ĐKXĐ: bạn tự tính nhé

PT tương đương: \(\frac{5}{x-1}-\frac{5}{x-3}=\frac{2}{x+1}-\frac{2}{x-4}\)

<=>\(\frac{5x-15}{\left(x-1\right)\left(x-3\right)}-\frac{5x-5}{\left(x-1\right)\left(x-3\right)}=\frac{2x-8}{\left(x+1\right)\left(x-4\right)}-\frac{2x+2}{\left(x+1\right)\left(x-4\right)}\)

<=>\(\frac{-10}{\left(x-1\right)\left(x-3\right)}=\frac{-10}{\left(x+1\right)\left(x-4\right)}\)

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

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

=>\(\left(x+1\right)\left(x-4\right)=\left(x-1\right)\left(x-3\right)\)

Còn lại bạn từ làm nhé:)

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

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

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

\(\left(x-1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)\)=0

\(x-1=0\)

\(x=1\)

Điều kiện: x khác 0

Đặt \(\frac{x^2+1}{x}=t\Rightarrow\frac{x}{x^2+1}=\frac{1}{t}\)

Khi đó: \(\frac{x^2+1}{x}+\frac{x}{x^2+1}=\frac{5}{2}\)

\(\Leftrightarrow t+\frac{1}{t}=\frac{5}{2}\)

\(\Leftrightarrow\frac{t^2+1}{t}=\frac{5}{2}\Rightarrow2t^2+2=5t\)

\(\Leftrightarrow2t^2-5t+2=0\Leftrightarrow\left(2t-1\right)\left(t-2\right)=0\Leftrightarrow\orbr{\begin{cases}t=\frac{1}{2}\\t=2\end{cases}}\)

Nếu \(t=\frac{1}{2}\Rightarrow\frac{x^2+1}{x}=\frac{1}{2}\Rightarrow2x^2+2=x\)

\(\Leftrightarrow2x^2-x+2=0\)

Mà \(2x^2-x+2=2\left(x-\frac{1}{4}\right)^2+\frac{15}{8}>0\forall x\)

Nên \(x\in\varnothing\)

Nếu \(t=2\Rightarrow\frac{x^2+1}{x}=2\Rightarrow x^2-2x+1=0\)

\(\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)(thỏa mãn ĐKXĐ)

Tập nghiệm của pt: \(S=\left\{1\right\}\)

\(\)

Theo BĐT AM-GM,ta có: \(x^2+1\ge2\left|x\right|\ge2x\Rightarrow\frac{x^2+1}{x}\ge2\)

Đặt \(\frac{x^2+t}{x}=t\left(t\ge2\right)\).Bài toán trở thành:

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

Áp dụng BĐT AM-GM: \(VT\ge1+\frac{3t}{4}\ge1+\frac{6}{4}=\frac{5}{2}\)

Mà \(VT=\frac{5}{2}\) .Dấu "=" xảy ra khi \(\frac{1}{t}=\frac{t}{4}\Leftrightarrow t=2\Leftrightarrow\frac{x^2+1}{x}=2\Leftrightarrow x^2+1=2x\Leftrightarrow x=1\)

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

Ai ngờ bài này có cách khác chính là dùng cách chứng minh của BĐT AM-GM cho hai số  chứ! -_-"

Bắt đầu từ chỗ: \(t+\frac{1}{t}=\frac{5}{2}\) (\(t\ge2\))

Có \(\left(\sqrt{\frac{1}{t}}-\sqrt{\frac{t}{4}}\right)^2\ge0\Rightarrow\frac{1}{t}+\frac{t}{4}\ge2.\frac{1}{2}=1\)

Suy ra \(t+\frac{1}{t}=\left(\frac{1}{t}+\frac{t}{4}\right)+\frac{3t}{4}\ge1+\frac{3t}{4}\ge1+\frac{3.2}{4}=\frac{5}{2}=VP\)

Dấu bằng xảy ra khi t = 2 tức là \(x^2+1=2x\Leftrightarrow x=1\)

Vậy ....

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

ĐK: x khác 0.

Đặt: \(\frac{x^2+1}{x}=t\ne0\)

Ta có phương trình ẩn t: \(t+\frac{1}{t}=\frac{5}{2}\Leftrightarrow2t^2-5t+2=0\Leftrightarrow\orbr{\begin{cases}t=2\\t=\frac{1}{2}\end{cases}}\)thỏa mãn

Với t = 2 ta có: \(\frac{x^2+1}{x}=2\Leftrightarrow x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)thỏa mãn

Với t =1/2 ta có: \(\frac{x^2+1}{x}=\frac{1}{2}\Leftrightarrow x^2-\frac{1}{2}x+1=0\Leftrightarrow\left(x^2-2.x.\frac{1}{4}+\frac{1}{16}\right)+\frac{15}{16}=0\)

<=> \(\left(x-\frac{1}{4}\right)^2+\frac{15}{16}=0\)phương trình vô nghiệm 

Vậy x = 1

\(\frac{x^2+1}{x}+\frac{x}{x^2+1}=\frac{5}{2}\)ĐKXĐ : \(x\ne0\)

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

Khử mẫu ta đc : \(2\left(x^2+1\right)^2+2x^2=5x\left(x^2+1\right)\)

\(2x^4+4x^2+2+2x^2=5x^3+5x\)

\(2x^4+6x^2+2=5x^3+5x\)

\(2x^4+6x^2+2-5x^3-5x=0\)

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

TH1 : \(2x^2-x+2=0\)

Ta có : \(\left(-1\right)^2-4.2.2=1-16=-15< 0\)

Nên phương trình vô nghiệm 

TH2 : \(\left(x-1\right)^2=0\Leftrightarrow x-1=0\Leftrightarrow x=1\)

Vậy nghiệm phương trình là 1 

ĐK:\(x\ne-1;-3;-5;-7;-9\)

\(pt\Leftrightarrow\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{2}{\left(x+3\right)\left(x+5\right)}+\frac{2}{\left(x+5\right)\left(x+7\right)}+\frac{2}{\left(x+7\right)\left(x+9\right)}=\frac{2}{5}\)

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-...-\frac{1}{x+9}=\frac{2}{5}\)

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+9}=\frac{2}{5}\)\(\Leftrightarrow\frac{8}{\left(x+1\right)\left(x+9\right)}=\frac{2}{5}\)

\(\Leftrightarrow2\left(x+1\right)\left(x+9\right)=40\)\(\Leftrightarrow x^2+10x-11=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x+11=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=1\\x=-11\end{cases}}\) (thoả)

Vậy....

xin lỗi nha, bài đó bằng có một cái 1/5 thôi, tại viết sai

ĐK : \(X\ne-1;-3;-7;-9\)

\(\frac{1}{x^2+4x+3}+\frac{1}{x^2+8x+15}+\frac{1}{x^2+12x+35}+\frac{1}{x^2+16x+63}=\frac{1}{5}\)

\(\frac{1}{\left(x+2\right)^2-1}+\frac{1}{\left(x+4\right)^2-1}+\frac{1}{\left(x+6\right)^2-1}+\frac{1}{\left(x-8\right)^2-1}=\frac{1}{5}\)

\(\frac{1}{\left(x+2-1\right)\left(x+2+1\right)}+\frac{1}{\left(x+4-1 \right)\left(x+4+1\right)}+\frac{1}{\left(x+6-1\right)\left(x+6+1\right)}+\frac{1}{\left(x+8-1\right)\left(x+8+1\right)}=\frac{1}{5}\)

\(\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+7\right)}+\frac{1}{\left(x+7\right)\left(x+9\right)}=\frac{1}{5}\)

\(\frac{1}{2}\cdot\left(\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+....-\frac{1}{x+9}\right)=\frac{1}{5}\)

\(\frac{1}{2}\cdot\left(\frac{1}{x+1}-\frac{1}{x+9}\right)=\frac{1}{5}\)

\(\frac{1}{x+1}-\frac{1}{x+9}=\frac{1}{5}:\frac{1}{2}=\frac{2}{5}\)

\(\frac{8}{\left(x+1\right)\left(x+9\right)}=\frac{2}{5}\)

\(2\left(x+1\right)\left(x+9\right)=40\)

\(2x^2+20x+18=40\Leftrightarrow x^2+10x+9=20\)

\(\Leftrightarrow x^2+10x-11=0\Leftrightarrow x^2+10x-10-1=0\)

\(\Leftrightarrow\left(x^2-1\right)+\left(10x-10\right)=0\Leftrightarrow\left(x-1\right)\left(x+1\right)+10\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+11\right)=0\)

\(\orbr{\begin{cases}x-1=0\\x++11=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-11\end{cases}}}\)( Thõa mãn ) 

Vậy ...............

x^2 + 4x + 3 = (x+1)(x+3)

x^2 + 8x + 15 = (x+3)(x+5)

x^2 + 12x + 35 = (x+5)(x+7)

x^2 + 16x + 63 = (x+7)(x+9)

Bạn phân tích ra quy luật rồi thì bạn giải tiếp sẽ có: 

             1/x+1 -1/x+9 = 2/5

             8/(x+1)(x+9) =2/5

             (x+1)(x+9) = 20

             x^2 +10x+9 = 20

             x^2 +10x -11 = 0

             (x-1)(x+11) = 0

Vậy x=1 hoặc x= -11(thỏa măn ĐKXĐ)          

2) đặt \(x^2+x+1=t\left(t>0\right)\)   ==> \(x^2+x+2=t+1\)

nên pt trên trở thành 

\(\left(\frac{1}{t}\right)^2+\left(\frac{1}{t+1}\right)^2=\frac{13}{36}\)

<=> \(\frac{1}{t^2}+\frac{1}{t^2+2t+1}=\frac{13}{36}\)

<=> \(13t^4+26t^3-59t^2-72t-36=0\)

<=> \(13t^4-26t^3+52t^3-104t^2+45t^2-90t+18t-36=0\)

<=> \(13t^3\left(t-2\right)+52t^2\left(t-2\right)+45t\left(t-2\right)+18\left(t-2\right)=0\)

<=>\(\left(t-2\right)\left(13t^3+52t^2+45t+18\right)=0\)

<=> \(\left(t-2\right)\left(t+3\right)\left(13t^2+13t+6\right)=0\)

<=> \(\orbr{\begin{cases}t=2\left(tmdk\right)\\t=-3\left(ktmdk\right)\end{cases}}\)

đến đây bạn thay vào làm nốt nhá

1.

Đặt \(a=\frac{x\left(5-x\right)}{x+1};b=x+\frac{5-x}{x+1}\)

Ta cần giải pt : \(a.b=6\)(1)

Ta có: \(a+b=\frac{x\left(5-x\right)}{x+1}+x+\frac{5-x}{x+1}=\frac{5x-x^2+x^2+x+5-x}{x+1}=5\)

\(\Rightarrow a=5-b\)

Thế \(a=5-b\)vào (1)

\(\Rightarrow\left(5-b\right)b=6\)

\(\Leftrightarrow b^2-5b+6=0\)

\(\Leftrightarrow\left(b-2\right)\left(b-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}b=2\\b=3\end{cases}\Leftrightarrow\orbr{\begin{cases}x+\frac{5-x}{x+1}=2\\x+\frac{5-x}{x+1}=3\end{cases}}}\)

Giải 2 pt trên, ta có nghiệm : \(x=1\)

Vừa lm xong mt bị sụp ... 

\(\frac{1}{x-1}+\frac{3}{3x+5}=\frac{2}{x+2}+\frac{1}{x+3}\)ĐKXĐ : \(x\ne1;-\frac{5}{3};-2;-3\)

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

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

Khử mẫu và rút gọn ta đc : \(-3x^3+2x^2+45x+52=0\)

Mời cao nhân giải tiếp.

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

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

\(\Leftrightarrow\frac{9x-1}{3x^2+2x-5}=\frac{3x+8}{x^2+5x+6}\)

\(\Rightarrow9x^3+44x^2+49x-6=9x^3+30x^2+x-40\)

\(\Leftrightarrow14x^2-48x+34=0\)

\(\Rightarrow14x^2-14x-34x+34=0\)

\(\Rightarrow\left(x-1\right)\left(14x-34\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\14x-34=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{17}{7}\end{cases}}}\)

Ngu nên làm dài dòng thôi