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a)Q=\(\dfrac{1+x}{x}\)
b)x không tính được hoặc đề sai
c)?
a: \(Q=\dfrac{1+x}{x\left(x+1\right)}\cdot\dfrac{x+1}{1}=\dfrac{x+1}{x}\)
b: Để Q=1 thì x+1=x(loại)
c: \(Q-\dfrac{1}{2}=\dfrac{x+1}{x}-\dfrac{1}{2}=\dfrac{2x+2-x}{2x}=\dfrac{x+2}{2x}\)
TH1: x>0 hoặc x<-2
=>Q>0
TH2: -2<x<0
=>Q<0
a) Ta có: \(B=\left(\dfrac{x}{3x-9}+\dfrac{2x-3}{3x-x^2}\right)\cdot\dfrac{3x^2-9x}{x^2+6x+9}\)
\(=\left(\dfrac{x}{3\left(x-3\right)}-\dfrac{2x-3}{x\left(x-3\right)}\right)\cdot\dfrac{3x\left(x-3\right)}{\left(x+3\right)^2}\)
\(=\left(\dfrac{x^2}{3x\left(x-3\right)}-\dfrac{3\left(2x-3\right)}{3x\left(x-3\right)}\right)\cdot\dfrac{3x\left(x-3\right)}{\left(x+3\right)^2}\)
\(=\dfrac{x^2-6x+9}{3x\left(x-3\right)}\cdot\dfrac{3x\left(x-3\right)}{\left(x+3\right)^2}\)
\(=\dfrac{x^2-6x+9}{x^2+6x+9}\)
b) Ta có: \(A=\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right):\dfrac{1}{x+2}\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right):\dfrac{1}{x+2}\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}\right):\dfrac{1}{x+2}\)
\(=\left(\dfrac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}\right):\dfrac{1}{x+2}\)
\(=\dfrac{-6}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{1}\)
\(=\dfrac{-6}{x-2}\)
Bài 2:
a, ĐKXĐ: \(x\ne\pm1;x\ne\dfrac{-1}{2}\)
\(P=\left(\dfrac{x-1}{x+1}-\dfrac{x}{x-1}-\dfrac{3x+1}{1-x^2}\right):\dfrac{2x+1}{x^2-1}\)
\(P=\left(\dfrac{x-1}{x+1}-\dfrac{x}{x-1}+\dfrac{3x+1}{x^2-1}\right).\dfrac{x^2-1}{2x+1}\)
\(P=\dfrac{\left(x-1\right)^2-x\left(x+1\right)+3x+1}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\)
\(P=\dfrac{x^2-2x+1-x^2-x+3x+1}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\)
\(P=\dfrac{2}{2x+1}\)
b, ĐKXĐ: \(x\ne\pm1;x\ne\dfrac{-1}{2}\)
Để \(P=\dfrac{3}{x-1}\Leftrightarrow\dfrac{2}{2x+1}=\dfrac{3}{x-1}\Leftrightarrow2\left(x-1\right)=3\left(2x+1\right)\)
\(\Leftrightarrow2x-2=6x+3\)\(\Leftrightarrow-4x=5\Leftrightarrow x=\dfrac{-5}{4}\)(TMĐK)
c, \(ĐKXĐ:x\ne\pm1;x\ne\dfrac{-1}{2}\)
Để \(P\in Z\Leftrightarrow\dfrac{2}{2x+1}\in Z\Leftrightarrow2x+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
+) Với \(2x+1=1\Leftrightarrow x=0\left(TMĐK\right)\)
+) Với \(2x+1=-1\Leftrightarrow x=-1\left(KTMĐK\right)\)
+) Với \(2x+1=2\Leftrightarrow x=\dfrac{1}{2}\left(TMĐK\right)\)
+) Với \(2x+1=-2\Leftrightarrow x=\dfrac{-3}{2}\left(TMĐK\right)\)
Vậy để \(P\in Z\Leftrightarrow x\in\left\{0;\dfrac{1}{2};\dfrac{-3}{2}\right\}\)
\(a,A=\dfrac{5-3}{5+2}=\dfrac{2}{7}\\ b,B=\dfrac{3x-9+2x+6-3x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ c,C=AB=\dfrac{x-3}{x+2}\cdot\dfrac{2}{x-3}=\dfrac{2}{x+2}\\ C=-\dfrac{1}{3}\Leftrightarrow x+2=-6\Leftrightarrow x=-8\left(tm\right)\)
1 ) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=-15\)
\(\Leftrightarrow x^3-6x^2+12x-8-\left(x^3-27\right)+6\left(x^2+2x+1\right)=-15\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=-15\)
\(\Leftrightarrow\left(x^3-x^3\right)+\left(6x^2-6x^2\right)+\left(12x+12x\right)+\left(27+6-8\right)=-15\)
\(\Leftrightarrow24x+25=-15\)
\(\Leftrightarrow24x=-40\)
\(\Leftrightarrow x=-\dfrac{5}{3}\)
Vậy \(x=-\dfrac{5}{3}\)
Câu 1 :
a) Rút gọn P :
\(P=\dfrac{x+1}{3x-x^2}:\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{12x^2}{x^2-9}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left[\dfrac{\left(3+x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{\left(3-x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{12x^2}{\left(3-x\right)\left(3+x\right)}\right]\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left(\dfrac{9+6x+x^2-9+6x-x^2-12x^2}{\left(3-x\right)\left(3+x\right)}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\dfrac{12x-12x^2}{\left(3-x\right)\left(x+3\right)}\)
\(P=\dfrac{x+1}{x\left(3-x\right)}.\dfrac{\left(3-x\right)\left(x+3\right)}{12x\left(1-x\right)}\)
\(P=\dfrac{\left(x+1\right)\left(x+3\right)}{12x^2\left(1-x\right)}\)
a) Điều kiện xác định :
x ≠ 3; x ≠ -3; x ≠ 0
M = \(\dfrac{x}{x^2-9}\) - \(\dfrac{1}{x+3}\): ( \(\dfrac{x}{x\left(x-3\right)}\) - \(\dfrac{x-3}{x\left(x-3\right)}\) )
M = \(\dfrac{x}{x^2-9}\) - \(\dfrac{1}{x+3}\) : ( \(\dfrac{x-x+3}{x\left(x-3\right)}\) )
M = \(\dfrac{x}{x^2-9}\) - \(\dfrac{1}{x+3}\) : \(\dfrac{3}{x\left(x-3\right)}\)
M = \(\dfrac{x}{x^2-9}\) - \(\dfrac{x\left(x-3\right)}{3\left(x+3\right)}\) = \(\dfrac{x}{\left(x-3\right)\left(x+3\right)}\) - \(\dfrac{x\left(x-3\right)}{3\left(x+3\right)}\)
M = \(\dfrac{3x}{3\left(x-3\right)\left(x+3\right)}\) - \(\dfrac{x\left(x-3\right)^2}{3\left(x-3\right)\left(x+3\right)}\)
M = \(\dfrac{3x-x\left(x-3\right)^2}{3\left(x-3\right)\left(x+3\right)}\) = \(\dfrac{3x-x\left(x^2-6x+9\right)}{3\left(x-3\right)\left(x+3\right)}\)
M = \(\dfrac{3x-x^3+6x^2-9x}{3\left(x-3\right)\left(x+3\right)}\) = \(\dfrac{-x^3+6x^2-6x}{3\left(x-3\right)\left(x+3\right)}\)
Mk đang mệt sai thì bạn thông cảm cho mk.
a: \(M=\dfrac{x}{\left(x-3\right)\left(x+3\right)}-\dfrac{1}{x+3}:\dfrac{x-x+3}{x\left(x-3\right)}\)
\(=\dfrac{x}{\left(x-3\right)\left(x+3\right)}-\dfrac{1}{x+3}\cdot\dfrac{x\left(x-3\right)}{3}\)
\(=\dfrac{x}{\left(x-3\right)\left(x+3\right)}-\dfrac{x\left(x-3\right)}{3\left(x+3\right)}\)
\(=\dfrac{3x-x\left(x^2-6x+9\right)}{3\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{3x-x^3+6x^2-9x}{3\left(x-3\right)\left(x+3\right)}=\dfrac{-x^3+6x^2-6x}{3\left(x-3\right)\left(x+3\right)}\)
b: Để M>1/2 thì M-1/2>0
=>\(\dfrac{-x^3+6x^2-6x}{3\left(x^2-9\right)}-\dfrac{1}{2}>0\)
=>\(\dfrac{-2x^3+12x^2-12x-3x^2+9}{6\left(x^2-9\right)}>0\)
=>\(\dfrac{-2x^3+9x^2-12x+9}{x^2-9}>0\)
TH1: \(\left\{{}\begin{matrix}-2x^3+9x^2-12x+9>0\\x^2-9>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 3\\\left[{}\begin{matrix}x>3\\x< -3\end{matrix}\right.\end{matrix}\right.\Leftrightarrow x< -3\)
TH2: \(\left\{{}\begin{matrix}-2x^3+9x^2-12x+9< 0\\x^2-9< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>3\\-3< x< 3\end{matrix}\right.\Leftrightarrow x\in\varnothing\)