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1,\(\frac{3}{2x+6}-\frac{x-6}{x\left(2x+6\right)}\)
=\(\frac{3x}{x\left(2x+6\right)}+\frac{x-6}{x\left(2x+6\right)}\)
=\(\frac{3x+x-6}{x\left(2x+6\right)}\)=\(\frac{4x-6}{x\left(2x+6\right)}=\frac{2\left(2x-3\right)}{x\left(2x+6\right)}\)
mk ko biết làm
xin lỗi bn nhae
xin lỗi vì đã ko giúp được bn
chcus bn học gioi!
nhae@@@
\(\frac{5x+10}{4x-8}\cdot\frac{4-2x}{x+2}\)( ĐKXĐ : \(x\ne\pm2\))
\(=\frac{5\left(x+2\right)}{2\left(2x-4\right)}\cdot\frac{-\left(2x-4\right)}{x+2}\)
\(=\frac{-5\left(x+2\right)\left(2x-4\right)}{2\left(2x-4\right)\left(x+2\right)}\)
\(=-\frac{5}{2}\)
\(\frac{x^2-36}{2x+10}\cdot\frac{3}{6-x}\)( ĐKXĐ : \(x\ne-5;x\ne6\))
\(=\frac{\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)}\cdot\frac{3}{-\left(x-6\right)}\)
\(=\frac{3\left(x-6\right)\left(x+6\right)}{-2\left(x+5\right)\left(x-6\right)}\)
\(=\frac{3\left(x+6\right)}{-2\left(x+5\right)}=\frac{3x+18}{-2x-10}=-\frac{3x+18}{2x+10}\)
a)
Điều kiện : \(\hept{\begin{cases}4x-8\ne0\\x+2\ne0\end{cases}}\)
\(\hept{\begin{cases}x\ne2\\x\ne-2\end{cases}}\)
\(=\frac{5\left(x+2\right)}{-2\left(4-2x\right)}\cdot\frac{4-2x}{x+2}\)
\(=\frac{-5}{2}\)
b)
Điều kiện : \(\hept{\begin{cases}2x+10\ne0\\6-x\ne0\end{cases}}\)
\(\hept{\begin{cases}x\ne-5\\x\ne6\end{cases}}\)
\(=\frac{\left(x-6\right)\left(x+6\right)}{2x+10}\cdot\frac{3}{6-x}\)
\(=\frac{-6\left(x+6\right)\cdot3}{2x+10}\)
\(=\frac{-9\left(x+6\right)}{x+5}\)
\(=\frac{-9x-54}{x+5}\)
\(=\frac{-9\left(x+5\right)-9}{x+5}\)
\(=-9-\frac{9}{x+5}\)
a) \(\dfrac{3}{4xy}+\dfrac{5x}{2x^2z}+\dfrac{7}{6yz^2}\) (MSC: \(12x^2yz^2\))
\(=\dfrac{3\cdot3xz^2}{4xy\cdot3xz^2}+\dfrac{5x\cdot6yz}{2x^2z\cdot6yz}+\dfrac{7\cdot2x^2}{6yz^2\cdot2x^2}\)
\(=\dfrac{9xz^2}{12x^2yz^2}+\dfrac{30xyz}{12x^2yz^2}+\dfrac{14x^2}{12x^2yz^2}\)
\(=\dfrac{9xz^2+30xyz+14x^2}{12x^2yz^2}\)
\(=\dfrac{x\left(9z^2+30yz+14x\right)}{12x^2yz^2}\)
\(=\dfrac{9z^2+30yz+14x}{12x^2yz^2}\)
b) \(\dfrac{x^2}{x^2+3x}+\dfrac{3}{x+3}+\dfrac{3}{x}\)
\(=\dfrac{x^2}{x\left(x+3\right)}+\dfrac{3}{x+3}+\dfrac{3}{x}\)
\(=\dfrac{x}{x+3}+\dfrac{3}{x+3}+\dfrac{3}{x}\)
\(=\dfrac{x+3}{x+3}+\dfrac{3}{x}\)
\(=1+\dfrac{3}{x}\)
\(=\dfrac{x}{x}+\dfrac{3}{x}\)
\(=\dfrac{x+3}{x}\)
a: \(=\dfrac{3\cdot3\cdot xz^2+5x\cdot6\cdot y+7\cdot x^2\cdot2}{12x^2yz^2}=\dfrac{9xz^2+30xy+14x^2}{12x^2yz^2}\)
\(=\dfrac{9z^2+30y+14x}{12xyz^2}\)
b: \(=\dfrac{x}{x+3}+\dfrac{3}{x+3}+\dfrac{3}{x}=1+\dfrac{3}{x}=\dfrac{x+3}{x}\)
a) \(\frac{3x+5}{2\left(x-1\right)}+\frac{4}{x-2}=\frac{\left(3x+5\right)\left(x-2\right)+4\cdot2\left(x-1\right)}{2\left(x-1\right)\left(x-2\right)}=\frac{3x^2-6x+5x-10+8x-8}{2\left(x-1\right)\left(x-2\right)}\)
\(=\frac{3x^2+7x-18}{2\left(x-1\right)\left(x-2\right)}\)
b) \(\frac{2x^2+1}{4x^2-2x}+\frac{3-3x}{1-2x}+\frac{3}{2x}=\frac{2x^2+1+4x\left(3-3x\right)+2\cdot3\left(1-2x\right)}{4x\left(1-2x\right)}=\frac{2x^2+1+12-12x+6-12x}{4x\left(1-2x\right)}\)\(=\frac{2x^2-24x+19}{4x\left(1-2x\right)}\)
Đề này... bạn xem lại đi. Chứ thế này thì dùng máy tính cũng không làm nổi T-T
a) \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x-6}{9x^2-4}\)
\(=\dfrac{3x+2-3x+2-3x+6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{-3x+10}{\left(3x-2\right)\left(3x+2\right)}\)
b) \(\dfrac{x+25}{2x^2-50}-\dfrac{x+5}{x^2-5x}-\dfrac{5-x}{2x^2+10x}\)
\(=\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}-\dfrac{x+5}{x\left(x-5\right)}+\dfrac{x-5}{2x\left(x+5\right)}\)
\(=\dfrac{x^2+25x-2\left(x+5\right)^2+\left(x-5\right)^2}{2x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{x^2+25x-2x^2-20x-50+x^2-10x+25}{2x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-5x-25}{2x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-5\left(x+5\right)}{2x\left(x-5\right)\left(x+5\right)}=\dfrac{-5}{2x\left(x-5\right)}\)
c) Ta có: \(\dfrac{1-2x}{2x}-\dfrac{4x}{2x-1}-\dfrac{3}{2x-4x^2}\)
\(=\dfrac{-\left(2x-1\right)^2-8x^2+3}{2x\left(2x-1\right)}\)
\(=\dfrac{-\left(4x^2-4x+1\right)-8x^2+3}{2x\left(2x-1\right)}\)
\(=\dfrac{-4x^2+4x-1-8x^2+3}{2x\left(2x-1\right)}\)
\(=\dfrac{-12x^2+4x+2}{2x\left(2x-1\right)}\)
a: \(\dfrac{4-x^2}{x-3}+\dfrac{2x-2x^2}{3-x}+\dfrac{5-4x}{x-3}\)
\(=\dfrac{4-x^2-2x+2x^2+5-4x}{x-3}=\dfrac{x^2-6x+9}{x-3}\)
=(x-3)^2/(x-3)
=x-3
b: \(\dfrac{2}{x+2}+\dfrac{-4}{2-x}+\dfrac{5x+2}{4-x^2}\)
\(=\dfrac{2}{x+2}-\dfrac{4}{x-2}-\dfrac{5x+2}{x^2-4}\)
\(=\dfrac{2x-4-4x-8-5x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{-7x-14}{\left(x-2\right)\left(x+2\right)}\)
=-7(x+2)/(x-2)(x+2)
=-7/(x-2)
\(\frac{5x-5}{2x+2}:\frac{x^2-x}{2x^2+4x+2}\)
\(=\frac{5\left(x-1\right)}{2\left(x+1\right)}.\frac{2\left(x+1\right)^2}{x\left(x-1\right)}\)
\(=\frac{5\left(x+1\right)}{x}\)