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Bài 1:
a: ĐKXĐ: \(x\notin\left\{0;2;-2;3\right\}\)
\(A+\left(\dfrac{4x}{x+2}-\dfrac{8x^2}{\left(x+2\right)\left(x-2\right)}\right):\left(\dfrac{x-1}{x\left(x-2\right)}-\dfrac{2}{x}\right)\)
\(=\dfrac{4x^2-8x-8x^2}{\left(x+2\right)\left(x-2\right)}:\dfrac{x-1-2x+4}{x\left(x-2\right)}\)
\(=\dfrac{-4x\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{x\left(x-2\right)}{-x+3}\)
\(=\dfrac{-4x}{-x+3}=\dfrac{4x}{x-3}\)
b: Để A<0 thi x/x-3<0
=>0<x<3
ĐKXĐ:\(x\ne\pm2;x\ne-3;x\ne0\)
\(P=1+\frac{x-3}{x^2+5x+6}\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\left[\frac{8x^2}{4x^2\left(x-2\right)}-\frac{3x}{3\left(x^2-4\right)}-\frac{1}{x+2}\right]\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\left(\frac{2}{x-2}-\frac{x}{x^2-4}-\frac{1}{x+2}\right)\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\left[\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right]\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\cdot\frac{2x+4-x-x+4}{\left(x-2\right)\left(x+2\right)}\)
\(=1+\frac{8\left(x-3\right)}{\left(x+2\right)^2\left(x+3\right)\left(x-2\right)}\)
Đề sai à ??
a) A = x2(m + 5) - x(m + 5)(x + 3/2) + (x - m)
A = mx2 + 5x2 - mx2 - 3/2mx - 5x2 - 15/2x + x - m
A = -3/2mx - m - 13/2x
b) Khi m = -1, ta có:
(-3/2).(-1).x - (-1) - 13/2x = 0
<=> 3/2x - 13/2x + 1 = 0
<=> 3/2x - 13/2x = 0 - 1
<=> 3/2x - 13/2x = -1
<=> 3x - 13x = -2
<=> -10x = -2
<=> x = -2/-10 = 1/5
a: \(A=\dfrac{x-2-2x-4+x}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-\left(x-2\right)\left(x+1\right)}{6\left(x+2\right)}\)
\(=\dfrac{-6}{\left(x+2\right)}\cdot\dfrac{-\left(x+1\right)}{6\left(x+2\right)}=\dfrac{\left(x+1\right)}{\left(x+2\right)^2}\)
b: A>0
=>x+1>0
=>x>-1
c: x^2+3x+2=0
=>(x+1)(x+2)=0
=>x=-2(loại) hoặc x=-1(loại)
Do đó: Khi x^2+3x+2=0 thì A ko có giá trị
B1: ĐXXĐ: \(x\ne\pm2;x\ne-1\)
\(=\left(\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}-\dfrac{2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{x}{\left(x+2\right)\left(x-2\right)}\right):\dfrac{-6\left(x+2\right)}{\left(x-2\right)\left(x+1\right)}\)
\(=\left(\dfrac{x-2-2x-2+x}{\left(x+2\right)\left(x-2\right)}\right):\dfrac{-6\left(x+2\right)}{\left(x-2\right)\left(x+1\right)}\)
\(=\dfrac{-4}{\left(x+2\right)\left(x-2\right)}:\dfrac{-6\left(x+2\right)}{\left(x-2\right)\left(x+1\right)}\)
\(=\dfrac{-4}{\left(x+2\right)\left(x-2\right)}.\dfrac{\left(x-2\right)\left(x+1\right)}{-6\left(x+2\right)}=\dfrac{2\left(x+1\right)}{3\left(x+2\right)^2}\)
b, \(A=\dfrac{2\left(x+1\right)}{3\left(x+2\right)^2}>0\)
\(\Leftrightarrow2x+2>0\) (vì \(3\left(x+2\right)^2\ge0\forall x\))
\(\Leftrightarrow x>-1\).
-Vậy \(x\in\left\{x\in Rlx>-1;x\ne2\right\}\) thì \(A>0\).
a) M = ( 2x + 3)(2x - 3) - 2(x + 5)2 - 2(x - 1)(x + 2)
= 4x2 - 9 - 2(x2 + 10x + 25) - 2(x2 + x - 2)
= 4x2 - 9 - 2x2 - 20x - 50 - 2x2 - 2x + 4
= -22x - 55 = -11(2x + 5)
b) M = -11(2x + 5) = - 11(2.\(\frac{-7}{3}\)+ 5) = \(\frac{-11}{3}\)
b) M = -11(2x + 5) = 0
\(\Rightarrow\)2x + 5 = 0
\(\Rightarrow\)x = \(\frac{-5}{2}\)
Ta có: M = (2x+3)(2x-3) - 2(x+5)2 - 2(x-1)(x+2) \(=\left(2x\right)^2-3^2-2\left(x^2+10x+25\right)-\) \(2\left(x^2+x-2\right)\)
\(=4x^2-9-2x^2-20x-50-2x^2-2x+4\) =\(\left(4x^2-2x^2-2x^2\right)-\left(20x+2x\right)-\left(50+9-4\right)\) \(=-22x-55\)
b, Với x = \(-2\frac{1}{3}=\frac{-7}{3}\)
\(\Rightarrow M=-22.\frac{-7}{3}-55=\frac{154}{3}-55=\frac{-11}{3}\)
c, Để M = 0 => -22x - 55 = 0 \(\Rightarrow-22x=55\Rightarrow x=\frac{-55}{22}=\frac{-5}{2}\)
Vậy \(x=\frac{-5}{2}\)
Ôi mình nhầm để giải lại:
a)đkxđ: x\(\ne\left\{-1;1;2\right\}\)
M=\(\dfrac{\left(x^2-3x+2\right)\left(x^2-4\right)}{\left(x^2-1\right)\left(x^2-4x+4\right)}=\dfrac{\left(x-1\right)\left(x-2\right)\left(x-2\right)\left(x+2\right)}{\left(x-1\right)\left(x+1\right)\left(x-2\right)^2}=\dfrac{x+2}{x+1}\)
b)Với x\(\ne\left\{-1;1;2\right\}\) thì M=\(\dfrac{x+2}{x+1}\)
Để M>0 thì \(\dfrac{x+2}{x+1}\)>0
<=> \(\left\{{}\begin{matrix}x+1>0\\x+2>0\end{matrix}\right.\)hoặc\(\left\{{}\begin{matrix}x+1< 0\\x+2< 0\end{matrix}\right.\)
<=>x>-1 hoặc x<-2
Vậy x>-1 hoặc x<-2 và x khác {1;2} thì M>0
M<0 <=>\(\dfrac{x+2}{x+1}\)<0
<=>\(\left\{{}\begin{matrix}x+1< 0\\x+2>0\end{matrix}\right.hoặc}\left\{{}\begin{matrix}x+1>0\\x+2< 0\end{matrix}\right.\)<=>\(\left\{{}\begin{matrix}x< -1\\x>-2\end{matrix}\right.hoặc}\left\{{}\begin{matrix}x>-1\\x< -2\end{matrix}\right.\)
Vậy -2<x<-1 thì M<0
M=0<=> \(\dfrac{x+2}{x+1}\)=0
=>x+2=0
<=>x=-2(TMĐKXĐ)
Vậy x=-2 thì M=0
M vô nghĩa khi M không xác định <=> x={-1;1;2}
\(\dfrac{\left(x^2-3x+2\right)\left(x^2-4\right)}{\left(x^2-1\right)\left(x^2-4x+4\right)}\)
\(\dfrac{\left(x^2-x-2x+2\right)\left(x-2\right)\left(x+2\right)}{\left(x-1\right)\left(x+1\right)\left(x^2-2x-2x+4\right)}\)
\(\dfrac{\left[x\left(x-1\right)-2\left(x-1\right)\right]\left(x-2\right)\left(x+2\right)}{\left(x-1\right)\left(x+1\right)\left[x\left(x-2\right)-2\left(x-2\right)\right]}\)
\(\dfrac{\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x-2\right)}=\dfrac{x+2}{x-1}\)