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a/ \(\dfrac{1}{x^2+x}+\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}\)
\(=\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+4}\)
Vậy..
b/ \(\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}\)
\(=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}\)
\(=\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}\)
\(=\dfrac{1}{x+1}-\dfrac{1}{x+5}\)
Vậy..
ĐK: \(3x\ne\pm y;x\ne0\)
A = \(\dfrac{3x}{3x+y}-\dfrac{x}{3x-y}+\dfrac{2x}{\left(3x-y\right)\left(3x+y\right)}\)
= \(\dfrac{3x\left(3x-y\right)-x\left(3x+y\right)+2x}{\left(3x-y\right)\left(3x+y\right)}=\dfrac{6x^2-4xy+2x}{\left(3x-y\right)\left(3x+y\right)}=\dfrac{2x\left(3x-2y+1\right)}{\left(3x-y\right)\left(3x+y\right)}\)
Thay x = 1; y=2, ta có:
A = \(\dfrac{2.1\left(3.1-2.2+1\right)}{\left(3.1-2\right)\left(3.1+2\right)}=0\)
\(M=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{x+5}\)\(\Rightarrow M=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{x+4}{\left(x+4\right)\left(x+5\right)}\)\(\Rightarrow M=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{x+4}\)\(\Rightarrow M=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{x+3}\)
\(\Rightarrow M=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{x+2}\)
\(\Rightarrow M=\dfrac{1}{x+1}\)
M= \(\dfrac{1}{(x+1)(x+2)}\)+ \(\dfrac{1}{(x+2)(x+3)}\)+ \(\dfrac{1}{(x+3)(x+4)}\)+ \(\dfrac{1}{(x+4)(x+5)}\)+ \(\dfrac{1}{x+5}\)
M= \(\dfrac{1}{x+1}\)- \(\dfrac{1}{x+2}\)+ \(\dfrac{1}{x+2}\)- \(\dfrac{1}{x+3}\)+ \(\dfrac{1}{x+3}\)- \(\dfrac{1}{x+4}\)+ \(\dfrac{1}{x+4}\)+ \(\dfrac{1}{x+5}\)
M= \(\dfrac{1}{x+1} + \dfrac{1}{x+5}\)
M= \(\dfrac{x+5}{(x+1)(x+5)} + \dfrac{x+1}{(x+1)(x+5)} \)
M= \(\dfrac{x+5+x+1}{(x+1)(x+5)}\)
M= \(\dfrac{2x+6}{(x+1)(x+5)}\)
M= \(\dfrac{2(x+3)}{(x+1)(x+5)}\)
a, Với x khác 1
\(A=\dfrac{x^2+x+1-3x^2+2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1-x}{\left(x-1\right)\left(x^2+x+1\right)}=-\dfrac{1}{x^2+x+1}\)
b, Ta có \(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\Rightarrow\dfrac{-1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}< 0\)
Vậy với x khác 1 thì bth A luôn nhận gtri âm
\(=\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+4}-\dfrac{1}{x+5}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+5}=\dfrac{x+5-x}{x\left(x+5\right)}=\dfrac{5}{x\left(x+5\right)}\)
\(B=\dfrac{1}{x\left(x+1\right)}+\)\(\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}\)
=>B=\(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}\)
=>B=1/x - 1/x+5=5/x(x+5)
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
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
a, ĐKXĐ: x≠±3
A=\(\left(\dfrac{3-x}{x+3}.\dfrac{x^2+6x+9}{x^2-9}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)
A=\(\left(\dfrac{3-x}{x+3}.\dfrac{\left(x+3\right)^2}{\left(x+3\right)\left(x-3\right)}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)
A=\(\left(\dfrac{3-x}{x-3}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)
A=\(\left(\dfrac{9-x^2}{x^2-9}+\dfrac{x^2-3x}{x^2-9}\right):\dfrac{3x^2}{x+3}\)
A=\(\left(\dfrac{-3}{x+3}\right):\dfrac{3x^2}{x+3}\)
A=\(\dfrac{-1}{x^2}\)
b, Thay x=\(-\dfrac{1}{2}\) (TMĐKXĐ) vào A ta có:
\(\dfrac{-1}{\left(-\dfrac{1}{2}\right)^2}\)=-4
c, A<0 ⇔ \(\dfrac{-1}{x^2}< 0\) ⇔ x2>0 (Đúng với mọi x)
Vậy để A<0 thì x đúng với mọi giá trị (trừ ±3)
\(M=\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}\)
\(M=\dfrac{1}{x-1}-\dfrac{1}{x-2}+\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-4}+\dfrac{1}{x-4}-\dfrac{1}{x-5}\)
\(M=\dfrac{1}{x-1}-\dfrac{1}{x-5}\)
\(M=\dfrac{x-5-x+1}{\left(x-5\right)\left(x-1\right)}=-\dfrac{4}{x^2-6x+5}\)