\(\frac{2004x}{2x^2+x+1}+\frac{2005x}{2x^2+x+1}=902\)

\(\Leftrightarrow\frac{2004x+2005x}{2x^2+x+1}=902\)

\(\Leftrightarrow\frac{4009x}{2x^2+x+1}=902\)

\(\Leftrightarrow4009x=902\left(2x^2+x+1\right)\)

\(\Leftrightarrow4009x=1804x^2+902x+902\)

\(\Leftrightarrow-1804x^2+3107x=902\)

Bn tự làm tiếp. Số to quá bn -.-

Đặt x² + 2x = a ta có

\(\frac{1}{a-3}\)+ \(\frac{18}{a+2}\)= \(\frac{18}{a+1}\)

<=> a² - 15a + 56 = 0

<=> a = (7;8)

Thế vô tìm được nghiệm 

\(ĐKXĐ:x\ne-1\)

Từ phương trình suy ra \(\frac{x^2-x+1}{x^3+1}+\frac{2x^2+1}{x^3+1}+\frac{2x^2\left(x-1\right)\left(x+1\right)}{x^3+1}=2x\)

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

\(\Leftrightarrow2x^4+x^2-x+2=2x^4+2x\)

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

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

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

Vậy tập nghiệm của phương trình \(S=\left\{1;2\right\}\)

\(\Leftrightarrow\frac{7x+4}{\sqrt{2\left(x-1\right)\left(x+1\right)}}+\frac{2\sqrt{2x+1}}{\sqrt{2\left(x+1\right)}}=3+\frac{3\sqrt{2x+1}}{\sqrt{x-1}}\)

\(\Leftrightarrow7x+4+2\sqrt{\left(2x+1\right)\left(x-1\right)}=3\sqrt{2\left(x-1\right)\left(x+1\right)}+3\sqrt{2\left(2x+1\right)\left(x+1\right)}\)

 \(\Leftrightarrow\left(7x+4+\sqrt{8x^2-4x-4}\right)^2=\left(\sqrt{18x^2-18}+\sqrt{36^2+54x+18}\right)^2\)

\(\Leftrightarrow\left(7x+4\right)^2+8x^2-4x-4+2\left(7x+4\right)\sqrt{8x^2-4x-4}\)\(=18x^2-18+36x^2+54x+18+2\sqrt{\left(18x^2-18\right)\left(36x^2+54x+18\right)}\)

\(\Leftrightarrow3x^2-2x+12+4\left(7x+4\right)\sqrt{\left(x-1\right)\left(2x+1\right)}=36\left(x+1\right)\sqrt{\left(x-1\right)\left(2x+1\right)}\)

\(\Leftrightarrow3x^2-2x+12=4\left(2x+5\right)\sqrt{\left(x-1\right)\left(2x+1\right)}\)

\(\Leftrightarrow\left(3x^2-2x+12\right)^2=16\left(2x+5\right)^2\left(x-1\right)\left(2x+1\right)\)

\(\Leftrightarrow119x^4+588x^3+1940x^2-672x-544=0\left(1\right)\)

Ta thấy x>1 => Vế trái (1) \(>119.1^4+588.1^3+1940.1^2-672.1-544=1431>0\)

=> pt vô nghiệm.

đặt \(a=2x^2-x+1\)
\(\Rightarrow\frac{1}{a}+\frac{3}{a+2}=\frac{10}{a+6}\)
Đến đây đơn giản r

\(2+\frac{2x^2-8x}{2x^2+8x}+\frac{2x^2+7x+23}{2x^2+7x-4}=\frac{2x+5}{2x-1}\)

\(\Leftrightarrow2+\frac{2x\left(x-4\right)}{2x\left(x+4\right)}+\frac{2x^2+7x+23}{\left(2x-1\right)\left(x+4\right)}=\frac{2x+5}{2x-1}\)

\(\Leftrightarrow2+\frac{x-4}{x+4}+\frac{2x^2+7x+23}{\left(2x-1\right)\left(x+4\right)}-\frac{2x+5}{2x-1}=0\)

\(\Leftrightarrow\frac{2\left(x+4\right)\left(2x-1\right)}{\left(x+4\right)\left(2x-1\right)}+\frac{\left(x-4\right)\left(2x-1\right)}{\left(x+4\right)\left(2x-1\right)}+\frac{2x^2+7x+23}{\left(2x-1\right)\left(x+4\right)}-\frac{\left(2x+5\right)\left(x+4\right)}{\left(2x-1\right)\left(x+4\right)}=0\)

\(\Leftrightarrow\frac{2\left(x+4\right)\left(2x-1\right)+\left(x-4\right)\left(2x-1\right)+2x^2+7x+23-\left(2x+5\right)\left(x+4\right)}{\left(x+4\right)\left(2x-1\right)}=0\)

\(\Leftrightarrow2\left(x+4\right)\left(2x-1\right)+\left(x-4\right)\left(2x-1\right)+2x^2+7x+23-\left(2x+5\right)\left(x+4\right)=0\)

\(\Leftrightarrow2\left(2x^2+7x-4\right)+\left(2x^2-9x+4\right)+2x^2+7x+23-\left(2x^2+13x+20\right)=0\)

\(\Leftrightarrow4x^2+14x-8+2x^2-9x+4+2x^2+7x+23-2x^2-13x-20=0\)

\(\Leftrightarrow6x^2+7x-1=0\)

\(\Leftrightarrow6\left(x^2+2.\frac{7}{12}.x+\frac{49}{144}\right)-\frac{193}{144}=0\)

\(\Leftrightarrow\left(x+\frac{7}{12}\right)^2=\frac{\frac{193}{144}}{6}=\frac{193}{864}\)

Bạn tự làm nốt.

Tương tự với Cb.

Có chắc là đề ổn không bạn?

Hoặc là xem bài mình hộ với; ngộ nhỡ mình sai. Chứ kết quả lẻ quá; chẳng đẹp gì :>