-2\right)}{6\left(x+2\right)\left(x-2\right)}-\dfrac{2x\left(x+2\right)}{6\left(x+2\right)\left(x-2\right)}-\dfrac{4x^2}{6x\left(x+2\right)\left(x-2\right)}\right).\dfrac{-2\left(x-2\right)}{13x+6}\)\(=\dfrac{6x^2-9x-6-2x^2-4x-4x^2}{6\left(x+2\right)\left(x-2\right)}.\dfrac{-12\left(x-2\right)}{13x+6}\)\(=\dfrac{-\left(13x+6\right)}{6\left(x+2\right)\left(x-2\right)}.\dfrac{-12\left(x-2\right)}{13x+6}\)
\(=\dfrac{2}{x+2}\)
NH
1
17 tháng 6 2022
\(A=\left(\dfrac{2x+1}{2\left(x+2\right)}-\dfrac{x}{3\left(x-2\right)}-\dfrac{2x^2}{3\left(x-2\right)\left(x+2\right)}\right)\cdot\dfrac{24-12x}{13x+6}\)
\(=\dfrac{3\left(2x+1\right)\left(x-2\right)-2x\left(x+2\right)-4x^2}{6\left(x-2\right)\left(x+2\right)}\cdot\dfrac{12\left(2-x\right)}{13x+6}\)
\(=\dfrac{3\left(2x^2-3x-2\right)-2x^2-4x-4x^2}{x+2}\cdot\dfrac{-2}{13x+6}\)
\(=\dfrac{6x^2-9x-6-6x^2-4x}{x+2}\cdot\dfrac{-2}{13x+6}=\dfrac{2}{x+2}\)
14 tháng 3 2022
a. \(A=\left(\dfrac{2-3x}{x^2+2x-3}-\dfrac{x+3}{1-x}-\dfrac{x+1}{x+3}\right):\dfrac{3x+12}{x^3-1}\left(ĐKXĐ:x\ne1;x\ne-3\right)\)
\(=\left(\dfrac{2-3x}{\left(x-1\right)\left(x+3\right)}+\dfrac{x+3}{x-1}-\dfrac{x+1}{x+3}\right):\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\left(\dfrac{2-3x}{\left(x-1\right)\left(x+3\right)}+\dfrac{\left(x+3\right)^2}{\left(x-1\right)\left(x+3\right)}-\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+3\right)}\right):\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{2-3x+x^2+6x+9-x^2+1}{\left(x-1\right)\left(x+3\right)}:\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{3x+12}{\left(x-1\right)\left(x+3\right)}:\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{3x+12}{\left(x-1\right)\left(x+3\right)}.\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{3x+12}=\dfrac{x^2+x+1}{x+3}\)
\(M=A.B=\dfrac{x^2+x+1}{x+3}.\dfrac{x^2+x-2}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^2+x-2}{x+3}\)
b. -Để M thuộc Z thì:
\(\left(x^2+x-2\right)⋮\left(x+3\right)\)
\(\Rightarrow\left(x^2+3x-2x-6+4\right)⋮\left(x+3\right)\)
\(\Rightarrow\left[x\left(x+3\right)-2\left(x+3\right)+4\right]⋮\left(x+3\right)\)
\(\Rightarrow4⋮\left(x+3\right)\)
\(\Rightarrow x+3\in\left\{1;2;4;-1;-2;-4\right\}\)
\(\Rightarrow x\in\left\{-2;-1;1;-4;-5;-7\right\}\)
c. \(A^{-1}-B=\dfrac{x+3}{x^2+x+1}-\dfrac{x^2+x-2}{x^3-1}\)
\(=\dfrac{x+3}{x^2+x+1}-\dfrac{x^2+x-2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{\left(x+3\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{x^2+x-2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^2-x+3x-3-x^2-x+2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)
\(=\dfrac{1}{x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}\)
\(Max=\dfrac{4}{3}\Leftrightarrow x=\dfrac{-1}{2}\)
13 tháng 11 2021
\(M=\dfrac{\left(x+2\right)\left(x+3\right)+x\sqrt{\left(3-x\right)\left(3+x\right)}}{x\left(3-x\right)+\left(x+2\right)\sqrt{\left(3-x\right)\left(3+x\right)}}:2\sqrt{\dfrac{3-x+2x}{3-x}}\left(-3\le x< 3;x\ne-1\right)\\ M=\dfrac{\sqrt{x+3}\left(x+2+x\sqrt{3-x}\right)}{\sqrt{3-x}\left[x+\left(x+2\right)\sqrt{3+x}\right]}:2\sqrt{\dfrac{x+3}{3-x}}\\ M=\dfrac{\sqrt{x+3}\left(x+2+x\sqrt{3-x}\right)}{\sqrt{3-x}\left[x+\left(x+2\right)\sqrt{3+x}\right]}\cdot\dfrac{3-x}{2\sqrt{\left(3-x\right)}\sqrt{\left(x+3\right)}}\)
\(M=\dfrac{x+2+x\sqrt{3-x}}{x+\left(x+2\right)\sqrt{3-x}}\cdot\dfrac{\sqrt{3-x}}{2\sqrt{3-x}}\\ M=\dfrac{\left(x+2\right)\sqrt{3-x}+x\left(3-x\right)}{2x\sqrt{3-x}+2\left(x+2\right)\sqrt{3-x}}\\ M=\dfrac{\sqrt{3-x}\left(2x+2\right)}{\sqrt{3-x}\left(2x+2x+4\right)}=\dfrac{2\left(x+1\right)}{4\left(x+1\right)}=\dfrac{1}{2}\)
