ĐKXĐ:\(x\ne1;2;3;4;5\)

\(\Leftrightarrow\frac{1}{x^2-x-2x+2}+\frac{1}{x^2-2x-3x+6}+\frac{1}{x^2-3x-4x+12}+\frac{1}{x^2-4x-5x+20}=\frac{1}{15}\)

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

\(\Leftrightarrow\frac{1}{x-1}-\frac{1}{x-2}+\frac{1}{x-2}-\frac{1}{x-3}+\frac{1}{x-3}-\frac{1}{x-4}+\frac{1}{x-4}-\frac{1}{x-5}=\frac{1}{15}\)

\(\Leftrightarrow\frac{1}{x-1}-\frac{1}{x-5}=\frac{1}{15}\)

\(\Leftrightarrow\frac{15\left(x-5\right)-15\left(x-1\right)}{15\left(x-1\right)\left(x-5\right)}=\frac{\left(x-1\right)\left(x-5\right)}{15\left(x-1\right)\left(x-5\right)}\)

\(\Rightarrow15x-75-15x+15=x^2-6x+5\)

\(\Leftrightarrow x^2-6x+65=0\)

\(\Leftrightarrow\left(x^2-6x+9\right)+56=0\)

\(\Leftrightarrow\left(x-3\right)^2=-56\) (Vô lý)

Vì bình phương một số không thể bằng âm

Vây \(S=\varnothing\)

tao đéo biết

gợi ý bn nhé:
1/(x²-3x+2) = 1/(x-2)  -   1/(x-1)

tương tự nhé

\(\Leftrightarrow\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}=\frac{1}{8}\)

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

\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+6}=\frac{1}{8}\)

\(\Leftrightarrow\frac{4}{\left(x+2\right)\left(x+6\right)}=\frac{4}{32}\)

\(\Rightarrow x^2+8x+12=32\)

\(\Leftrightarrow x^2+8x-20=0\)

Đến đây đơn giản rồi nhé

\(\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}=\frac{1}{8}\)

\(\Leftrightarrow\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}=\frac{1}{8}\)

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

\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+6}=0\)

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

\(\Leftrightarrow x+6=x+2\)

\(\Leftrightarrow x-x=2-6\)

\(\Leftrightarrow0x=-4\)

=> PT vô nghiệm

@ Harley @ ơi sai rồi, phải quy đồng như bạn Vân Nhi chứ không thể rút tử như thế

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

ĐK:\(x\ne-2;-3;-4;-5\)

MTC:\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right).6\)

Quy đồng khử mẫu:

ĐK: \(x\ne-2;-3;-4;-5;-6\)

\(\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}=\frac{1}{8}\)

\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}=\frac{1}{8}\)

\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+6}=\frac{1}{8}\)

\(\Leftrightarrow\frac{4}{\left(x+2\right)\left(x+6\right)}=\frac{1}{8}\Leftrightarrow\left(x+2\right)\left(x+6\right)=32\)

\(\Leftrightarrow x^2+8x-20=0\Rightarrow\left[{}\begin{matrix}x=2\\x=-10\end{matrix}\right.\)

\(...\Leftrightarrow\frac{1}{\left(x+2\right) \left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}=\frac{1}{8}\)

\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}=\frac{1}{8}\)

\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+6}=\frac{1}{18}\Leftrightarrow\frac{x+6}{\left(x+2\right)\left(x+6\right)}-\frac{x+2}{\left(x+2\right)\left(x+6\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{x+6-x-2}{\left(x+2\right)\left(x+6\right)}=\frac{1}{18}\Rightarrow\frac{4}{\left(x+2\right)\left(x+6\right)}=\frac{1}{18}\)

\(\Rightarrow\left(x+2\right)\left(x+6\right)=72\)

=> \(x^2+8x-60=0\)

Phân tich đa thức thành nhân tử để tìm x

Đk x khác -2;-3;-4;-5

pt <=> 1/(x+2).(x+3) + 1/(x+3).(x+4) + 1/(x+4).(x+5) = 1/6

<=> 1/x+2 - 1/x+3 + 1/x+3 - 1/x+4 + 1/x+4 - 1/x+5 = 1/6

<=> 1/x+2 - 1/x+5 = 1/6

<=> x+5-x-2/(x+2).(x+5) = 1/6

<=> 3/(x+2).(x+5) = 1/6

<=> (x+2).(x+5) = 3 : 1/6 = 18

<=> x^2+7x+10 = 18

<=> x^2+7x-8=0

<=> (x-1).(x+8) = 0

<=> x1=0 hoặc x+8=0

<=> x=1 hoặc x=-8

k mk nha

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

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

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

         \(=\frac{\left(x-6\right)-\left(x-2\right)}{\left(x-2\right)\left(x-6\right)}\)

          \(=\frac{4}{\left(x-2\right)\left(x-6\right)}\)

pt <=> 1/(x+2).(x+3) + 1/(x+3).(x+4) + 1/(x+4).(x+5) + 1/(x+5).(x+6) = 1/8

<=> 1/x+2 - 1/x+3 + 1/x+3 - 1/x+4 + 1/x+4 - 1/x+5 + 1/x+5 - 1/x+6 = 1/8

<=> 1/x+2 - 1/x+6 = 1/8

<=> (x+6-x-2)/(x+2).(x+6) = 1/8

<=> 4/(x+2).(x+6) = 1/8

<=>(x+2).(x+6) = 4 : 1/8 = 32

<=>x^2 + 8x + 12 = 32

<=> x^2+8x+12-32=0

<=>x^2+8x-20=0

<=>(x-2).(x+10)=0

<=> x-2 =0 hoặc x+10 = 0

<=> x=2 hoặc x=-10

giang sinh an lanh $%###Xuyen gam cu chuoi###%$