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HQ
7
15 tháng 2 2017
\(\frac{3}{\left(a-2\right)\left(a-3\right)}\). minh khong chac dau nha. neu sai thi thoi.
1 tháng 7 2020
a) \(A=\frac{1}{a^2+a}+\frac{1}{a^2+3a+2}+\frac{1}{a^2+5a+6}+\frac{1}{a^2+7a+12}+\frac{1}{a^2+9a+20}\)
\(A=\frac{1}{a\left(a+1\right)}+\frac{1}{\left(a+1\right)\left(a+2\right)}+\frac{1}{\left(a+2\right)\left(a+3\right)}+\frac{1}{\left(a+3\right)\left(a+4\right)}+\frac{1}{\left(a+4\right)\left(a+5\right)}\)
\(A=\frac{1}{a}-\frac{1}{a+1}+\frac{1}{a+1}-\frac{1}{a+2}+\frac{1}{a+2}-\frac{1}{a+3}+\frac{1}{a+3}-\frac{1}{a+4}+\frac{1}{a+4}-\frac{1}{a+5}\)
\(A=\frac{1}{a}-\frac{1}{a+5}=\frac{a+5-a}{a\left(a+5\right)}=\frac{5}{a^2+5a}\)
b) Điều kiện: \(a\ne0;-1;-2;-3;-4;-5\)
\(A>\frac{5}{6}\) \(\Leftrightarrow\frac{5}{a^2+5a}>\frac{5}{6}\) \(\Leftrightarrow\frac{5}{a^2+5a}-\frac{5}{6}>0\) \(\Leftrightarrow\frac{30-5a^2-25a}{30\left(a^2+5a\right)}>0\)
\(\Leftrightarrow\left[{}\begin{matrix}-6< a< -5\\0< a< 1\end{matrix}\right.\)
Kết luận: ....
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 7 2020
ĐKXĐ: ...
a/ \(A=\frac{1}{a\left(a+1\right)}+\frac{1}{\left(a+1\right)\left(a+2\right)}+\frac{1}{\left(a+2\right)\left(a+3\right)}+\frac{1}{\left(a+3\right)\left(a+4\right)}+\frac{1}{\left(a+4\right)\left(a+5\right)}\)
\(=\frac{1}{a}-\frac{1}{a+1}+\frac{1}{a+1}-\frac{1}{a+2}+...+\frac{1}{a+4}-\frac{1}{a+5}\)
\(=\frac{1}{a}-\frac{1}{a+5}=\frac{5}{a\left(a+5\right)}\)
\(A>\frac{5}{6}\Rightarrow\frac{5}{a\left(a+5\right)}>\frac{5}{6}\)
\(\Leftrightarrow\frac{1}{a\left(a+5\right)}-\frac{1}{6}>0\Leftrightarrow\frac{6-a^2-5a}{a\left(a+5\right)}>0\)
\(\Leftrightarrow\frac{\left(1-a\right)\left(a+6\right)}{a\left(a+5\right)}>0\Rightarrow\left[{}\begin{matrix}-6< a< -5\\0< a< 1\end{matrix}\right.\)
N
24 tháng 6 2019
\(P=\frac{1}{a^2-a}+\frac{1}{a^2-3a+2}+\frac{1}{a^2-5a+6}+\frac{1}{a^2-7a+12}+\frac{1}{a^2-9a+20}\)
\(=\frac{1}{a.\left(a-1\right)}+\frac{1}{\left(a-1\right).\left(a-2\right)}+\frac{1}{\left(a-2\right).\left(a-3\right)}+\frac{1}{\left(a-3\right).\left(a-4\right)}+\frac{1}{\left(a-4\right).\left(a-5\right)}\)
a) ĐKXĐ: \(a\ne0;1;2;3;4;5;6\)
b) \(P=\frac{1}{a-1}-\frac{1}{a}+\frac{1}{a-2}-\frac{1}{a-1}+\frac{1}{a-3}-\frac{1}{a-2}+\frac{1}{a-4}-\frac{1}{a-3}+\frac{1}{a-5}-\frac{1}{a-4}\)
\(A=\frac{1}{a-5}-\frac{1}{a}=\frac{a-\left(a-5\right)}{a.\left(a-5\right)}=\frac{5}{a.\left(a-5\right)}\)
c) \(a^3-a^2+2=0\)
\(\Leftrightarrow a^3+a^2-2a^2-2a+2a+2=0\)
\(\Leftrightarrow a^2.\left(a+1\right)-2a.\left(a+1\right)+2.\left(a+1\right)=0\)
\(\Leftrightarrow\left(a+1\right).\left(a^2-2a+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a+1=0\\a^2-2a+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}a=-1\\\left(a-1\right)^2=-1\left(loai\right)\end{cases}}}\)
Thay a=-1 vào P
\(P=\frac{5}{a.\left(a-5\right)}=\frac{5}{-1.\left(-1-5\right)}=\frac{5}{6}\)
MT
3
1 tháng 3 2017
\(A=\dfrac{1}{\left(a-2\right)\left(a-3\right)}+\dfrac{1}{\left(a-3\right)\left(a-4\right)}+\dfrac{1}{\left(a-4\right)\left(a-5\right)}\)\(A=\dfrac{1}{a-2}-\dfrac{1}{a-3}+\dfrac{1}{a-3}+\dfrac{1}{a-4}-\dfrac{1}{a-4}+\dfrac{1}{a-4}-\dfrac{1}{a-5}\)\(A=\dfrac{1}{a-2}-\dfrac{1}{a-5}=\dfrac{-3}{\left(a-2\right)\left(a-5\right)}\)
1 tháng 3 2017
\(\dfrac{1}{a^2-5a+6}+\dfrac{1}{a^2-7a+12}+\dfrac{1}{a^2-9a+20}\)
\(\Leftrightarrow\dfrac{1}{\left(a-2\right)\left(a-3\right)}+\dfrac{1}{\left(a-3\right)\left(a-4\right)}+\dfrac{1}{\left(a-4\right)\left(a-5\right)}\)
\(\Leftrightarrow\dfrac{\left(a-4\right)\left(a-5\right)}{\left(a-2\right)\left(a-3\right)\left(a-4\right)\left(a-5\right)}+\dfrac{\left(a-2\right)\left(a-5\right)}{\left(a-2\right)\left(a-3\right)\left(a-4\right)\left(a-5\right)}+\dfrac{\left(a-2\right)\left(a-3\right)}{\left(a-2\right)\left(a-3\right)\left(a-4\right)\left(a-5\right)}\)
\(\Leftrightarrow\dfrac{\left(a-4\right)\left(a-5\right)+\left(a-2\right)\left(a-5\right)+\left(a-2\right)\left(a-3\right)}{\left(a-2\right)\left(a-3\right)\left(a-4\right)\left(a-5\right)}\)
\(\Leftrightarrow\dfrac{\left(a-4\right)\left(a-5\right)+\left(a-2\right)\left[\left(a-3\right)+\left(a-5\right)\right]}{\left(a-2\right)\left(a-3\right)\left(a-4\right)\left(a-5\right)}\)
\(\Leftrightarrow\dfrac{\left(a-4\right)\left(a-5\right)+\left(a-2\right)\left(a-4\right)2}{\left(a-2\right)\left(a-3\right)\left(a-4\right)\left(a-5\right)}\)
\(\Leftrightarrow\dfrac{\left(a-4\right)\left[\left(a-5\right)+2\left(a-2\right)\right]}{\left(a-2\right)\left(a-3\right)\left(a-4\right)\left(a-5\right)}\)
\(\Leftrightarrow\dfrac{3a-9}{\left(x-2\right)\left(x-3\right)\left(x-5\right)}\)
\(\Leftrightarrow\dfrac{3\left(a-3\right)}{\left(a-2\right)\left(a-3\right)\left(a-5\right)}\)
\(\Leftrightarrow\dfrac{3}{\left(a-2\right)\left(a-5\right)}\)
NT
1
G
19 tháng 3 2018
\(M=\frac{1}{a^2-5a+6}+\frac{1}{a^2-7a+12}+\frac{1}{a^2-9a+20}+\frac{1}{a^2-11a+30}\)
\(M=\frac{1}{\left(a-2\right)\left(a-3\right)}+\frac{1}{\left(a-3\right)\left(a-4\right)}+\frac{1}{\left(a-4\right)\left(a-5\right)}+\frac{1}{\left(a-5\right)\left(a-6\right)}\)
\(M=\frac{1}{a-2}-\frac{1}{a-3}+\frac{1}{a-3}-\frac{1}{a-4}+\frac{1}{a-4}-\frac{1}{a-5}+\frac{1}{a-5}-\frac{1}{a-6}\)
\(M=\frac{1}{a-2}-\frac{1}{a-6}\)
25 tháng 1 2017
Bài 1:
\(B=\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}+\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}\)
\(=\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{-\left(0,625-0,5+\frac{5}{11}+\frac{5}{12}\right)}+\frac{3\left(0,5+\frac{1}{3}-0,25\right)}{5\left(0,5+\frac{1}{3}-0,25\right)}\)
\(=\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{-\left[5\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)\right]}+\frac{3}{5}\)
\(=\frac{-3}{5}+\frac{3}{5}\)
\(=0\)
Bài 2:
b) Giải:
Ta có: \(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^6}{b^6}=\frac{c^6}{d^6}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a^6}{b^6}=\frac{c^6}{d^6}=\frac{3a^6}{3b^6}=\frac{c^6}{d^6}=\frac{3a^6+c^6}{3b^6+d^6}\) (1)
\(\frac{a}{b}=\frac{c}{d}=\frac{a+b}{b+d}\)
\(\Rightarrow\left(\frac{a}{b}\right)^6=\left(\frac{a+c}{b+d}\right)^6=\frac{a^6}{b^6}=\frac{\left(a+c\right)^6}{\left(b+d\right)^6}\) (2)
Từ (1) và (2) \(\Rightarrow\frac{3a^6+c^6}{3b^6+d^6}=\frac{\left(a+c\right)^6}{\left(b+d\right)^6}\left(đpcm\right)\)