a.\(\Leftrightarrow\left(x+3\right)\left(x^2-x-2-2x^2+3x+5\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(-x^2+2x+3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=3\\x=-1\end{matrix}\right.\)

(x-2)(x+1)(x+3)=(x+3)(x+1)(2x-58)

\(x^3+2x^2-5x-6\)=\(2x^3+3x^2-14x-15\)

\(-x^3-x^2+9x+9=0\)

\(-x^2\left(x+1\right)+9\left(x+1\right)=0\)

\(\left(x+1\right)\left(9-x^2\right)\)=0

(x+1)(3-x)(3+x)=0

*x+1=0 =>x=-1

*3-x=0=>x=3

*3+x=0=>x=-3

Câu này tớ giải hơn 10 lần rồi cậu ( ko xàm :) 

\(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)

\(\Leftrightarrow\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)=\left(\frac{x+6}{94}+1\right)+\left(\frac{x+8}{92}+1\right)\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)

Vì \(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0\)

Do đó \(x+100=0\Leftrightarrow x=-100\)

Vậy pt có nghiệm : x=-100

Ta có : \(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)

=> \(\frac{x+2}{98}+1+\frac{x+4}{96}+1=\frac{x+6}{94}+1+\frac{x+8}{92}+1\)

=> \(\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)

=> \(\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)

=> \(\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)

Vì \(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0\)

=> x + 100 = 0

=> x = - 100

Vậy x = - 100

\(\Leftrightarrow\frac{x+2}{98}+1+\frac{x+4}{96}+1=\frac{x+6}{94}+1+\frac{x+8}{98}+1\)

\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}\right)=\left(x+100\right)\left(\frac{1}{94}+\frac{1}{92}\right)\)

Do \(\frac{1}{98}+\frac{1}{96}\ne\frac{1}{94}+\frac{1}{92}\)\(\Rightarrow x+100=0\)

                                                    \(\Leftrightarrow x=-100\)

           Vậy \(x=-100\)

~Hok Tốt~

\(a,⇔\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}-\frac{x-23}{27}=0\)

\(⇔(x-23)(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27})=0\)

\(⇔x-23=0\) (vì \(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}>0\))

\(⇔x=23\)

\(b,⇔\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}+\frac{x+100}{95}=0\)

\(⇔(x+100)(\frac{1}{98}+\frac{1}{97}+\frac{1}{96}+\frac{1}{95})=0\)

\(⇔x+100=0\) (vì \(\frac{1}{98}+\frac{1}{97}+\frac{1}{96}+\frac{1}{95}>0\))

\(⇔x=-100\)

\(c,⇔(\frac{x+1}{2012}+1)+(\frac{x+2}{2011}+1)=(\frac{x+3}{2010}+1)+(\frac{x+4}{2009}+1)\)

\(⇔\frac{x+2013}{2012}+\frac{x+2013}{2011}-\frac{x+2013}{2010}-\frac{x+2013}{2009}=0\)

\(⇔(x+2013)(\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009})=0\)

\(⇔x+2013=0\) (vì \(\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}<0\))

\(⇔x=-2013\)

Không biết bạn có gõ đúng đề cả 2 câu không ? Câu 2 không có nghiệm nguyên dương nhé bạn. Bạn xem lại.

có đúng đề không bạn

1/Tôi chỉ bt 1 câu thui thông cảm :)

P=\(\frac{x}{x-1}+\frac{4}{x+1}+\frac{4-6x}{x^2-1}\)       ĐK:\(\hept{\begin{cases}x-1\ne0\\x+1\ne\\x^2-1\ne0\end{cases}1}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne-1\\x\ne1\end{cases}}\)

P=\(\frac{x\left(x+1\right)+4\left(x-1\right)+4-6x}{\left(x-1\right).\left(x+1\right)}\) 

=\(\frac{x^2+x+4x-4+4-6x}{\left(x-1\right)\left(x+1\right)}=\frac{x^2-x}{\left(x-1\right)\left(x+1\right)}\)

=\(\frac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{x}{x+1}\)

^^ học tốt!

1/

\(đkxđ\Leftrightarrow x\ne\pm1\)

\(P=\frac{x}{x-1}+\frac{4}{x+1}+\frac{4-6x}{x^2-1}\)

\(=\frac{x}{x-1}+\frac{4}{x+1}+\frac{4-6x}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{4\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}+\frac{4-6x}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{x^2+x+4x-4+4-6x}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{x^2-x}{\left(x-1\right)\left(x+1\right)}=\frac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{x}{x+1}\)

2/

Kẻ \(DE//AB\left(E\in AC\right)\)

\(\Rightarrow\frac{DE}{AB}=\frac{EC}{AC}\)

\(\Delta ADE\)đều (vì .............)\(\Rightarrow AD=AE=DE\)

\(\Rightarrow\frac{AD}{AB}=\frac{AC-AE}{AC}\)mà \(AE=AD\)

\(\Rightarrow\frac{AB}{AB}=1-\frac{AD}{AC}\)

\(\Rightarrow\frac{AD}{AB}+\frac{AD}{AC}=1\)

\(\Rightarrow AD\left(\frac{1}{AB}+\frac{1}{AC}\right)=1\)

\(\Rightarrow\frac{1}{AB}+\frac{1}{AC}=\frac{1}{AD}\left(ĐPCM\right)\)

\(ĐK:x\ne0;x\ne-6\)

⇔ \(\frac{720\left(x+6\right)}{6x\left(x+6\right)}=\frac{6x\left(x+6\right)}{6x\left(x+6\right)}+\frac{x\left(x+6\right)}{6x\left(x+6\right)}+\frac{6x\left(120-x\right)}{6x\left(x+6\right)}\)

\(\Rightarrow720x+4320=6x^2+36x+x^2+6x+720x-6x^2\)

\(\Leftrightarrow6x^2+36x+x^2+6x+720x-6x^2-720x-4320=0\)

\(\Leftrightarrow x^2+42x-4320=0\)

\(\Leftrightarrow x^2+90x-48x-4320=0\)

\(\Leftrightarrow\left(x+90\right)\left(x-48\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+90=0\\x-48=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-90\\x=48\end{matrix}\right.\) ( tm )

\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\Leftrightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1+\frac{x+4}{96}+1=0\)

\(\Leftrightarrow\frac{x+10}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)

mà \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\)\(\Rightarrow x+100=0\Leftrightarrow x=-100\)

Ta có:\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)

\(\Rightarrow\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+4}{96}+1\right)=0\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)

\(\Rightarrow\left(x+100\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)

Vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\Rightarrow x+100=0\)

\(\Rightarrow x=-100\)

\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)

\(\Rightarrow\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+4}{96}+1\right)=-4+4\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)

\(\Rightarrow\left(x+100\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)

Mà \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}>0\)

\(\Rightarrow x+100=0\)

\(\Rightarrow x=0-100\)

\(\Rightarrow x=-100\)

Chúc bạn học tốt !!!! 

\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)

\(\Leftrightarrow\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+4}{96}+1\right)=0\) 

\(\Leftrightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\) 

\(\Leftrightarrow x+100=0\) 

\(\Leftrightarrow x=-100\) 

Vậy...

\(\Rightarrow\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+4}{96}+1\right)=-4+1+1+1+1\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)

\(\Rightarrow\left(x+100\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)

Vì 1/99+1/98+1/97+1/96>0

\(\Rightarrow x+100=0\Rightarrow x=-100\)