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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 \(P\in Z\Leftrightarrow x\in\left\{0;\dfrac{1}{2};\dfrac{-3}{2}\right\}\)
\(a,A=\dfrac{5-3}{5+2}=\dfrac{2}{7}\\ b,B=\dfrac{3x-9+2x+6-3x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ c,C=AB=\dfrac{x-3}{x+2}\cdot\dfrac{2}{x-3}=\dfrac{2}{x+2}\\ C=-\dfrac{1}{3}\Leftrightarrow x+2=-6\Leftrightarrow x=-8\left(tm\right)\)
\(=\dfrac{\left(x+2\right)^2}{x}\cdot\dfrac{x+2-x^2}{x+2}-\dfrac{x^2+6x+4}{x}\)
\(=\dfrac{\left(x+2\right)\left(-x^2+x+2\right)-x^2-6x-4}{x}\)
\(=\dfrac{-x^3+x^2+2x-2x^2+2x+4-x^2-6x-4}{x}\)
\(=\dfrac{-x^3-2x^2-2x}{x}=-x^2-2x-2\)
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: \(\left\{{}\begin{matrix}-2x^3+9x^2-12x+9< 0\\x^2-9< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>3\\-3< x< 3\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
\(a,=\dfrac{\left(x-2\right)^2-\left(x+2\right)^2}{\left(x-2\right)^2\left(x+2\right)^2}:\dfrac{x-2+x+2}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{-8x}{\left(x-2\right)^2\left(x+2\right)^2}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{2x}=\dfrac{-4}{\left(x-2\right)\left(x+2\right)}\)
\(b,=\dfrac{5x^2+26xy+5y^2+5x^2-26xy+5y^2}{x\left(x-5y\right)\left(x+5y\right)}\cdot\dfrac{\left(x-5y\right)\left(x+5y\right)}{x^2+y^2}\\ =\dfrac{10\left(x^2+y^2\right)}{x\left(x^2+y^2\right)}=\dfrac{10}{x}\)
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
\(a.\dfrac{2x-1}{x-1}+\dfrac{x}{x^2-3x+2}=\dfrac{6x-2}{x-2}\left(x\ne2;x\ne1\right)\)
\(\Leftrightarrow\dfrac{\left(2x-1\right)\left(x-2\right)+x}{\left(x-1\right)\left(x-2\right)}=\dfrac{\left(6x-2\right)\left(x-1\right)}{\left(x-1\right)\left(x-2\right)}\)
\(\Leftrightarrow2x^2-4x-x+2+x=6x^2-6x-2x+2\)
\(\Leftrightarrow2x^2-5x+2=6x^2-8x+2\)
\(\Leftrightarrow4x^2-3x=0\)
\(\Leftrightarrow x\left(4x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=\dfrac{3}{4}\left(TM\right)\end{matrix}\right.\)
KL........
\(b.A=\sqrt{x^2-x+1\dfrac{1}{4}}-2016=\sqrt{x^2-2.\dfrac{1}{2}x+\dfrac{1}{4}+1}-2016=\sqrt{\left(x-\dfrac{1}{2}\right)^2+1}-2016\ge1-2016=-2015\)
\(\Rightarrow A_{Min}=-2015."="\Leftrightarrow x=\dfrac{1}{2}\)
Bài 1:
a: \(P=\left(\dfrac{x-2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+2}{\left(x+1\right)^2}\right)\cdot\dfrac{\left(x-1\right)^2\cdot\left(x+1\right)^2}{4}\)
\(=\dfrac{x^2-x-2-x^2-x+2}{\left(x-1\right)\left(x+1\right)^2}\cdot\dfrac{\left(x-1\right)^2\cdot\left(x+1\right)^2}{4}\)
\(=\dfrac{-2x}{1}\cdot\dfrac{x-1}{4}=-\dfrac{x\left(x-1\right)}{2}\)
b: Để \(\dfrac{P-4}{5}=x\) thì P-4=5x
=>P=5x+4
\(\Leftrightarrow-\dfrac{x\left(x-1\right)}{2}=5x+4\)
=>-x2+x=10x+8
=>x2-x=-10x-8
=>x2+9x+8=0
=>x=-8(nhận) hoặc x=-1(loại)