# Bài 8: Phép chia các phân thức đại số

17 tháng 2 lúc 20:47

$P=\left(\dfrac{2+x}{2-x}+\dfrac{4x^2}{x^2-4}-\dfrac{2-x}{2+x}\right):\dfrac{x^2-3x}{2x^2-x^3}$

$=\left(\dfrac{-x-2}{x-2}+\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{x+2}\right):\dfrac{x\left(x-3\right)}{x^2\left(2-x\right)}$

a) Đkxđ:  $x-2\ne0\Leftrightarrow x\ne2$

$x+2\ne0\Leftrightarrow x\ne-2$

$2-x\ne0\Leftrightarrow x\ne2$

$x-3\ne0\Leftrightarrow x\ne3$

b) $P=\left(\dfrac{-x-2}{x-2}+\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{x+2}\right):\dfrac{x\left(x-3\right)}{x^2\left(2-x\right)}$

$=\left(\dfrac{-\left(x+2\right)\left(x+2\right)+4x^2+\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\right).\dfrac{x^2\left(2-x\right)}{x\left(x-3\right)}$

$=\dfrac{-x\left(-x^2-4x-4+4x^2+x^2-4x+4\right)}{\left(x+2\right)\left(x-3\right)}$

$=\dfrac{-x\left(4x^2-8x\right)}{\left(x+2\right)\left(x-3\right)}$

Chắc là mình làm sai rồi. Mình không làm nữa :v. Sorry bạn nha :<<

Bình luận (1)
17 tháng 2 lúc 22:48

Bài 8:

a) ĐKXĐ: $x\notin\left\{0;2;-2;3\right\}$

b) Ta có: $P=\left(\dfrac{2+x}{2-x}+\dfrac{4x^2}{x^2-4}-\dfrac{2-x}{2+x}\right):\dfrac{x^2-3x}{2x^2-x^3}$

$=\left(\dfrac{\left(2+x\right)^2}{\left(2-x\right)\left(2+x\right)}-\dfrac{4x^2}{\left(2-x\right)\left(2+x\right)}-\dfrac{\left(2-x\right)^2}{\left(2+x\right)\left(2-x\right)}\right):\dfrac{x\left(x-3\right)}{x^2\left(2-x\right)}$

$=\dfrac{x^2+4x+4-4x^2-\left(x^2-4x+4\right)}{\left(2-x\right)\left(2+x\right)}:\dfrac{x-3}{x\left(2-x\right)}$

$=\dfrac{-3x^2+4x+4-x^2+4x-4}{\left(2-x\right)\left(2+x\right)}\cdot\dfrac{x\left(2-x\right)}{x-3}$

$=\dfrac{-4x^2+8x}{2+x}\cdot\dfrac{x}{x-3}$

$=\dfrac{x\left(-4x^2+8x\right)}{\left(x+2\right)\left(x-3\right)}$

c) Ta có: |x-5|=2

$\Leftrightarrow\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\left(nhận\right)\\x=3\left(loại\right)\end{matrix}\right.$

Thay x=7 vào biểu thức $P=\dfrac{x\left(-4x^2+8x\right)}{\left(x+2\right)\left(x-3\right)}$, ta được:

$P=\dfrac{7\cdot\left(-4\cdot7^2+8\cdot7\right)}{\left(7+2\right)\left(7-3\right)}=\dfrac{7\cdot\left(-4\cdot28+56\right)}{9\cdot4}$

$\Leftrightarrow P=\dfrac{7\cdot\left(-112+56\right)}{36}=\dfrac{7\cdot\left(-56\right)}{36}=\dfrac{-98}{9}$

Vậy: Với |x-5|=2 thì $P=-\dfrac{98}{9}$

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24 tháng 4 lúc 20:56

A = ($\dfrac{x}{\left(x-2\right)\left(x+2\right)}$ + $\dfrac{1}{x+2}$ - $\dfrac{2}{x-2}$) : (1 - $\dfrac{x}{x+2}$)

= ($\dfrac{x}{\left(x-2\right)\left(x+2\right)}$ + $\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}$ - $\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}$) : ($\dfrac{x+2}{x+2}$ - $\dfrac{x}{x+2}$)

$\dfrac{x+x-2-2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}$ : $\dfrac{x+2-x}{x+2}$

$\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}$ : $\dfrac{2}{x+2}$

$\dfrac{-6}{\left(x-2\right)\left(x+2\right)}$ . $\dfrac{x+2}{2}$

$\dfrac{-3}{x-2}$

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2 tháng 1 lúc 9:53

Ta có: $\left(\dfrac{3x}{1-2x}+\dfrac{2x}{2x+1}\right):\dfrac{2x^2+5x}{1-4x+4x^2}$

$=\left(\dfrac{3x\left(2x+1\right)}{\left(1-2x\right)\left(1+2x\right)}+\dfrac{2x\left(1-2x\right)}{\left(1-2x\right)\left(1+2x\right)}\right):\dfrac{2x^2+5x}{\left(2x-1\right)^2}$

$=\dfrac{6x^2+3x+2x-4x^2}{\left(1-2x\right)\left(1+2x\right)}:\dfrac{2x^2+5x}{\left(1-2x\right)^2}$

$=\dfrac{2x^2+5x}{\left(1-2x\right)\left(1+2x\right)}:\dfrac{2x^2+5x}{\left(1-2x\right)^2}$

$=\dfrac{x\left(2x+5\right)}{\left(1-2x\right)\left(1+2x\right)}\cdot\dfrac{\left(1-2x\right)^2}{x\left(2x+5\right)}$

$=\dfrac{1-2x}{1+2x}$

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23 tháng 12 2020 lúc 21:33

Ta có: $x^3-5x-1:x+2$

$=\dfrac{x^3+2x^2-2x^2-4x-x-2+3}{x+2}$

$=\dfrac{x^2\left(x+2\right)-2x\left(x+2\right)-\left(x+2\right)+3}{x+2}$

$=\dfrac{\left(x+2\right)\left(x^2-2x-1\right)}{x+2}+\dfrac{3}{x+2}$

$=x^2-2x-1+\dfrac{3}{x+2}$

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23 tháng 12 2020 lúc 21:37

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11 tháng 12 2020 lúc 18:27

$\dfrac{3-3x}{\left(1+x\right)^2}:\dfrac{6x^2-6}{x+1}$

$=\dfrac{3\left(1-x\right)}{\left(x+1\right)^2}:\dfrac{6\left(x^2-1\right)}{x+1}$

$=\dfrac{-3\left(x-1\right)}{\left(x+1\right)^2}:\dfrac{6\left(x+1\right)\left(x-1\right)}{x+1}$

$=\dfrac{-3\left(x-1\right)}{\left(x+1\right)^2}\cdot\dfrac{x+1}{6\left(x+1\right)\left(x-1\right)}$

$=\dfrac{-3\left(x-1\right)\left(x+1\right)}{6\left(x+1\right)^3\left(x-1\right)}=\dfrac{-3\left(x+1\right)}{6\left(x+1\right)\left(x+1\right)^2}=\dfrac{-3}{6\left(x+1\right)^2}=\dfrac{-1}{2\left(x+1\right)^2}$

b) Bạn có thể viết kiểu latex được không ạ ?

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28 tháng 11 2020 lúc 16:56

áp dụng định lý Bơzon

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28 tháng 11 2020 lúc 20:35

Nãy nhầm :))

$f\left(x\right)⋮\left(x-2\right)\left(x-3\right)$

$\Leftrightarrow\left\{{}\begin{matrix}f\left(2\right)=4+2a-b=0\\f\left(3\right)=9+3a-b=0\end{matrix}\right.$

$\Leftrightarrow\left\{{}\begin{matrix}2a-b=-4\\3a-b=-9\end{matrix}\right.$

$\Leftrightarrow\left\{{}\begin{matrix}a=-5\\2a-b=-4\end{matrix}\right.$

$\Leftrightarrow a=-5;b=-6$

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12 tháng 8 2020 lúc 10:39

Bài 1:

a)

b)

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