Cho biểu thức A=(3+x/3-x - 3 -x/3+x - 4x^2/x^2-9) : (5/3-x - 4x+2/3x-x^2) a. Rút gọn A b. Tìm x để A=3 c.tìm x để A>4x
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a ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
\(P=\dfrac{x^2-4x-4}{4-x^2}+\dfrac{3x+9}{x+2}\)
\(=\dfrac{-x^2+4x+4}{\left(x-2\right)\left(x+2\right)}+\dfrac{3x+9}{\left(x+2\right)}\)
\(=\dfrac{-x^2+4x+4+\left(3x+9\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-x^2+4x+4+3x^2-6x+9x-18}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{2x^2+7x-14}{\left(x-2\right)\left(x+2\right)}\)
b: khi x=8 thì \(P=\dfrac{2\cdot8^2+7\cdot8-14}{\left(8-2\right)\left(8+2\right)}=\dfrac{2\cdot64+56-14}{64-4}=\dfrac{17}{6}\)
\(a,A=\dfrac{9-3x+x^2+10x+25-x^2+1}{\left(x-1\right)\left(x+5\right)}\\ A=\dfrac{7x+35}{\left(x-1\right)\left(x+5\right)}=\dfrac{7\left(x+5\right)}{\left(x-1\right)\left(x+5\right)}=\dfrac{7}{x-1}\\ b,A\in Z\\ \Leftrightarrow x-1\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\\ \Leftrightarrow x\in\left\{-6;0;2;8\right\}\left(tm\right)\\ b,A< 0\Leftrightarrow x-1< 0\left(7>0\right)\\ \Leftrightarrow x< 1;x\ne-5\\ c,\left|A\right|=3\Leftrightarrow\dfrac{7}{\left|x-1\right|}=3\Leftrightarrow\left|x-1\right|=\dfrac{7}{3}\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}+1=\dfrac{10}{3}\left(tm\right)\\x=-\dfrac{7}{3}+1=-\dfrac{4}{3}\left(tm\right)\end{matrix}\right.\)
a) đk: x khác 0;2;-2;3
A = \(\left(\dfrac{2+x}{2-x}-\dfrac{4x^2}{x^2-4}-\dfrac{2-x}{2+x}\right):\dfrac{x^2-3x}{2x^2-x^3}\)
= \(\left(\dfrac{2+x}{2-x}+\dfrac{4x^2}{\left(2-x\right)\left(2+x\right)}-\dfrac{2-x}{2+x}\right):\dfrac{x-3}{2x-x^2}\)
= \(\left(\dfrac{\left(x+2\right)^2+4x^2-\left(2-x\right)^2}{\left(2-x\right)\left(2+x\right)}\right):\dfrac{x-3}{x\left(2-x\right)}\)
= \(\dfrac{x^2+4x+4+4x^2-x^2+4x-4}{\left(2-x\right)\left(2+x\right)}.\dfrac{x\left(2-x\right)}{x-3}\)
= \(\dfrac{4x^2+8x}{\left(2-x\right)\left(2+x\right)}.\dfrac{x\left(2-x\right)}{x-3}\)
= \(\dfrac{4x\left(x+2\right)}{\left(2-x\right)\left(2+x\right)}.\dfrac{x\left(2-x\right)}{x-3}=\dfrac{4x^2}{x-3}\)
b) Có \(\left|x-5\right|=2\)
<=> \(\left[{}\begin{matrix}x-5=2< =>x=7\left(Tm\right)\\x-5=-2< =>x=3\left(L\right)\end{matrix}\right.\)
Thay x = 7 vào A, ta có:
\(A=\dfrac{4.7^2}{7-3}=49\)
c) A = \(\dfrac{4x^2}{x-3}⋮4\left(\forall x\right)\)
Bài 1 :
a, \(\left(x+3\right)^2+\left(x-3\right)^2+2\left(x^2-9\right)\)
\(=x^2+6x+9+x^2-6x+9+2x^2-18\)
\(=4x^2\)
b, \(\left(4x-1\right)^3-\left(4x-3\right)\left(16x^2+3\right)\)
\(=64x^3-32x^2+4x-16x^2+8x-1-64x^3-12x+48x^2+9=8\)