tớ ko bt lm abc , tớ lm d thôi nha , thứ lỗi 

\(\frac{5}{2x-3}-\frac{1}{x+2}=\frac{5}{x-6}-\frac{7}{2x-1}\)

\(\frac{3x+13}{2x^2+x-6}=\frac{5}{x-6}+\frac{7}{1-2x}\)

\(\frac{3x+13}{\left(x+2\right)\left(2x-3\right)}=\frac{3x+37}{\left(x-6\right)\left(2x-1\right)}\)

\(\frac{10-9x}{-4x^3+32x^2-51x+18}=0\)

\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{10}{9}\end{cases}}\)

Xét: \(\sqrt{1+n^2+\frac{n^2}{\left(n+1\right)^2}}=\sqrt{\frac{\left(n+1\right)^2+n^2\left(n+1\right)^2+n^2}{\left(n+1\right)^2}}\) (với \(n\inℕ\))

\(=\sqrt{\frac{n^2+2n+1+n^4+2n^3+n^2+n^2}{\left(n+1\right)^2}}\)

\(=\sqrt{\frac{n^4+n^2+1+2n^3+2n^2+2n}{\left(n+1\right)^2}}\)

\(=\sqrt{\frac{\left(n^2+n+1\right)^2}{\left(n+1\right)^2}}=\frac{n^2+n+1}{n+1}=n+\frac{1}{n+1}\)

Áp dụng vào ta tính được: \(\sqrt{1+2015^2+\frac{2015^2}{2016^2}}+\frac{2015}{2016}=2015+\frac{1}{2016}+\frac{2015}{2016}\)

\(=2015+1=2016\)

Khi đó: \(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=2016\)

\(\Leftrightarrow\left|x-1\right|+\left|x-2\right|=2016\)

Đến đây xét tiếp các TH nhé, ez rồi:))

chẳng biết đúng ko,mới lớp 5

\(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=\sqrt{1+2015^2+\frac{2015^2}{2016^2}}+\frac{2015}{2016}\)

\(\sqrt{x^2}-\sqrt{2x}+\sqrt{1}+\sqrt{x^2}-\sqrt{4x}+\sqrt{4}=\sqrt{1}+\sqrt{2015^2}+\sqrt{\frac{2015^2}{2016^2}}+\frac{2015}{2016}\)

\(\sqrt{x^2}-\sqrt{6x}+3=1+2015+\frac{2015}{2016}+\frac{2015}{2016}\)

\(x-\sqrt{6x}=1+\frac{2015}{1+2016+2016}-3\)

\(x-\sqrt{6x}=2-\frac{2015}{4033}\)

\(x-\sqrt{6x}=\frac{6051}{4033}\)

pt <=>\(\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-2\right)^2}=\sqrt{1+2.2015+2015^2-2.2015+\frac{2015^2}{2016^2}}+\frac{2015}{2016}\)

\(\Leftrightarrow\left(x-1\right)+\left(x-2\right)=\sqrt{2016^2-2.2016.\frac{2015}{2016}+\frac{2015^2}{2016^2}}+\frac{2015}{2016}\)

\(\Leftrightarrow2x-3=\sqrt{\left(2016-\frac{2015}{2016}\right)^2}+\frac{2015}{2016}\)

\(\Leftrightarrow2x-3=2016-\frac{2015}{2016}+\frac{2015}{2016}\)

\(\Leftrightarrow2x-3=2016\)

\(\Leftrightarrow2x=2019\)

\(\Leftrightarrow x=\frac{2019}{2}\)

\(\frac{2x-4}{2014}+\frac{2x-2}{2016}\) và \(\frac{2x-1}{2017}+\frac{2x-3}{2015}\)

VT = \(\frac{2x-4}{2014}+\frac{2x-2}{2016}\)

= \(\frac{2x-4}{2014}+1+\frac{2x-2}{2016}+1\)

= \(\frac{2x-2018}{2014}+\frac{2x-2018}{2016}\)

VP = \(\frac{2x-1}{2017}+\frac{2x-3}{2015}\)

= \(\frac{2x-1}{2017}+1+\frac{2x-3}{2015}+1\)

= \(\frac{2x-2018}{2017}+\frac{2x-2018}{2015}\)

Mà \(\frac{2x-2018}{2014}>\frac{2x-2018}{2015}\) và \(\frac{2x-2018}{2016}>\frac{2x-2018}{2017}\)

nên \(\frac{2x-4}{2014}+\frac{2x-2}{2016}\) > \(\frac{2x-1}{2017}+\frac{2x-3}{2015}\)

Chúc bn học tốt!!

1)

a) \(\frac{x+5}{3x-6}-\frac{1}{2}=\frac{2x-3}{2x-4}< =>\frac{2\left(x+5\right)}{2\left(3x-6\right)}-\frac{3x-6}{2\left(3x-6\right)}=\frac{3\left(2x-3\right)}{3\left(2x-4\right)}.\)

(đk:x khác \(\frac{1}{2}\))

\(\frac{2x+10}{6x-12}-\frac{3x-6}{6x-12}=\frac{6x-9}{6x-12}< =>2x+10-3x+6=6x-9< =>x=\frac{25}{7}\)

Vậy x=\(\frac{25}{7}\)

b) /7-2x/=x-3 \(x\ge\frac{7}{2}\)

(đk \(x\ge3,\frac{7}{2}< =>x\ge\frac{7}{2}\))

\(\Rightarrow\orbr{\begin{cases}7-2x=x-3\\7-2x=-\left(x-3\right)\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{10}{3}\left(< \frac{7}{2}\Rightarrow l\right)\\x=4\left(tm\right)\end{cases}}}\)

Vậy x=4

2)

\(\frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4}>\frac{x-4}{5}+\frac{x-5}{6}\)

\(\Leftrightarrow\frac{30\left(x-1\right)}{60}+\frac{20\left(x-2\right)}{60}+\frac{15\left(x-3\right)}{60}-\frac{12\left(x-4\right)}{60}-\frac{10\left(x-5\right)}{60}>0\)

\(\Leftrightarrow30x-30+20x-40+15x-45-12x+48-10x+50>0\Leftrightarrow43x-17>0\Leftrightarrow x>\frac{17}{43}\)

\(\frac{x}{3} + \frac{{2x + 1}}{6} = \frac{{4\left( {x - 2} \right)}}{5}\)

\(\frac{{10x}}{{3.10}} + \frac{{\left( {2x + 1} \right).5}}{{6.5}} = \frac{{6.4\left( {x - 2} \right)}}{{5.6}}\)

\(\frac{{10x}}{{30}} + \frac{{10x + 5}}{{30}} = \frac{{24x - 48}}{{30}}\)

\(10x + 10x + 5 = 24x - 48\)

\(10x + 10x - 24x =  - 5 - 48\)

\( - 4x =  - 53\)

\(x = \left( { - 53} \right):\left( { - 4} \right)\)

\(x = \frac{{53}}{4}\)

Vậy phương trình có nghiệm là \(x = \frac{{53}}{4}\).  

a) \(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)

\(\Rightarrow\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1\)

\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}=\frac{x+10}{7}+\frac{x+10}{6}\)

\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)

\(\Rightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)

Mà \(\left(\frac{1}{9}< \frac{1}{8}< \frac{1}{7}< \frac{1}{6}\right)\)nên \(\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)< 0\)

\(\Rightarrow x+10=0\Rightarrow x=-10\)

Vậy x = -10

b) \(\frac{x}{2012}+\frac{x+1}{2013}+\frac{x+2}{2014}+\frac{x+3}{2015}+\frac{x+4}{2016}=5\)

\(\Rightarrow\frac{x}{2012}-1+\frac{x+1}{2013}-1+\frac{x+2}{2014}-1\)

\(+\frac{x+3}{2015}-1+\frac{x+4}{2016}-1=0\)

\(\Rightarrow\frac{x-2012}{2012}+\frac{x-2012}{2013}+\frac{x-2012}{2014}\)\(+\frac{x-2012}{2015}+\frac{x-2012}{2016}=0\)

\(\Rightarrow\left(x-2012\right)\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)=0\)

Mà \(\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)>0\)nên x - 2012 = 0

Vậy x = 2012

a, (x+1)/9 +1 + (x+2)/8  =  (x+3)/7 + 1 + (x+4)/6 + 1

<=> (x+10)/9 +(x+10)/8 = (x+10)/7 + (x+10)/6

<=> (x+10). (1/9 +1/8 - 1/7 -1/6) =0

vì 1/9 +1/8 -1/7 - 1/6 khác 0

=> x+10=0

=> x=-10