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a) \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(b-c\right)\left(c-a\right)}+\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)
\(=\dfrac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
b) \(\dfrac{\left(a^2-\left(b+c\right)^2\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(a^2+c^2-2ac-b^2\right)}\)
\(=\dfrac{\left(a-b-c\right)\left(a+b+c\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(\left(a-c\right)^2-b^2\right)}\)
\(=\dfrac{\left(a-c-b\right)\left(a-c+b\right)}{\left(a-c-b\right)\left(a-c+b\right)}=1\)
c) \(\dfrac{x-1}{x^3}-\dfrac{x+1}{x^3-x^2}+\dfrac{3}{x^3-2x^2+x}\)
\(=\dfrac{x-1}{x^3}-\dfrac{x+1}{x^2\left(x-1\right)}+\dfrac{3}{x\left(x-1\right)^2}\)
\(=\dfrac{\left(x-1\right)^3-x\left(x+1\right)\left(x-1\right)+3x^2}{x^3\left(x-1\right)^2}\)
\(=\dfrac{x^3-3x^2+3x-1-x^3+x+3x^2}{x^3\left(x-1\right)^2}\)
\(=\dfrac{4x-1}{x^3\left(x-1\right)^2}\)
d) \(\left(\dfrac{x^2-y^2}{xy}-\dfrac{1}{x+y}\left(\dfrac{x^2}{y}-\dfrac{y^2}{x}\right)\right):\dfrac{x-y}{x}\)
\(=\left(\dfrac{\left(x-y\right)\left(x+y\right)}{xy}-\dfrac{1}{x+y}.\dfrac{x^3-y^3}{xy}\right):\dfrac{x-y}{x}\)
\(=\left(\dfrac{\left(x-y\right)\left(x+y\right)}{xy}-\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{xy\left(x+y\right)}\right):\dfrac{x-y}{x}\)
\(=\dfrac{\left(x-y\right)\left(x^2+2xy+y^2-x^2-xy-y^2\right)}{xy\left(x+y\right)}.\dfrac{x}{x-y}\)
\(=\dfrac{x}{x+y}\)
a: \(=\dfrac{1}{x-y}-\dfrac{3xy}{\left(x-y\right)\left(x^2+xy+y^2\right)}+\dfrac{x-y}{x^2+xy+y^2}\)
\(=\dfrac{x^2+xy+y^2-3xy+x^2-2xy+y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{2x^2-4xy+2y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}=\dfrac{2\left(x-y\right)}{x^2+xy+y^2}\)
d: \(=\dfrac{x^3-1}{x-1}-\dfrac{x^2-1}{x+1}\)
\(=x^2+x+1-x+1=x^2+2\)
a) \(\left(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}\right):\dfrac{4x}{10x-5}\)
\(=\left(\dfrac{\left(2x+1\right)\left(2x+1\right)}{2x^2-1}-\dfrac{\left(2x-1\right)\left(2x-1\right)}{2x^2-1}\right):\dfrac{4x}{10x-5}\)
\(=\left(\dfrac{\left(2x+1\right)^2-\left(2x-1\right)^2}{2x^2-1}\right):\dfrac{4x}{10x-5}\)
\(=\left(\dfrac{\left(2x+1-2x-1\right)\left(2x+1+2x-1\right)}{2x^2-1}\right):\dfrac{4x}{10x-5}\)
\(=\dfrac{4x}{2x^2-1}.\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{5}{2x+1}\)
b) \(\left(\dfrac{1}{x^2+1}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\left(\dfrac{1}{x^2+1}-\dfrac{x\left(2-x\right)}{x\left(x+1\right)}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\left(\dfrac{1-2x+x^2}{x^2+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\left(\dfrac{1-2x+x^2}{x^2+1}\right):\left(\dfrac{1}{x}+\dfrac{x^2}{x}-\dfrac{2x}{x}\right)\)
\(=\left(\dfrac{1-2x+x^2}{x^2+1}\right):\left(\dfrac{x^2-2x+1}{x}\right)\)
\(=\dfrac{\left(x-1\right)^2}{x^2+1}.\dfrac{x}{\left(x-1\right)^2}\)
\(=\dfrac{x}{x^2+1}\)
c) d) Tự làm đi mình làm biếng quass >.< ^^
a: \(=\dfrac{4x^2+4x+1-\left(4x^2-4x+1\right)}{\left(2x-1\right)\left(2x+1\right)}\cdot\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{8x}{2x+1}\cdot\dfrac{5}{4x}=\dfrac{10}{2x+1}\)
c: \(=\dfrac{1}{x-1}-\dfrac{x\left(x-1\right)\left(x+1\right)}{x^2+1}\cdot\left(\dfrac{x+1-x+1}{\left(x-1\right)^2\cdot\left(x+1\right)}\right)\)
\(=\dfrac{1}{x-1}-\dfrac{x}{x^2+1}\cdot\dfrac{2}{\left(x-1\right)}=\dfrac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}=\dfrac{x-1}{x^2+1}\)
Bài 1:
\(a,=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+2y^2}{2\left(x-y\right)\left(x+y\right)}=\dfrac{2y\left(x+y\right)}{2\left(x-y\right)\left(x+y\right)}=\dfrac{y}{x-y}\\ b,Sửa:\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\\ =\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}:\dfrac{3x-9-x^2}{3x\left(x+3\right)}=\dfrac{x^2+3x+9}{x\left(x-3\right)\left(x+3\right)}\cdot\dfrac{-3x\left(x+3\right)}{x^2-3x+9}\\ =\dfrac{-3}{x-3}\)
Bài 2:
\(a,\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\\ b,\Leftrightarrow x^3+x^2+x+a=\left(x+1\right)\cdot a\left(x\right)\\ \text{Thay }x=-1\Leftrightarrow-1+1-1+a=0\Leftrightarrow a=1\)
1: \(MTC=2\left(x-y\right)\left(x+y\right)\)
\(\dfrac{x-y}{2x^2-4xy+2y^2}=\dfrac{x-y}{2\left(x-y\right)^2}=\dfrac{1}{2\left(x-y\right)}=\dfrac{1\cdot\left(x+y\right)}{2\left(x-y\right)\left(x+y\right)}=\dfrac{x+y}{2\left(x-y\right)\left(x+y\right)}\)
\(\dfrac{x+y}{2x^2+4xy+2y^2}\)
\(=\dfrac{x+y}{2\left(x^2+2xy+y^2\right)}\)
\(=\dfrac{x+y}{2\left(x+y\right)^2}=\dfrac{1}{2\left(x+y\right)}=\dfrac{x-y}{2\left(x+y\right)\left(x-y\right)}\)
\(\dfrac{1}{x^2-y^2}=\dfrac{2}{2\left(x^2-y^2\right)}=\dfrac{2}{2\left(x-y\right)\left(x+y\right)}\)
2: \(\dfrac{1}{x^2+8x+15}=\dfrac{1}{\left(x+3\right)\left(x+5\right)}=\dfrac{x+3}{\left(x+3\right)^2\cdot\left(x+5\right)}\)
\(\dfrac{1}{x^2+6x+9}=\dfrac{1}{\left(x+3\right)^2}=\dfrac{x+5}{\left(x+3\right)^2\cdot\left(x+5\right)}\)
3: \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}=\dfrac{1\cdot\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\dfrac{a-c}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(\dfrac{1}{\left(c-b\right)\left(c-a\right)}=\dfrac{1}{\left(b-c\right)\left(a-c\right)}=\dfrac{a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(\dfrac{1}{\left(b-a\right)\left(a-c\right)}=\dfrac{-1}{\left(a-b\right)\left(a-c\right)}=\dfrac{-\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
Giúp mình nhé mọi người !
\(1.\)
\(a.\)
\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{1\left(x-1\right)\left(x^2+3\right)}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{x^3-x^2+3x-3}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^3+x^2+3x+3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^2\left(x+1\right)+3\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{\left(x^2+3\right)\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=x-1\)
\(b.\)
\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)
\(=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2\left(x^2-y^2\right)}-\dfrac{\left(x-y\right)^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2}{2\left(x^2-y^2\right)}-\dfrac{x^2-2xy+y^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4xy+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4y\left(x+y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2y}{\left(x-y\right)}\)
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