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a: \(=-\dfrac{1}{x\left(x-1\right)}+\dfrac{-1}{\left(x-1\right)\left(x-2\right)}+\dfrac{-1}{\left(x-2\right)\left(x-3\right)}+...+-\dfrac{1}{\left(x-4\right)\left(x-5\right)}+\dfrac{1}{x-5}\)
\(=\dfrac{1}{x}-\dfrac{1}{x-1}+\dfrac{1}{x-1}-\dfrac{1}{x-2}+\dfrac{1}{x-2}-\dfrac{1}{x-3}+...+\dfrac{1}{x-4}-\dfrac{1}{x-5}+\dfrac{1}{x-5}\)
=1/x
b: \(=\dfrac{1}{x}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+9}+\dfrac{1}{x+9}\)
=1/x
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)
a ) \(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+4}\)
\(=\dfrac{4}{x\left(x+4\right)}\)
b ) \(\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}\)
\(=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}\)
\(=\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}\)
\(=\dfrac{1}{x+1}-\dfrac{1}{x+5}\)
\(=\dfrac{4}{\left(x+1\right)\left(x+5\right)}\)
:D
a) \(x^3-\dfrac{1}{4}x=0\)
⇔ \(x.\left(x^2-\dfrac{1}{4}\right)=0\)
⇔ \(x\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)=0\)
⇔ x = 0 hoặc \(x=\dfrac{1}{2}\) hoặc \(x=\dfrac{-1}{2}\)
b) (2x - 1)2 - (x + 3)2 = 0
⇔ (2x - 1 - x - 3)(2x - 1 + x + 3) = 0
⇔ (x - 4)(3x +2) = 0
⇔ x = 4 hoặc \(x=\dfrac{-2}{3}\)
c) 2x2 - x - 6 = 0
⇔ 2x2 - 4x + 3x - 6 = 0
⇔ 2x(x - 2) + 3(x - 2) = 0
⇔ (x - 2) (2x + 3) = 0
⇔ x = 2 hoặc \(x=\dfrac{-3}{2}\)
2)a.
\(B=\left(\dfrac{x}{x^2-36}-\dfrac{x-6}{x^2+6x}\right):\dfrac{2x-6}{x^2+6x}\\ =\left(\dfrac{x\left(x^2+6x\right)-\left(x-6\right)\left(x^2-36\right)}{\left(x^2-36\right)\left(x^2+6x\right)}\right).\dfrac{x^2+6x}{2x-6}\\ =\dfrac{x^2\left(x+6\right)-\left(x-6\right)^2.\left(x+6\right)}{x^2-36}.\dfrac{1}{2x-6}\\ =\dfrac{\left(x+6\right)\left(x^2-\left(x-6\right)^2\right)}{x^2-36}.\dfrac{1}{2x-6}\\ =\dfrac{\left(x-x+6\right)\left(x+x-6\right)}{x-6}.\dfrac{1}{2x-6}\\ =\dfrac{6.\left(2x-6\right)}{x-6}.\dfrac{1}{2x-6}\\ =\dfrac{6}{x-6}\)
b)
\(x=2\Leftrightarrow B=\dfrac{6}{x-6}=\dfrac{6}{2-6}=\dfrac{6}{-4}=-\dfrac{3}{2}\)
a: \(B=\dfrac{21+x^2-x-12-x^2+4x-3}{\left(x+3\right)\left(x-3\right)}:\dfrac{x+3-1}{x+3}\)
\(=\dfrac{3x+6}{\left(x+3\right)\left(x-3\right)}\cdot\dfrac{x+3}{x+2}\)
\(=\dfrac{3}{x-3}\)
b: |2x+1|=5
=>2x+1=5 hoặc 2x+1=-5
=>x=-3(loại) hoặc x=2(nhận)
Khi x=2 thì \(B=\dfrac{3}{2-3}=-3\)
c: Để B=-3/5 thì x-3=-5
=>x=-2(loại)
d: Để B<0 thì x-3<0
=>x<3
1) điều kiện xác định : \(x\notin\left\{-1;-2;-3;-4\right\}\)
ta có : \(\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\) \(\Leftrightarrow\dfrac{\left(x+3\right)\left(x+4\right)+\left(x+1\right)\left(x+4\right)+\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)\(\Leftrightarrow\dfrac{x^2+7x+12+x^2+5x+4+x^2+3x+2}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{3x^2+15x+18}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)
\(\Leftrightarrow6\left(3x^2+15x+18\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18\left(x^2+5x+6\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18\left(x+2\right)\left(x+3\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18=\left(x+1\right)\left(x+4\right)\) ( vì điều kiện xác định )
\(\Leftrightarrow18=x^2+5x+4\Leftrightarrow x^2+5x-14=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+7=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-7\end{matrix}\right.\left(tmđk\right)\)
vậy \(x=2\) hoặc \(x=-7\) mấy câu kia lm tương tự nha bn
Ta có : \(P=\dfrac{1}{x^2+x}+\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}\)
=\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}\)
=\(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-...+\dfrac{1}{x+4}-\dfrac{1}{x+5}\)
= \(\dfrac{1}{x}-\dfrac{1}{x+5}=\dfrac{5}{x\left(x+5\right)}\)
a, Với x=\(\dfrac{\sqrt{29}-5}{2}\Rightarrow A=\dfrac{5}{\dfrac{\sqrt{29}-5}{2}\left(\dfrac{\sqrt{29}-5}{2}+5\right)}\)
Mấy cái còn lại tương tự , bạn tự làm nha