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a)Vì |4x - 2| = 6 <=> 4x - 2 ϵ {6,-6} <=> x ϵ {2,-1}
Thay x = 2, ta có B không tồn tại
Thay x = -1, ta có B = \(\dfrac{1}{3}\)
b)ĐKXĐ:x ≠ 2,-2
Ta có \(A=\dfrac{5}{x+2}+\dfrac{3}{2-x}-\dfrac{15-x}{4-x^2}=\dfrac{10-5x+3x+6}{\left(x+2\right)\left(2-x\right)}-\dfrac{15-x}{4-x^2}=\dfrac{16-2x}{\left(x+2\right)\left(2-x\right)}-\dfrac{15-x}{4-x^2}=\dfrac{2x-16}{\left(x+2\right)\left(x-2\right)}-\dfrac{15-x}{4-x^2}=\dfrac{2x-16}{x^2-4}+\dfrac{15-x}{x^2-4}=\dfrac{x-1}{x^2-4}\)c)Từ câu b, ta có \(A=\dfrac{x-1}{x^2-4}\)\(\Rightarrow\dfrac{2A}{B}=\dfrac{\dfrac{\dfrac{2x-2}{x^2-4}}{2x+1}}{x^2-4}=\dfrac{2x-2}{2x+1}< 1\) với mọi x
Do đó không tồn tại x thỏa mãn đề bài
a) điều kiện \(x\ne\pm2\)
\(A=\left(\dfrac{4}{x+2}+\dfrac{2}{x-2}+\dfrac{5x-6}{4-x^2}\right):\dfrac{1}{3x-2x^2-6}\)
\(A=\left(\dfrac{4}{x+2}+\dfrac{2}{x-2}-\dfrac{5x-6}{x^2-4}\right):\dfrac{1}{3x-2x^2-6}\)
\(A=\left(\dfrac{4}{x+2}+\dfrac{2}{x-2}-\dfrac{5x-6}{\left(x-2\right)\left(x+2\right)}\right):\dfrac{1}{3x-2x^2-6}\)
\(A=\dfrac{4\left(x-2\right)+2\left(x+2\right)-\left(5x-6\right)}{\left(x+2\right)\left(x-2\right)}:\dfrac{1}{3x-2x^2-6}\)
\(A=\dfrac{4x-8+2x+4-5x+6}{\left(x+2\right)\left(x-2\right)}:\dfrac{1}{3x-2x^2-6}\)
\(A=\dfrac{x+2}{\left(x+2\right)\left(x-2\right)}:\dfrac{1}{3x-2x^2-6}\)
\(A=\dfrac{1}{x-2}.\dfrac{3x-2x^2-6}{1}=\dfrac{3x-2x^2-6}{x-2}\)
b) ta có : \(3x-2x^2-6=-2x^2+3x-6=-\left(2x^2-3x+6\right)\)
\(=\left(\left(\sqrt{2}x\right)^2-2.\sqrt{2}x.\dfrac{3}{2\sqrt{2}}+\left(\dfrac{3}{2\sqrt{2}}\right)^2\right)+\dfrac{39}{8}\)
\(=\left(\sqrt{2}x-\dfrac{3}{2\sqrt{2}}\right)^2+\dfrac{39}{8}\ge\dfrac{39}{8}>0\)
\(\Rightarrow A\le0\) \(\Leftrightarrow x-2\le0\) (mà đk : \(x\ne2\) \(\Rightarrow x-2\ne0\))
vậy \(A\le0\Leftrightarrow A< 0\) \(\Leftrightarrow x-2< 0\Leftrightarrow x< 2\) vậy \(x< 2\)
a: ĐKXĐ: \(x\notin\left\{0;-4;-2;2\right\}\)
b: \(B=\dfrac{1}{x+2}-\dfrac{x^2-4}{x+4}\cdot\left(\dfrac{4x^2+x^2+4x+4}{4x^2\left(x+2\right)^2}\right)\)
\(=\dfrac{1}{x+2}-\dfrac{\left(x-2\right)}{x+4}\cdot\dfrac{5x^2+4x+4}{4x^2\left(x+2\right)}\)
\(=\dfrac{4x^3+16x^2-\left(x-2\right)\left(5x^2+4x+4\right)}{4x^2\left(x+4\right)\left(x+2\right)}\)
\(=\dfrac{4x^3+16x^2-5x^3-4x^2-4x+10x^2+8x+8}{4x^2\left(x+4\right)\left(x+2\right)}\)
\(=\dfrac{-x^3+22x^2+4x+8}{4x^2\left(x+4\right)\left(x+2\right)}\)
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. \(A^{-1}-B=\dfrac{x+3}{x^2+x+1}-\dfrac{x^2+x-2}{x^3-1}\)
\(=\dfrac{x+3}{x^2+x+1}-\dfrac{x^2+x-2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{\left(x+3\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{x^2+x-2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^2-x+3x-3-x^2-x+2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)
\(=\dfrac{1}{x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}\)
\(Max=\dfrac{4}{3}\Leftrightarrow x=\dfrac{-1}{2}\)
a: \(=\dfrac{x^4+15x+7}{x^4+15x+7}\cdot\dfrac{x}{14x^2+1}\cdot\dfrac{4x^3+4}{2x^3+2}=\dfrac{2x}{14x^2+1}\)
b: \(=\dfrac{x^7+3x^2+2}{x^7+3x^2+2}\cdot\dfrac{x^2+x+1}{x^3-1}\cdot\dfrac{3x}{x+1}\)
\(=\dfrac{1}{x-1}\cdot\dfrac{3x}{x+1}=\dfrac{3x}{x^2-1}\)
`a)M=(x^4+2)/(x^6+1)+(x^2-1)/(x^4-x^2+1)-(x^2+3)/(x^4+4x^2+3)`
`=(x^4+2)/(x^6+1)+(x^2-1)/(x^4-x^2+1)-(x^2+3)/((x^2+1)(x^2+3))`
`=(x^4+2)/(x^6+1)+((x^2-1)(x^2+1))/(x^6+1)-1/(x^2+1)`
`=(x^4+2+x^4-1-x^4+x^2-1)/(x^2+1)`
`=(x^4+x^2)/(x^2+1)`
`=(x^2(x^2+1))/(x^2+1)`
`=x^2`
`b)` tìm gtnn chứ?
`M=x^2>=0`
Dấu '=" `<=>x=0`