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\(a,=\dfrac{2y}{x}\\ b,=\dfrac{3\left(x+4\right)}{4\left(x-4\right)}\cdot\dfrac{-2\left(x-4\right)}{x+4}=\dfrac{-3}{2}\\ c,=\dfrac{\left(x+2\right)^2}{2\left(x-3\right)}\cdot\dfrac{x-3}{x+2}=\dfrac{x+2}{2}\\ d,=\dfrac{x+4+2x-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{3x}{x^2-4}\)
a, \(\dfrac{2y}{x}\)
b, \(\dfrac{3\left(x+4\right)}{4\left(x-4\right)}.\dfrac{-2\left(x-4\right)}{x+4}=\dfrac{-3}{2}\)
c, \(\dfrac{\left(x+2\right)^2}{2\left(x-3\right)}.\dfrac{x-3}{x+2}=\dfrac{x+2}{2}\)
d, \(\dfrac{x+4}{\left(x-2\right)\left(x+2\right)}+\dfrac{2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{x+4+2x-4}{\left(x+2\right)\left(x-2\right)}=\dfrac{3x}{x^2-4}\)
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a: \(=\dfrac{x^3+2x+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^3-x^2+3x-3}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^2+3}{x^2+x+1}\)
b: \(=\dfrac{x^2-2x-3+x^2+2x-3+2x-2x^2}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{2x-6}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x+3}\)
c: \(=\dfrac{6-7+x}{3\left(x-1\right)}=\dfrac{x-1}{3\left(x-1\right)}=\dfrac{1}{3}\)
d: \(=\dfrac{x^3+2x+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^3-x^2+3x-3}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^2+3}{x^2+x+1}\)
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a: Xét ΔABC có BM/BC=BD/BA
nên MD//AC
=>MM' vuông góc AB
=>M đối xứngM' qua AB
b: Xét tứ giác AMBM' có
D là trung điểm chung của AB và MM'
MA=MB
Do đó: AMBM' là hình thoi
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a: ĐKXĐ: x<>2; x<>-3
b: \(P+\dfrac{x^2-4-5-x-3}{\left(x-2\right)\left(x+3\right)}=\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=\dfrac{x-4}{x-2}\)
c: Để P=-3/4 thì x-4/x-2=-3/4
=>4x-8=-3x+6
=>7x=14
=>x=2(loại)
e: x^2-9=0
=>x=3 (nhận) hoặc x=-3(loại)
Khi x=3 thì \(P=\dfrac{3-4}{3-2}=-1\)
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Đề như này đúng chưa ạ?: (x-2)(x2 + 2x+4) - 128 + x3
=x3 - 23 - 128 + x3
= 2x3 -136
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Gọi vận tốc ca nô là x ( x > 0 )
Theo bài ra ta có pt \(\dfrac{72}{x+3}+\dfrac{54}{x-3}=6\Rightarrow x=21\left(tm\right)\)
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\(a,A=\dfrac{5x-15+2x+6-3x^2+2x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{-3x^2+9x}{\left(x-3\right)\left(x+3\right)}\\ A=\dfrac{-3x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{-3x}{x+3}\\ b,\left|x-2\right|=1\Leftrightarrow\left[{}\begin{matrix}x-2=1\\2-x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\\ \Leftrightarrow A=\dfrac{-3\cdot1}{1+3}=\dfrac{-3}{4}\\ c,A=\dfrac{-3\left(x+3\right)+9}{x+3}=-3+\dfrac{9}{x+3}\in Z\\ \Leftrightarrow x+3\inƯ\left(9\right)=\left\{-9;-3;-1;1;3;9\right\}\\ \Leftrightarrow x\in\left\{-12;-6;-4;-2;0;6\right\}\left(tm\right)\)