\(\dfrac{6\left(4-4x+x^2\right)}{x^3y}:\dfrac{x-2}{x^3y^2}\)

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21 tháng 4 2017

Giải bài 42 trang 54 Toán 8 Tập 1 | Giải bài tập Toán 8

10 tháng 2 2018

d. ĐKXĐ: x khác 1, x khác 3

\(\dfrac{x+5}{x-1}=\dfrac{x+1}{\left(x-3\right)}-\dfrac{8}{x^2-4x+3}\)

\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+5\right)}{\left(x-1\right)\left(x-3\right)}=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x-3\right)}-\dfrac{8}{\left(x-1\right)\left(x-3\right)}\) \(\Leftrightarrow x^2+2x-15=x^2-1-8\)

\(\Leftrightarrow2x-15+1+8=0\)

\(\Leftrightarrow2x-6=0\)

\(\Leftrightarrow x=3\) (loại)

Vậy pt vô nghiệm

14 tháng 7 2017

a) \(\left(2x^2-3y^2\right)^2=\left(2x^2\right)^2-2.2x^2.3y^2+\left(3y^2\right)^2\)

\(=4x^4-12x^2y^2+9y^4\)

b) \(\left(4x-3y\right)^2-\left(4x+3y\right)^2\)

\(=\left(4x-3y-4x-3y\right)\left(4x-3y+4x+3y\right)\)

\(=-6y.8x\)

\(=-48xy\)

c) \(\left(x+\dfrac{1}{4}\right)^2=x^2+2x.\dfrac{1}{4}+\left(\dfrac{1}{4}\right)^2\)

\(=x^2+\dfrac{1}{2}x+\dfrac{1}{16}\)

d) \(\left(5x-y\right)^2\)

\(=25x^2-10xy+y^2\).

6 tháng 10 2017

a)\(\dfrac{32x^5\left(3y-7\right)^5}{-4x\left(7-3y\right)^4}=\dfrac{-4x.\left(-8x^4\right)\left(3y-7\right)^4\left(3y-7\right)}{-4x\left(3y-7\right)^4}\)

\(=\dfrac{\left(-8x^4\right)\left(3y-7\right)}{1}=\left(-8x^4\right)\left(3y-7\right)\)

\(=-32x^4y+56x^4\)

b) \(\dfrac{12x^3\left(3x-5\right)^2}{4x\left(3x-5\right)^2}-\dfrac{2x\left(x+7\right)}{\left(x+7\right)^3}=\dfrac{12x^3}{4x}-\dfrac{2x}{\left(x+7\right)^2}\)

\(=3x^2-\dfrac{2x}{\left(x+7\right)^2}\)

\(\)

6 tháng 10 2017

bạn ơi bạn ghi nhầm đề bài câu b r

28 tháng 6 2017

Phép trừ các phân thức đại số

27 tháng 11 2017

1) \(\dfrac{x^2}{x+1}+\dfrac{2x}{x^2-1}-\dfrac{1}{1-x}+1\)

\(=\dfrac{x^2}{x+1}+\dfrac{2x}{x^2-1}+\dfrac{1}{x-1}+1\)

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

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

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

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

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

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

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

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

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

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

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

\(=\dfrac{x}{x\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{1}{\left(x-1\right)\left(x+1\right)}\)

7 tháng 12 2018

\(\dfrac{\left(x+y\right)2}{x2+xy}+\dfrac{\left(x-y\right)2}{x2-xy}=-\left(\dfrac{\left(x-y\right)2}{x2-xy}\right)+\dfrac{\left(x-y\right)2}{x2-xy}=0\)

24 tháng 11 2022

b: \(\dfrac{x^2-4x}{xy-4x-3y+12}+\dfrac{x-2}{y-4}\)

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

\(=\dfrac{x^2-4x+x^2-5x+6}{\left(y-4\right)\left(x-3\right)}=\dfrac{2x^2-9x+6}{\left(y-4\right)\left(x-3\right)}\)

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

\(=\dfrac{y^2+2y-10}{\left(y-5\right)\left(x+1\right)}\)

30 tháng 9 2018

a ) \(\left(\dfrac{20x}{3y^2}\right):\left(\dfrac{4x^3}{5y}\right)=\dfrac{20x}{3y^2}.\dfrac{5y}{4x^3}=\dfrac{100xy}{12x^3y^2}=\dfrac{25}{3x^2y}\)

b ) Đ/k : \(x\ne-4\)

Ta có : \(\dfrac{4x+12}{\left(x+4\right)^2}:\dfrac{3\left(x+3\right)}{x+4}\)

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

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

\(=\dfrac{4}{3\left(x+4\right)}\)

\(=\dfrac{4}{3x+12}\)