Rút gọn biểu thức:
A= \(\frac{1}{a^2-5a+6}+\frac{1}{a^2-7a+12}+\frac{1}{a^2-9a+20}\)
K
Khách
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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
\(P=\frac{5}{a.\left(a-5\right)}=\frac{5}{-1.\left(-1-5\right)}=\frac{5}{6}\)
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/ \(A=\frac{1}{a\left(a+1\right)}+\frac{1}{\left(a+1\right)\left(a+2\right)}+\frac{1}{\left(a+2\right)\left(a+3\right)}+\frac{1}{\left(a+3\right)\left(a+4\right)}+\frac{1}{\left(a+4\right)\left(a+5\right)}\)
\(=\frac{1}{a}-\frac{1}{a+1}+\frac{1}{a+1}-\frac{1}{a+2}+...+\frac{1}{a+4}-\frac{1}{a+5}\)
\(=\frac{1}{a}-\frac{1}{a+5}=\frac{5}{a\left(a+5\right)}\)
\(A>\frac{5}{6}\Rightarrow\frac{5}{a\left(a+5\right)}>\frac{5}{6}\)
\(\Leftrightarrow\frac{1}{a\left(a+5\right)}-\frac{1}{6}>0\Leftrightarrow\frac{6-a^2-5a}{a\left(a+5\right)}>0\)
\(\Leftrightarrow\frac{\left(1-a\right)\left(a+6\right)}{a\left(a+5\right)}>0\Rightarrow\left[{}\begin{matrix}-6< a< -5\\0< a< 1\end{matrix}\right.\)
IM
10 tháng 8 2016
Đăỵ tổng là A
\(\Rightarrow A=\frac{1}{a^2-5a-4+10}+\frac{1}{a^2-7a-16+28}+\frac{1}{a^2-9a-25+45}+\frac{1}{a^2-11a-36+66}\)
\(\Rightarrow A=\frac{1}{\left(a^2-4\right)-\left(5a-10\right)}+\frac{1}{\left(a^2-16\right)-\left(7a-28\right)}+\frac{1}{\left(a^2-25\right)-\left(9a-45\right)}+\frac{1}{\left(a^2-36\right)-\left(11a-66\right)}\)
\(\Rightarrow A=\frac{1}{\left(a+2\right)\left(a-2\right)-5\left(a-2\right)}+\frac{1}{\left(a+4\right)\left(a-4\right)-7\left(a-4\right)}+\frac{1}{\left(a-5\right)\left(a+5\right)-9\left(a-5\right)}+\frac{1}{\left(a-6\right)\left(a+6\right)-11\left(a-6\right)}\)
\(\Rightarrow A=\frac{1}{\left(a-2\right)\left(a-3\right)}+\frac{1}{\left(a-4\right)\left(a-3\right)}+\frac{1}{\left(a-5\right)\left(a-4\right)}+\frac{1}{\left(a-6\right)\left(a-5\right)}\)
\(\Rightarrow A=\frac{1}{a-3}-\frac{1}{a-2}+\frac{1}{a-4}-\frac{1}{a-3}+\frac{1}{a-5}-\frac{1}{a-4}+\frac{1}{a-6}-\frac{1}{a-5}\)
\(\Rightarrow A=\frac{1}{a-6}-\frac{1}{a-2}\)
\(\Rightarrow A=\frac{\left(a-2\right)-\left(a-6\right)}{\left(a-6\right)\left(a-2\right)}=\frac{4}{\left(a-6\right)\left(a-2\right)}\)
NT
1
G
19 tháng 3 2018
\(M=\frac{1}{a^2-5a+6}+\frac{1}{a^2-7a+12}+\frac{1}{a^2-9a+20}+\frac{1}{a^2-11a+30}\)
\(M=\frac{1}{\left(a-2\right)\left(a-3\right)}+\frac{1}{\left(a-3\right)\left(a-4\right)}+\frac{1}{\left(a-4\right)\left(a-5\right)}+\frac{1}{\left(a-5\right)\left(a-6\right)}\)
\(M=\frac{1}{a-2}-\frac{1}{a-3}+\frac{1}{a-3}-\frac{1}{a-4}+\frac{1}{a-4}-\frac{1}{a-5}+\frac{1}{a-5}-\frac{1}{a-6}\)
\(M=\frac{1}{a-2}-\frac{1}{a-6}\)
TA
0
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
\(\dfrac{1}{a^2-5a+6}+\dfrac{1}{a^2-7a+12}+\dfrac{1}{a^2-9a+20}\)
\(\Leftrightarrow\dfrac{1}{\left(a-2\right)\left(a-3\right)}+\dfrac{1}{\left(a-3\right)\left(a-4\right)}+\dfrac{1}{\left(a-4\right)\left(a-5\right)}\)
\(\Leftrightarrow\dfrac{\left(a-4\right)\left(a-5\right)}{\left(a-2\right)\left(a-3\right)\left(a-4\right)\left(a-5\right)}+\dfrac{\left(a-2\right)\left(a-5\right)}{\left(a-2\right)\left(a-3\right)\left(a-4\right)\left(a-5\right)}+\dfrac{\left(a-2\right)\left(a-3\right)}{\left(a-2\right)\left(a-3\right)\left(a-4\right)\left(a-5\right)}\)
\(\Leftrightarrow\dfrac{\left(a-4\right)\left(a-5\right)+\left(a-2\right)\left(a-5\right)+\left(a-2\right)\left(a-3\right)}{\left(a-2\right)\left(a-3\right)\left(a-4\right)\left(a-5\right)}\)
\(\Leftrightarrow\dfrac{\left(a-4\right)\left(a-5\right)+\left(a-2\right)\left[\left(a-3\right)+\left(a-5\right)\right]}{\left(a-2\right)\left(a-3\right)\left(a-4\right)\left(a-5\right)}\)
\(\Leftrightarrow\dfrac{\left(a-4\right)\left(a-5\right)+\left(a-2\right)\left(a-4\right)2}{\left(a-2\right)\left(a-3\right)\left(a-4\right)\left(a-5\right)}\)
\(\Leftrightarrow\dfrac{\left(a-4\right)\left[\left(a-5\right)+2\left(a-2\right)\right]}{\left(a-2\right)\left(a-3\right)\left(a-4\right)\left(a-5\right)}\)
\(\Leftrightarrow\dfrac{3a-9}{\left(x-2\right)\left(x-3\right)\left(x-5\right)}\)
\(\Leftrightarrow\dfrac{3\left(a-3\right)}{\left(a-2\right)\left(a-3\right)\left(a-5\right)}\)
\(\Leftrightarrow\dfrac{3}{\left(a-2\right)\left(a-5\right)}\)
H9
HT.Phong (9A5)
CTVHS
8 tháng 7 2023
a) \(\sqrt{9a^4}=\sqrt{\left(3a^2\right)^2}=\left|3a^2\right|=3a^2\)
b) \(2\sqrt{a^2}-5a=2\left|a\right|-5a=-2a-5a=-7a\)
c) \(\sqrt{16\left(1+4x+4x^2\right)}=\sqrt{\left[4\left(1+2x\right)\right]^2}=\left|4\left(1+2x\right)\right|=4\left(1+2x\right)\)
Để mk giải cho
\(\frac{3}{\left(a-2\right)\left(a-3\right)}\). minh khong chac dau nha. neu sai thi thoi.