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A
14 tháng 12 2018
Thặc vler .V
A/\(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}\)
\(=\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}\)
\(=\left[\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}\right]+\left[\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}\right]\)
\(=\left[\frac{x+3}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}+\frac{x+1}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}\right]+\left[\frac{x+5}{\left(x+3\right)\left(x+4\right)\left(x+5\right)}+\frac{x+3}{\left(x+3\right)\left(x+4\right)\left(x+5\right)}\right]\)
\(=\frac{2x+4}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}+\frac{2x+8}{\left(x+3\right)\left(x+4\right)\left(x+5\right)}\)
\(=\frac{2\left(x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}+\frac{2\left(x+4\right)}{\left(x+3\right)\left(x+4\right)\left(x+5\right)}\)
\(=\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{2}{\left(x+3\right)\left(x+5\right)}\)
\(=\frac{2x+10}{\left(x+1\right)\left(x+3\right)\left(x+5\right)}+\frac{2x+2}{\left(x+1\right)\left(x+3\right)\left(x+5\right)}\)
\(=\frac{4x+12}{\left(x+1\right)\left(x+3\right)\left(x+5\right)}\)
\(=\frac{4\left(x+3\right)}{\left(x+1\right)\left(x+3\right)\left(x+5\right)}\)
\(=\frac{4}{\left(x+1\right)\left(x+5\right)}\)
B/\(\frac{x-1}{x-2}+\frac{1}{2-x}\)
\(=\frac{x-1}{x-2}-\frac{1}{x-2}\)
\(=\frac{x-1-1}{x-2}\)
\(=\frac{x-2}{x-2}\)
\(=1\)
11 tháng 12 2019
\(a)=\frac{-2\left(x+3\right)}{x\left(1-3x\right)}.\frac{1-3x}{x\left(x+3\right)}\)
\(=\frac{-2}{x^2}\)
\(b)=\frac{\left(x+3\right)\left(x-3\right)}{x\left(x-3\right)}-\frac{x^2}{x\left(x-3\right)}+\frac{9}{x\left(x-3\right)}\)
\(=\frac{x^2-3x+3x-9-x^2+9}{x\left(x-3\right)}\)
\(=x\left(x-3\right)\)
\(c)=\frac{x+3}{\left(x-1\right)\left(x+1\right)}-\frac{1}{x\left(x+1\right)}\)
\(=\frac{\left(x+3\right).x}{x\left(x-1\right)\left(x+1\right)}-\frac{1.\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2+3x-x+1}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x\left(x+3\right)-\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x+3}{x+1}\)
# Sắp ik ngủ nên làm vậy hoi, ko chắc phần kq câu b và c đâu nha
KT
12 tháng 1 2018
\(ĐKXĐ:\)\(x\ne\left\{0;1;2;3;4;5\right\}\)
\(P=\frac{1}{x^2-x}+\frac{1}{x^2-3x+2}+\frac{1}{x^2-5x+6}+\frac{1}{x^2-7x+12}+\frac{1}{x^2-9x+20}\)
\(=\frac{1}{x\left(x-1\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}\)
\(=\frac{1}{x-1}-\frac{1}{x}+\frac{1}{x-2}-\frac{1}{x-1}+\frac{1}{x-3}-\frac{1}{x-2}+\frac{1}{x-4}-\frac{1}{x-3}+\frac{1}{x-5}-\frac{1}{x-4}\)
\(=\frac{1}{x-5}-\frac{1}{x}\)
\(=\frac{5}{x\left(x-5\right)}\)
Ta có: \(x^3-x^2+2=0\)
\(\Leftrightarrow\)\(\left(x+1\right)\left(x^2-2x+2\right)=0\)
Xét: \(x^2-2x+2=\left(x-1\right)^2+1\)\(>0\)
\(\Rightarrow\)\(x+1=0\)
\(\Leftrightarrow\)\(x=-1\)(t/m)
Vậy tại \(x=-1\) thì:
\(P=\frac{5}{-1\left(-1-5\right)}=\frac{5}{6}\)
12 tháng 1 2018
ĐKXĐ \(x\ne0,1,2,3,4,5\)
\(P=\frac{1}{x\left(x-1\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}\)
\(P=\frac{1}{x-1}-\frac{1}{x}+\frac{1}{x-2}-\frac{1}{x-1}+...+\frac{1}{x-5}-\frac{1}{x-4}\)
\(P=\frac{1}{x-5}-\frac{1}{x}\)
\(P=\frac{5}{x\left(x-5\right)}\)
4 tháng 5 2019
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 \(S=\varnothing\)
Vẫn tách như trên nha cậu .
\(=\frac{1}{x-1}-\frac{1}{x}+\frac{1}{x-2}-\frac{1}{x-1}+...+\frac{1}{x-5}-\frac{1}{x-4}\)
\(=-\frac{1}{x}+\frac{1}{x-5}=\frac{x-x+5}{x\left(x-5\right)}=\frac{5}{x\left(x-5\right)}\)
Vậy \(D=\frac{5}{x\left(x-5\right)}\)
Tách như bước ở dấu "=" thứ hai nhé bước mà có \(\frac{1}{x\left(x-1\right)}\)nhé