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a) điều kiện xác định : \(x\ne0\)
ta có : \(A=\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{3}{x\left(x^4+x^2+1\right)}\)
\(\Leftrightarrow A=\dfrac{\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(x^2-x+1\right)}=\dfrac{3}{x\left(x^4+x^2+1\right)}\)
\(\Leftrightarrow A=\dfrac{x^3-x^2+x+x^2-x+1-\left(x^3+x^2+x-x^2-x-1\right)}{x^4-x^3+x^2+x^3-x^2+x+x^2-x+1}=\dfrac{3}{x\left(x^4+x^2+1\right)}\) \(\Leftrightarrow A=\dfrac{x^3-x^2+x+x^2-x+1-x^3-x^2-x+x^2+x+1}{x^4+x^2+1}=\dfrac{3}{x\left(x^4+x^2+1\right)}\) \(\Leftrightarrow A=\dfrac{2}{x^4+x^2+1}=\dfrac{3}{x\left(x^4+x^2+1\right)}\Leftrightarrow\left(x^4+x^2+1\right)A=2=\dfrac{3}{x}\) \(\Leftrightarrow2x=3\Leftrightarrow x=\dfrac{3}{2}\left(tmđk\right)\) vậy \(x=\dfrac{3}{2}\)
b) điều kiện : \(x\notin\left\{-4;-5;-6;-7\right\}\)
\(B=\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\)
\(\Leftrightarrow B=\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow B=\dfrac{\left(x+6\right)\left(x+7\right)+\left(x+4\right)\left(x+7\right)+\left(x+4\right)\left(x+5\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow B=\dfrac{x^2+13x+42+x^2+11x+28+x^2+9x+20}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow B=\dfrac{3x^2+33x+90}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\) \(\Leftrightarrow B=\dfrac{3\left(x+5\right)\left(x+6\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\) \(\Leftrightarrow B=\dfrac{3}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\) \(\Leftrightarrow54=\left(x+4\right)\left(x+7\right)\)\(\Leftrightarrow54=x^2+11x+28\Leftrightarrow x^2+11x+28-54=0\)
\(\Leftrightarrow x^2+11x-26=0\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+13=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\left(tmđk\right)\)
vậy \(x=2;x=-13\)
a: \(=\dfrac{4xy+x^2-2xy+y^2}{2\left(x+y\right)\left(x-y\right)}\cdot\dfrac{2x}{x+y}-\dfrac{y}{x-y}\)
\(=\dfrac{x}{x-y}-\dfrac{y}{x-y}=1\)
b: \(=\dfrac{x^2+x-2-2x^2-2x}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{3\left(x+1\right)}{x}+\dfrac{4x^2+x+7}{x\left(x-1\right)}\)
\(=\dfrac{-x^2-x-2}{\left(x-1\right)}\cdot\dfrac{3}{x}+\dfrac{4x^2+x+7}{x\left(x-1\right)}\)
\(=\dfrac{4x^2+x+7-3x^2-3x-6}{x\left(x-1\right)}=\dfrac{x^2-2x+1}{x\left(x-1\right)}=\dfrac{x-1}{x}\)
c: \(=\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}\)
\(=\dfrac{x+7-x-4}{\left(x+7\right)\left(x+4\right)}=\dfrac{3}{\left(x+4\right)\left(x+7\right)}\)
a) Ta có: \(\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\left(\dfrac{1}{x\left(x+1\right)}+\dfrac{x+2}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\dfrac{x^2+2x+1}{x\left(x+1\right)}:\dfrac{x^2-2x+1}{x}\)
\(=\dfrac{\left(x+1\right)^2}{x\left(x+1\right)}\cdot\dfrac{x}{\left(x-1\right)^2}\)
\(=\dfrac{x+1}{\left(x-1\right)^2}\)
b) Ta có: \(\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{1-6x+9x^2}\)
\(=\dfrac{3x\left(3x+1\right)+2x\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}:\dfrac{2x\left(3x+5\right)}{\left(1-3x\right)^2}\)
\(=\dfrac{9x^2+3x+2x-6x^2}{\left(1-3x\right)\left(1+3x\right)}:\dfrac{2x\left(3x+5\right)}{\left(1-3x\right)^2}\)
\(=\dfrac{3x^2+5x}{\left(1-3x\right)\left(1+3x\right)}\cdot\dfrac{\left(1-3x\right)^2}{2x\left(3x+5\right)}\)
\(=\dfrac{x\left(3x+5\right)}{1+3x}\cdot\dfrac{1-3x}{2x\left(3x+5\right)}\)
\(=\dfrac{2\left(1-3x\right)}{3x+1}\)
c) Ta có: \(\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)
\(=\left(\dfrac{9}{x\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x\left(x+3\right)}-\dfrac{x}{3\left(x+3\right)}\right)\)
\(=\dfrac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\dfrac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)
\(=\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}\cdot\dfrac{3x\left(x+3\right)}{3x-9-x^2}\)
\(=\dfrac{x^2-3x+9}{x-3}\cdot\dfrac{3}{-\left(x^2-3x+9\right)}\)
\(=\dfrac{-3}{x-3}\)
$ĐKXĐ:x \neq -4;-5;-6;-7$
$pt⇔\dfrac{1}{x^2+4x+5x+20}+\dfrac{1}{x^2+5x+6x+30}+\dfrac{1}{x^2+6x+7x+42}=\dfrac{1}{18}$
$⇔\dfrac{1}{(x+4)(x+5)}+\dfrac{1}{(x+5)(x+6)}+\dfrac{1}{(x+6)(x+7)}=\dfrac{1}{18}$
$⇔\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}$
$⇔\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}$
$⇔\dfrac{3}{(x+4)(x+7)}=\dfrac{1}{18}$
$⇔x^2+11x+28=54$
$⇔x^2+11x-26=0$
$⇔x^2-2x+13x-26=0$
$⇔(x-2)(x+13)=0$
$⇔$ \(\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)(t/m)
Vậy phương trình đã cho có tập nghiệm $S=(2;-13)$
Ta có:
\(x^2+9x+2x=\left(x+4\right)\left(x+5\right)\)
\(x^2+11x+30=\left(x+6\right)\left(x+5\right)\)
\(x^2+13x+42=\left(x+6\right)\left(x+7\right)\)
ĐK: \(\left\{{}\begin{matrix}x\ne-4\\x\ne-5\\x\ne-6\\x\ne-7\end{matrix}\right.\)
pt \(\Leftrightarrow\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{18\left(x+7\right)}{18\left(x+4\right)\left(x+7\right)}-\dfrac{18\left(x+4\right)}{18\left(x+4\right)\left(x+7\right)}=\dfrac{\left(x+4\right)\left(x+7\right)}{18\left(x+4\right)\left(x+7\right)}\)
\(\Rightarrow18\left(x+7\right)-18\left(x+4\right)=\left(x+4\right)\left(x+7\right)\)
\(\Leftrightarrow\left(x+13\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+13=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-13\\x=2\end{matrix}\right.\) (tm)
b) \(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x^2+4x+5x+20}+\dfrac{1}{x^2+5x+6x+30}+\dfrac{1}{x^2+6x+7x+42}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x\left(x+4\right)+5\left(x+4\right)}+\dfrac{1}{x\left(x+5\right)+6\left(x+5\right)}+\dfrac{1}{x\left(x+6\right)+7\left(x+6\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{x+7}{\left(x+4\right)\left(x+7\right)}-\dfrac{x+4}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{3}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow\left(x+4\right)\left(x+7\right)=54\)
\(\Leftrightarrow x^2+11x+28-54=0\)
\(\Leftrightarrow x^2-2x+13x-26=0\)
\(\Leftrightarrow x\left(x-2\right)+13\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)
\(\Leftrightarrow\) x - 2 = 0 hoặc x + 13 = 0
\(\Leftrightarrow\) x = 2 hoặc x = -13
Vậy x = 2 hoặc x = -13.
a) 1/x(x + 1) + 1/(x + 1)(x + 2) + 1/(x + 2)(x + 3) + 1/(x + 3)(x + 4)
( 1/x - 1/x+1) + (1/x+1 - 1/x+2) + (1/x+2 - 1/ x+3) + 1/(x+3 - 1/x+4)
(1/x +1/x+4) - ( 1/x+2 - 1/x+2) - ( 1/x+3 - 1/x+3)
1/x +1/x+4
2x+4/x(x+4)
d: ĐKXĐ: x<>-4; x<>-5; x<>-6; x<>-7
\(PT\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
=>\(\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
=>\(\dfrac{x+7-x-4}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)
=>x^2+11x+28=54
=>x^2+11x-26=0
=>(x+13)(x-2)=0
=>x=2 hoặc x=-13
e: \(\dfrac{x-241}{17}+\dfrac{x-220}{19}+\dfrac{x-195}{21}+\dfrac{x-166}{23}=10\)
\(\Leftrightarrow\left(\dfrac{x-241}{17}-1\right)+\left(\dfrac{x-220}{19}-2\right)+\left(\dfrac{x-195}{21}-3\right)+\left(\dfrac{x-166}{23}-4\right)=0\)
=>x-258=0
=>x=258
a: \(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
=>\(\dfrac{x+7-x-4}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)
=>(x+4)(x+7)=54
=>x^2+11x+28-54=0
=>(x+13)(x-2)=0
=>x=-13 hoặc x=2
b: \(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-...+\dfrac{1}{x+4}-\dfrac{1}{x+5}=\dfrac{1}{3}\)
=>\(\dfrac{x+5-x-1}{\left(x+5\right)\left(x+1\right)}=\dfrac{1}{3}\)
=>x^2+6x+5=12
=>x^2+6x-7=0
=>(x+7)(x-1)=0
=>x=-7 hoặc x=1
Bài 58:
a, \(\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{x+1-x}{x\left(x+1\right)}=\dfrac{1}{x\left(x+1\right)}\)
b, \(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{n\left(n+1\right)}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\)
\(=1-\dfrac{1}{n+1}=\dfrac{n+1-1}{n+1}=\dfrac{n}{n+1}\)
Vậy...
\(B=\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}\)
\(=\dfrac{1}{x^2+4x+5x+20}+\dfrac{1}{x^2+5x+6x+30}+\dfrac{1}{x^2+6x+7x+42}\)
\(=\dfrac{1}{x\left(x+4\right)+5\left(x+4\right)}+\dfrac{1}{x\left(x+5\right)+6\left(x+5\right)}+\dfrac{1}{x\left(x+7\right)+6\left(x+7\right)}\)
\(=\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}\)
\(=\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}\)
\(=\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{x+7-x-4}{\left(x+4\right)\left(x+7\right)}\)
\(=\dfrac{3}{x^2+11x+28}\)
Vậy...
58,
\(\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{x+1}{x\left(x+1\right)}-\dfrac{x}{x\left(x+1\right)}=\dfrac{x+1-x}{x\left(x+1\right)}=\dfrac{1}{x\left(x+1\right)}\)b,
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n+1\right)}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\)\(=1-\dfrac{1}{n\left(n+1\right)}=\dfrac{n^2+n-1}{n\left(n+1\right)}\)
\(B=\dfrac{1}{\left(x^2+9x+20\right)}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}\)\(=\dfrac{1}{\left(x^2+4x\right)+\left(5x+20\right)}+\dfrac{1}{\left(x^2+5x\right)+\left(6x+30\right)}+\dfrac{1}{\left(x^2+6x\right)+\left(7x+42\right)}\)\(=\dfrac{1}{x\left(x+4\right)+5\left(x+4\right)}+\dfrac{1}{x\left(x+5\right)+6\left(x+5\right)}+\dfrac{1}{x\left(x+6\right)+7\left(x+6\right)}\)\(=\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}\)\(=\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x-7}\)\(=\dfrac{1}{x+4}-\dfrac{1}{x+7}\)