\(\frac{1}{x^2-2x+2}-1+\frac{2}{x^2-2x+3}-1+2-\frac{6}{x^2-2x+4}=0\)

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

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

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

Xét (1), đặt \(a=x^2-2x+3\) pt trở thành:

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

\(\Leftrightarrow\frac{a-3}{a^2-1}=\frac{1}{a}\Leftrightarrow a^2-3a=a^2-1\Leftrightarrow3a=1\Rightarrow a=\frac{1}{3}\)

\(\Rightarrow x^2-2x+3=\frac{1}{3}\Leftrightarrow x^2-2x+1+\frac{5}{3}=0\)

\(\Leftrightarrow\left(x-1\right)^2+\frac{5}{3}=0\) (vô nghiệm)

Vậy \(x=1\)

1) Nhìn cái pt hết ham, nhưng bấm nghiệm đẹp v~`~

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

\(\Leftrightarrow\left(\sqrt{2}+2\right)\left(x\sqrt{2}-1\right)-2x\sqrt{2}+\sqrt{2}=0\)

\(\Leftrightarrow2x-\sqrt{2}+2x\sqrt{2}-2-2x\sqrt{2}+\sqrt{2}=0\)

\(\Leftrightarrow2x-2=0\Leftrightarrow2x=2\Rightarrow x=1\)

Mấy bài kia sao cái phương trình dài thê,s giải sao nổi

#muon roi ma sao con 

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

\(\Leftrightarrow\frac{6x-9-2x+2}{6}=\frac{x+2}{6}\)

\(\Rightarrow4x-7=x+2\Leftrightarrow3x=9\Leftrightarrow x=3\)

Vậy tập nghiệm của phương trình là S = { 3 } 

\(M=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)

\(=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)\)

\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)

\(=\left(x^2+5x\right)^2-36\ge-36\)

Dấu "=" xảy ra khi \(x\in\left\{0;-5\right\}\)

Giải PT \(\frac{x-6}{2010}+\frac{x-603}{471}+\frac{x-1}{403}=9\)

\(\Leftrightarrow\frac{x-6}{2010}+\frac{x-603}{471}+\frac{x-1}{403}-9=0\)

\(\Leftrightarrow\left(\frac{x-6}{2010}-1\right)+\left(\frac{x-603}{471}-3\right)+\left(\frac{x-1}{403}-5\right)=0\)

\(\Leftrightarrow\frac{x-2016}{2010}+\frac{x-2016}{471}+\frac{x-2016}{403}=0\)

\(\Leftrightarrow\left(x-2016\right)\left(\frac{1}{2010}+\frac{1}{471}+\frac{1}{403}\right)=0\)

Mà \(\left(\frac{1}{2010}+\frac{1}{471}+\frac{1}{403}\right)\ne0\)

\(\Leftrightarrow x-2016=0\Leftrightarrow x=2016\)

Vậy x=2016

b) \(M=\left(x-1\right)\left(x+2\right).\left(x+3\right)\left(x+6\right)\)

\(M=\left[\left(x-1\right)\left(x+6\right)\right].\left[\left(x+2\right).\left(x+3\right)\right]\)

\(M=\left(x^2+5x-6\right).\left(x^2+5x+6\right)=\left(x^2+5x\right)^2-36\)

Các bạn tự làm tiếp được rồi nhé

\(\frac{x-6}{2010}+\frac{x-603}{471}+\frac{x-1}{403}=9\)

\(\Rightarrow\frac{x-6}{2010}-1+\frac{x-603}{471}-3+\frac{x-1}{403}-5=9\)

\(\Rightarrow\frac{x-2016}{2010}+\frac{x-2016}{471}+\frac{x-2016}{403}=0\)

\(\Rightarrow\left(x-2016\right)\left(\frac{1}{2010}+\frac{1}{471}+\frac{1}{403}\right)=0\)

Mà \(\left(\frac{1}{2010}+\frac{1}{471}+\frac{1}{403}\right)>0\)nên x - 2016 = 0

Vậy x= 2016

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

ĐKXĐ \(x\ne-2,-3,-4\)

=> \(\frac{1}{x+2}+\frac{2}{x+3}-\frac{6}{x+4}=0\)

=> \(\frac{3x+7}{\left(x+2\right)\left(x+3\right)}-\frac{6}{x+4}=0\)

=> \(\frac{\left(3x+7\right)\left(x+4\right)-6\left(x+2\right)\left(x+3\right)}{\left(x+2\right)\left(x+3\right)\left(x+4\right)}=0\)

=> (3x + 7)(x + 4) - 6(x² + 5x + 6) = 0

=> 3x² + 19x + 28 - 6x² - 30x - 36 = 0

=> -3x² - 11x - 8 = 0

=> -3x² - 3x - 8x - 8 = 0

=> -3x(x + 1) - 8(x + 1) = 0

=> (x + 1)(-3x - 8) = 0

=> \(\orbr{\begin{cases}x=-1\\x=-\frac{8}{3}\end{cases}}\)

Vậy ...

b) Thiếu dữ liệu cuả đề 

c) \(\frac{6x+22}{x+2}-\frac{2x+7}{x+3}=\frac{x+4}{x^2+5x+6}\)

ĐKXĐ \(x\ne-2;-3\)

=> \(\frac{\left(6x+22\right)\left(x+3\right)-\left(x+2\right)\left(2x+7\right)}{\left(x+2\right)\left(x+3\right)}=\frac{x+4}{\left(x+2\right)\left(x+3\right)}\)

=> \(6x^2+40x+66-x\left(2x+7\right)-2\left(2x+7\right)=x+4\)

=> \(6x^2+40x+66-2x^2-7x-4x-14=x+4\)

=> 4x² + 29x + 52 = x + 4

=> 4x² + 29x + 52 - x - 4 = 0

=> 4x² + 28x + 48 = 0

=> 4(x² + 7x + 12) = 0

=> x² + 7x +12 = 0

=> x² + 3x + 4x + 12 = 0

=> x(x + 3) + 4(x + 3) = 0

=> (x + 3)(x + 4) = 0

=> \(\orbr{\begin{cases}x=-3\\x=-4\end{cases}}\) 

Mà \(x\ne-2,-3\)nên x = -3 loại

Vậy x = -4

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

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

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

<=> 5(x + 1)(2x - 1) - 2(x - 2)(2x - 1) = -3(x - 2)(x + 3)(x + 1)

<=> 6x² + 15x - 9 = -3x³ - 6x² + 15x + 18

<=> 6x² - 9 = -3x³ - 6x² + 18

<=> 6x² - 9 + 3x³ + 6x² - 18 = 0

<=> 12x² - 27 + 3x³ = 0

<=> 3(4x² - 9 + x³) = 0

<=> 3(x² + x - 3)(x + 3) = 0

<=> \(\orbr{\begin{cases}x=-3\\x=\frac{-1\pm\sqrt{13}}{2}\end{cases}}\)

DKXD \(x\ne\frac{1}{2};2;-1;3,;-3\)  

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

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

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

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

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

<=> \(x^2-x-2=1-2x\)

<=> \(x^2+x-3=0\)

<=> \(\orbr{\begin{cases}x=\frac{-1+\sqrt{13}}{2}\\x=\frac{-1-\sqrt{13}}{2}\end{cases}}\)

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