\(x\ne\left\{-4;-3;-2;-1\right\}\)

\(\Leftrightarrow\frac{x^2+x+1}{x+1}-1+\frac{x^2+2x+2}{x+2}-1=\frac{x^2+3x+3}{x+3}-1+\frac{x^2+4x+4}{x+4}-1\)

\(\Leftrightarrow\frac{x^2}{x+1}+\frac{x^2+x}{x+2}-\frac{x^2+2x}{x+3}-\frac{x^2+3x}{x+4}=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\frac{1}{x+3}-\frac{1}{x+1}=\frac{1}{x+2}-\frac{1}{x+4}\left(1\right)\end{matrix}\right.\)

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

\(\Leftrightarrow\left(x+2\right)\left(x+4\right)+\left(x+1\right)\left(x+3\right)=0\)

\(\Leftrightarrow2x^2+10x+11=0\Rightarrow x=\frac{-5\pm\sqrt{3}}{2}\)

a) ĐKXĐ: x≠0

Ta có: \(\frac{9}{x}+2=-6\)

⇔\(\frac{9}{x}+2+6=0\)

⇔\(\frac{9}{x}+8=0\)

⇔\(\frac{9}{x}+\frac{8x}{x}=0\)

⇔9+8x=0

⇔8x=-9

hay \(x=-\frac{9}{8}\)

Vậy: \(x=-\frac{9}{8}\)

b) ĐKXĐ: x≠0;x≠-1;x≠-3

Ta có: \(\frac{7}{x+1}+\frac{-18x}{x\left(x^2+4x+3\right)}=\frac{-4}{x+3}\)

⇔\(\frac{7}{x+1}+\frac{-18x}{x\left(x+1\right)\left(x+3\right)}-\frac{-4}{x+3}=0\)

⇔\(\frac{7x\left(x+3\right)}{\left(x+1\right)\cdot x\cdot\left(x+3\right)}+\frac{-18x}{\left(x+1\right)\cdot x\cdot\left(x+3\right)}-\frac{-4x\left(x+1\right)}{\left(x+3\right)\cdot x\cdot\left(x+1\right)}=0\)

⇔\(7x^2+21x-18x+4x\left(x+1\right)=0\)

\(\Leftrightarrow7x^2+21x-18x+4x^2+4x=0\)

⇔\(11x^2+7x=0\)

\(\Leftrightarrow x\left(11x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\11x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\11x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=\frac{-7}{11}\end{matrix}\right.\)

Vậy: \(x=\frac{-7}{11}\)

c) ĐKXĐ: x≠1; x≠-3

Ta có: \(\frac{3x-1}{x-1}-1=\frac{2x+5}{x+3}+\frac{4}{x^2-2x+3}\)

⇔\(\frac{3x-1}{x-1}-1-\frac{2x+5}{x+3}-\frac{4}{\left(x-1\right)\left(x+3\right)}=0\)

⇔\(\frac{\left(3x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{\left(x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{\left(2x+5\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}-\frac{4}{\left(x-1\right)\left(x+3\right)}=0\)

⇔\(\left(3x-1\right)\left(x+3\right)-\left(x-1\right)\left(x+3\right)-\left(2x+5\right)\left(x-1\right)-4=0\)

\(\Leftrightarrow3x^2+9x-x-3-\left(x^2+3x-x-3\right)-\left(2x^2-2x+5x-5\right)-4=0\)

\(\Leftrightarrow3x^2+8x-3-\left(x^2+2x-3\right)-\left(2x^2+3x-5\right)-4=0\)

\(\Leftrightarrow3x^2+8x-3-x^2-2x+3-2x^2-3x+5-4=0\)

\(\Leftrightarrow3x+1=0\)

\(\Leftrightarrow3x=-1\)

hay \(x=\frac{-1}{3}\)

Vậy: \(x=\frac{-1}{3}\)

ĐKXĐ: \(x\ne1;x\ne2\)

\(\frac{2x-1}{x-1}+\frac{3x-2}{x-2}=\frac{x^2+4x+5}{x^2-3x+2}+4\\ \Leftrightarrow\frac{2x^2-5x+2}{\left(x-1\right)\left(x-2\right)}+\frac{3x^2-5x+2}{\left(x-1\right)\left(x-2\right)}=\frac{x^2+4x+5}{\left(x-1\right)\left(x-2\right)}+\frac{4x^2-12x+8}{\left(x-1\right)\left(x-2\right)}\\ \Rightarrow2x^2-5x+2+3x^2-5x+2=x^2+4x+5+4x^2-12x+8\\ \Leftrightarrow2x^2+3x^2-x^2-4x^2-5x-5x-4x+12x=5-2-2\\ \Leftrightarrow-2x=1\\ \Leftrightarrow x=\frac{-1}{2}\left(tm\right)\)Vậy tập nghiệm của phương trình là: \(S=\left\{-\frac{1}{2}\right\}\)

ĐKXĐ: x∉{1;2}

Ta có: \(\frac{2x-1}{x-1}+\frac{3x-2}{x-2}=\frac{x^2+4x+5}{x^2-3x+2}+4\)

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

Suy ra: \(\left(2x-1\right)\left(x-2\right)+\left(3x-2\right)\left(x-1\right)=x^2+4x+5+4\left(x^2-3x+2\right)\)

\(\Leftrightarrow2x^2-4x-x+2+3x^2-3x-2x+2=x^2+4x+5+4x^2-12x+8\)

\(\Leftrightarrow5x^2-10x+4=5x^2-8x+13\)

\(\Leftrightarrow5x^2-10x+4-5x^2+8x-13=0\)

\(\Leftrightarrow-2x-9=0\)

\(\Leftrightarrow-2x=9\)

hay \(x=\frac{-9}{2}\)(tm)

Vậy: \(S=\left\{-\frac{9}{2}\right\}\)

tae tae ơi khó quá hổng hiểu j hết trơn

mình làm câu cuối thôi nhé , những câu còn lại bạn tự làm đi , dễ mà :)))) chỉ cần quy đồng mẫu lên là được 

\(=\frac{x+1}{58}+1+\frac{x+2}{57}+1=\frac{x+3}{56}+1+\frac{x+4}{55}\)

\(=\frac{x+59}{58}+\frac{x+59}{57}=\frac{x+59}{56}+\frac{x+59}{55}\)

\(=\frac{x+59}{58}+\frac{x+59}{57}-\frac{x+59}{56}-\frac{x+59}{55}=0\)

\(=\left(x+59\right)\left(\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}\right)=0\)

Vì \(\left(\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}\right)\) luôn khác 0 

<=> x + 59 = 0 

<=> x=-59 

1,(3x-2)(4x+5)=0

\(\Leftrightarrow\left\{{}\begin{matrix}3x-2=0\\4x+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=2\\4x=-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{2}{3}\\x=\frac{-5}{4}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là ...

2,\(5\left(2x-3\right)-4\left(5x-7\right)=19-2\left(x+11\right)\)

\(\Leftrightarrow10x-15-20x+28=19-2x-22\)

\(\Leftrightarrow10x-20x+2x=15-28+19-22\)

\(\Leftrightarrow-8x=-16\)

=> x= 2

vậy..

3,\(\left(x^2-2x+1\right)-4=0\)

\(\Leftrightarrow\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+1\right)-4=0\)

\(\Leftrightarrow\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}\right)+\frac{3}{4}-4=0\)

\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2-\frac{13}{4}=0\) ( vô nghiệm )

(vì \(\left(x-\frac{1}{2}\right)^2\ge0\Rightarrow\left(x-\frac{1}{2}\right)^2-\frac{13}{4}\ge0\) )

từ đó suy ra phương trình vô nghiệm

5,\(\frac{4x+3}{2}-2+3x=\frac{2x-1}{10}+\frac{19x+2}{5}-1\)

\(\Leftrightarrow\frac{5\left(4x+3\right)}{10}-\frac{10\left(2-3x\right)}{10}=\frac{2x-1}{10}+\frac{2\left(19x+2\right)}{10}-\frac{10}{10}\)

\(\Leftrightarrow\frac{20x+15}{10}-\frac{20-30x}{10}=\frac{2x-1}{10}+\frac{38x+4}{10}-\frac{10}{10}\)

\(\Rightarrow20x+15-20+30x=2x-1+38x+4-10\)

\(\Leftrightarrow20x+30x-2x-38x=-15+20-1+4-10\)

\(\Leftrightarrow10x=-2\)

\(\Leftrightarrow x=-5\)

Vậy ....

p/s : thực ra mk cx chỉ ms học th nên giải bài tập về phương trình vẫn còn nhiều chỗ sai nữa,có gì mong mn giúp đỡ :)

a) \(\frac{5x-2}{2-2x}+\frac{2x-1}{2}+\frac{x^2+x-3}{1-x}=1\)

ĐK: x≠1

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

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

<=>\(\frac{5x-2+2x-1-2x^2+x+2x^2+2x-6}{2\left(1-x\right)}=1\)

<=>\(\frac{10x-9}{2\left(1-x\right)}=1\)

<=> 10x-9=2(1-x)

<=>10x-9=2-2x

<=> 10x+2x= 2+9

<=> 12x=11

<=> x= \(\frac{11}{12}\left(tm\right)\)

b) \(\frac{6x-1}{2-x}+\frac{9x+4}{x+2}=\frac{3x^2-2x+1}{x^2-4}\)

ĐK: x≠2, x≠-2

<=>\(\frac{6x-1}{-\left(x-2\right)}+\frac{9x+4}{x+2}-\frac{3x^2-2x+1}{\left(x-2\right)\left(x+2\right)}=0\)

<=> -(x+2).(6x-1)+(x-2).(9x+4)-(3x²-2x+1)=0

<=> -(6x²-x+12x-2)+9x²+4x-18x-8-3x²+2x-1 = 0

<=> -6x²-11x+2+9x²+4x-18x-8-3x²+2x-1=0

<=> -23x-7=0

<=> -23x=7

<=> x= \(\frac{-7}{23}\left(tm\right)\)

tham khảo câu d trong

https://hoc24.vn/hoi-dap/question/919967.html

c) \(\frac{1}{x-1}\)+\(\frac{2x^2-5}{x^3-1}\)=\(\frac{4}{x^2+x+1}\) (ĐKXĐ:x≠1)

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

⇒x²+x+1+2x²-5=4x-4

⇔3x²-3x=0

⇔3x(x-1)=0

⇔x=0 (TMĐK) hoặc x=1 (loại)

Vậy tập nghiệm của phương trình đã cho là:S={0}

Bài 2 :

a, Ta có : \(\left(x+4\right)\left(x-1\right)=0\)

=> \(\left[{}\begin{matrix}x+4=0\\x-1=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=-4\\x=1\end{matrix}\right.\)

b, Ta có : \(\left(3x-2\right)\left(4x-7\right)=0\)

=> \(\left[{}\begin{matrix}3x-2=0\\4x-7=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}3x=2\\4x=7\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{2}{3}\\x=\frac{7}{4}\end{matrix}\right.\)

c, Ta có : \(\left(x+5\right)\left(x^2+1\right)=0\)

=> \(\left[{}\begin{matrix}x+5=0\\x^2+1=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=-5\\x^2+1=0\left(VL\right)\end{matrix}\right.\)

d, Ta có : \(x\left(x-1\right)\left(x^2+4\right)=0\)

=> \(\left[{}\begin{matrix}x=0\\x-1=0\\x^2+4=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=0\\x=1\\x^2+4=0\left(VL\right)\end{matrix}\right.\)

e, Ta có : \(\left(3x+2\right)\left(x+\frac{1}{2}\right)=0\)

=> \(\left[{}\begin{matrix}3x+2=0\\x+\frac{1}{2}=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=-\frac{2}{3}\\x=-\frac{1}{2}\end{matrix}\right.\)

f, Ta có : \(\left(x+2\right)\left(x+3\right)\left(x^2+7\right)=0\)

=> \(\left[{}\begin{matrix}x+2=0\\x-3=0\\x^2+7=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=-2\\x=3\\x^2+7=0\left(VL\right)\end{matrix}\right.\)

Bài 1 :

a, Ta có : \(1-\frac{x+3}{4}-\frac{x-2}{6}=0\)

=> \(\frac{12}{12}-\frac{3\left(x+3\right)}{12}-\frac{2\left(x-2\right)}{12}=0\)

=> \(12-3\left(x+3\right)-2\left(x-2\right)=0\)

=> \(12-3x-9-2x+4=0\)

=> \(-5x=-7\)

=> \(x=\frac{7}{5}\)