a) ĐKXĐ : \(\hept{\begin{cases}x\ne0\\x\ne2\\x\ne-4\end{cases}}\)

\(A=\frac{3}{x+4}-\frac{x\left(x-1\right)}{x+4}\times\frac{2x-5}{x\left(x-2\right)\left(x+4\right)}-\frac{17}{\left(x+4\right)^2}\)

\(=\frac{3\left(x+4\right)}{\left(x+4\right)^2}-\frac{x\left(x-1\right)\left(2x-5\right)}{\left(x+4\right)x\left(x-2\right)\left(x+4\right)}-\frac{17}{\left(x+4\right)^2}\)

\(=\frac{3x+12}{\left(x+4\right)^2}-\frac{\left(x-1\right)\left(2x-5\right)}{\left(x+4\right)^2\left(x-2\right)}-\frac{17}{\left(x+4\right)^2}\)

\(=\frac{\left(3x+12\right)\left(x-2\right)}{\left(x+4\right)^2\left(x-2\right)}-\frac{2x^2-7x+5}{\left(x+4\right)^2\left(x-2\right)}-\frac{17\left(x-2\right)}{\left(x+4\right)^2\left(x-2\right)}\)

\(=\frac{3x^2+6x-24-2x^2+7x-5-17x+34}{\left(x+4\right)^2\left(x-2\right)}\)

\(=\frac{x^2-4x+5}{\left(x+4\right)^2\left(x-2\right)}=\frac{x^2-4x+5}{x^3+6x^2-32}\)

b) \(18A=1\)

<=> \(18\times\frac{x^2-4x+5}{x^3+6x^2-32}=1\)( ĐK : \(\hept{\begin{cases}x\ne0\\x\ne2\\x\ne-4\end{cases}}\))

<=> \(\frac{x^2-4x+5}{x^3+6x^2-32}=\frac{1}{18}\)

<=> 18( x² - 4x + 5 ) = x³ + 6x² - 32

<=> 18x² - 72x + 90 = x³ + 6x² - 32

<=> x³ + 6x² - 32 - 18x² + 72x - 90 = 0

<=> x³ - 12x² + 72x - 122 = 0

Rồi đến đây chịu á :) 

Ý lộn == là \(\frac{x^2-2x}{x+4}\)ạ ==

a)

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

\(\Leftrightarrow\frac{x-1}{2017}+\frac{x-2}{2016}-\frac{x-3}{2015}-\frac{x-4}{2014}=0\)

\(\Leftrightarrow\left(\frac{x-1}{2017}-1\right)+\left(\frac{x-2}{2016}-1\right)-\left(\frac{x-3}{2015}-1\right)-\left(\frac{x-4}{2014}-1\right)=0\)

\(\Leftrightarrow\left(x-2018\right)\left(\frac{1}{2017}+\frac{1}{2016}-\frac{1}{2015}-\frac{1}{2014}\right)=0\)

\(\Leftrightarrow x=2018\)

Bài của Nguyễn Phương Hà ấy.

b)\(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\left(1\right)\)

Ta thấy: \(\frac{1}{a}-\frac{1}{b}=\frac{1}{ab}\)

\(\Rightarrow\left(1\right)\Leftrightarrow\frac{1}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Rightarrow x=1\)

Đề sai ! Sửa nhé :

a) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne\pm2\end{cases}}\)

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

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

\(\Leftrightarrow A=\frac{2\left(x+2\right)-4}{\left(x+2\right)^2}:\frac{2-\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow A=\frac{2x+4-4}{\left(x+2\right)^2}.\frac{\left(x+2\right)\left(x-2\right)}{-x}\)

\(\Leftrightarrow A=\frac{2x\left(x-2\right)}{-x\left(x+2\right)}\)

\(\Leftrightarrow A=-\frac{2\left(x-2\right)}{x+2}\)

b) Để \(A\le-2\)

\(\Leftrightarrow-\frac{2\left(x-2\right)}{x+2}\le-2\)

\(\Leftrightarrow\frac{2\left(x-2\right)}{x+2}\ge2\)

\(\Leftrightarrow\frac{x-2}{x+2}\ge1\)

\(\Leftrightarrow x-2\ge x+2\)

\(\Leftrightarrow-2\ge2\)(ktm)

Vậy để \(A\le-2\Leftrightarrow x\in\varnothing\)

a.

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

\(A=\left(\frac{2.\left(x^2+8\right)}{\left(x+2\right).\left(x^2+8\right)}-\frac{4\left(x+2\right)}{\left(x+2\right)\left(x^2+8\right)}\right):\left(\frac{2}{\left(x-2\right)\left(x+2\right)}+\frac{1}{2-x}\right)\)

\(A=\left(\frac{2x^2+8-4x+8}{\left(x+2\right)\left(x^2+8\right)}\right):\left(\frac{2}{\left(x-2\right)\left(x+2\right)}+\frac{-1}{x-2}\right)\)

\(A=\left(\frac{2x\left(x-2\right)+16}{\left(x+2\right)\left(x^2+8\right)}\right):\left(\frac{2}{\left(x-2\right)\left(x+2\right)}+\frac{-x-2}{\left(x-2\right)\left(x+2\right)}\right)\)

\(A=\left(\frac{2x\left(x-2\right)+16}{\left(x+2\right)\left(x^2+8\right)}\right):\left(\frac{2-x-2}{\left(x-2\right)\left(x+2\right)}\right)\)

\(A=\left(\frac{\left(2x\left(x-2\right)+16\right)\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x^2+8\right)\left(-x\right)}\right)\)

\(A=\frac{\left(2x\left(x-2\right)+16\right)\left(x-2\right)}{\left(x^2+8\right)\left(-x\right)}\)

\(A=\frac{\left(2x^2-4x+16\right)\left(x-2\right)}{\left(x^2+8\right)\left(-x\right)}\)

\(A=\frac{\left(2x^3-4x-4x-4x^2+8x+16x-32\right)}{-x^3+8}\)

\(A=\frac{2x^3-4x^2+16x-32}{-x^3+8}\)

Em bị nhầm ở chỗ:

\(\frac{x-2}{x+2}\ge1\)

<=> \(\frac{x-2}{x+2}-1\ge0\)

<=> \(\frac{x-2-x-2}{x+2}\ge0\)

<=> \(-\frac{4}{x+2}\ge0\)

<=> x + 2 < 0

<=> x < - 2

a) \(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)

=> \(\left(\frac{x-6}{7}+1\right)+\left(\frac{x-7}{8}+1\right)+\left(\frac{x-8}{9}+1\right)=\left(\frac{x-9}{10}+1\right)+\left(\frac{x-10}{11}+1\right)+\left(\frac{x-11}{12}+1\right)\)

=> \(\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}+\frac{x+1}{12}=0\)

=> \(\left(x+1\right)\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)=0\)

=>  x + 1 = 0

=> x = -1

b) \(\frac{x-1}{2020}+\frac{x-2}{2019}-\frac{x-3}{2018}=\frac{x-4}{2017}\)

=> \(\left(\frac{x-1}{2020}-1\right)+\left(\frac{x-2}{2019}-1\right)-\left(\frac{x-3}{2018}-1\right)=\left(\frac{x-4}{2017}-1\right)\)

=> \(\frac{x-2021}{2020}+\frac{x-2021}{2019}-\frac{x-2021}{2018}=\frac{x-2021}{2017}\)

=> \(\left(x-2021\right)\left(\frac{1}{2020}+\frac{1}{2019}-\frac{1}{2018}-\frac{1}{2017}\right)=0\)

=> x - 2021 = 0

=> x = 2021

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

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

=> \(-\frac{1}{12}x+6=7\)

=> \(-\frac{1}{12}x=1\)

=> x = -12

phá ngoặc tính BT , nên kết quả sẽ ko ra con số nhận định !!! tui thử thui nha bà  !

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

\(x+\frac{1}{2}+x+\frac{1}{3}+y-5+x+\frac{1}{4}=\frac{1}{4}\)

\(3x+y-\frac{47}{12}=\frac{1}{4}\)

\(3x+y=\frac{25}{6}\)

\(3x=\frac{25}{6}-y\)

\(x=\frac{25-25y}{18}\)

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

\(x+\frac{1}{2}+x+\frac{1}{3}+y-5+x+\frac{1}{4}=\frac{1}{4}\)

\(3x+y-\frac{47}{12}=\frac{1}{4}\)

\(3x+y=\frac{25}{6}\)

\(y=\frac{25}{6}-3x\)

Vậy \(x=\frac{25-25y}{18}\)

\(y=\frac{25}{6}-3x\)

Ta có:

 \(|x+\frac{1}{2}|\ge x+\frac{1}{2}\forall x;|x+\frac{1}{3}|\ge x+\frac{1}{3}\forall x;|y-5|\ge y-5\forall y;|x+\frac{1}{4}|\ge x+\frac{1}{4}\forall x\)

\(\Rightarrow|x+\frac{1}{2}|+|x+\frac{1}{3}|+|y-5|+|x+\frac{1}{4}|\ge x+\frac{1}{2}+x+\frac{1}{3}+y-5+x+\frac{1}{4}\)

Mà \(|x+\frac{1}{2}|+|x+\frac{1}{3}|+|y-5|+|x+\frac{1}{4}|=\frac{1}{4}\)

\(\Rightarrow\frac{1}{4}\ge x+\frac{1}{2}+x+\frac{1}{3}+y-5+x+\frac{1}{4}\)

\(\Rightarrow\frac{1}{4}\ge3x+y-\frac{47}{12}\)

\(\Rightarrow3x+y\le\frac{25}{6}\)

\(\Rightarrow x\le\frac{\frac{25}{6}-y}{3}\)

Thay vào tính y

Làm phiền bạn Quỳnh

Bạn bảo thay vào tính y? Vậy bạn trình bày cho mình phần cuối. 

Bạn đang nhầm vấn đề nhé. (2 dòng cuối) 

Vd: a + b = 5 

<=> a = 5 - b

Thay a vào: 5 - b + b = 5 

<=> 0 = 0 ???

Bạn tính ra r bạn đưa lên bth bên trên??? Mình thấy bạn làm vậy thì k thể tìm ra a, b. Có thể mình k hiểu ý bạn. Nếu v bạn trình bày giúp mình phần cuối vì mình không hiểu ý bạn. Thankbạn. 

a) \(-ĐKXĐ:x\ne\pm2;1\)

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

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

\(=\left[\frac{x-2}{\left(x-2\right)\left(x+2\right)}+\frac{\left(-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x}{\left(x-2\right)\left(x+2\right)}\right]\)\(.\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)

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

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

b) \(A>0\Leftrightarrow\frac{x+1}{\left(x+2\right)^2}>0\Leftrightarrow\orbr{\begin{cases}x+1< 0;\left(x+2\right)^2< 0\left(voly\right)\\x+1>0;\left(x+2\right)^2>0\end{cases}}\)

\(\Leftrightarrow x>1;x>-2\Leftrightarrow x>1\)

Vậy với mọi x thỏa mãn x>1 thì A > 0

c) Ta có : \(x^2+3x+2=0\Leftrightarrow x^2+x+2x+2=0\)

\(\Leftrightarrow x\left(x+1\right)+2\left(x+1\right)=0\Leftrightarrow\left(x+1\right)\left(x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-2\end{cases}}\)

Vậy x = -1;-2