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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

\(P=\frac{5}{a.\left(a-5\right)}=\frac{5}{-1.\left(-1-5\right)}=\frac{5}{6}\)

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
1 tháng 7 2020

ĐKXĐ: ...

a/ \(A=\frac{1}{a\left(a+1\right)}+\frac{1}{\left(a+1\right)\left(a+2\right)}+\frac{1}{\left(a+2\right)\left(a+3\right)}+\frac{1}{\left(a+3\right)\left(a+4\right)}+\frac{1}{\left(a+4\right)\left(a+5\right)}\)

\(=\frac{1}{a}-\frac{1}{a+1}+\frac{1}{a+1}-\frac{1}{a+2}+...+\frac{1}{a+4}-\frac{1}{a+5}\)

\(=\frac{1}{a}-\frac{1}{a+5}=\frac{5}{a\left(a+5\right)}\)

\(A>\frac{5}{6}\Rightarrow\frac{5}{a\left(a+5\right)}>\frac{5}{6}\)

\(\Leftrightarrow\frac{1}{a\left(a+5\right)}-\frac{1}{6}>0\Leftrightarrow\frac{6-a^2-5a}{a\left(a+5\right)}>0\)

\(\Leftrightarrow\frac{\left(1-a\right)\left(a+6\right)}{a\left(a+5\right)}>0\Rightarrow\left[{}\begin{matrix}-6< a< -5\\0< a< 1\end{matrix}\right.\)

AH
Akai Haruma
Giáo viên
2 tháng 12 2019

Bài 1:

\(A=\frac{1}{a-b}+\frac{1}{a+b}+\frac{2a}{a^2+b^2}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)

\(=\frac{a+b+a-b}{(a-b)(a+b)}+\frac{2a}{a^2+b^2}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}=\frac{2a}{a^2-b^2}+\frac{2a}{a^2+b^2}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)

\(=(2a).\frac{a^2+b^2+a^2-b^2}{(a^2-b^2)(a^2+b^2)}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)

\(=\frac{4a^3}{a^4-b^4}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)

\(=4a^3.\frac{a^4+b^4+a^4-b^4}{(a^4-b^4)(a^4+b^4)}+\frac{8a^7}{a^8+b^8}=\frac{8a^7}{a^8-b^8}+\frac{8a^7}{a^8+b^8}=8a^7.\frac{a^8+b^8+a^8-b^8}{(a^8-b^8)(a^8+b^8)}\)

\(=\frac{16a^{15}}{a^{16}-b^{16}}\)

--------------

\(B=\frac{1}{a(a+1)}+\frac{1}{(a+1)(a+2)}+\frac{1}{(a+2)(a+3)}=\frac{(a+1)-a}{a(a+1)}+\frac{(a+2)-(a+1)}{(a+1)(a+2)}+\frac{(a+3)-(a+2)}{(a+2)(a+3)}\)

\(=\frac{1}{a}-\frac{1}{a+1}+\frac{1}{a+1}-\frac{1}{a+2}+\frac{1}{a+2}-\frac{1}{a+3}\)

\(=\frac{1}{a}-\frac{1}{a+3}=\frac{3}{a(a+3)}\)

AH
Akai Haruma
Giáo viên
2 tháng 12 2019

Bài 2:

Bạn tham khảo lời giải tương tự tại link sau:

Câu hỏi của Law Trafargal - Toán lớp 8 | Học trực tuyến

\(=\left[\dfrac{\left(a-1\right)^2}{a^2+a+1}+\dfrac{2a^2-4a-1}{\left(a-1\right)\left(a^2+a+1\right)}+\dfrac{1}{a-1}\right]:\dfrac{2a}{3}\)

\(=\dfrac{a^3-3a^2+3a-1+2a^2-4a-1+a^2+a+1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{3}{2a}\)

\(=\dfrac{a^3-1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{3}{2a}=\dfrac{3}{2a}\)

22 tháng 5 2020

Bó tay!!! 🐷

22 tháng 5 2020

chuẩn

4 tháng 1 2017

a/ đk: a\(\ne b\), b\(\ne0,a\ne-b\)

= \(\frac{a\left(a-b\right)-a^2-b^2}{a-b}.\frac{a+b+2b}{b\left(a+b\right)}\)

= \(\frac{a^2-ab-a^2-b^2}{a-b}.\frac{a+3b}{b\left(a+b\right)}\)

= \(\frac{-ab-b^2}{a-b}.\frac{a+3b}{b\left(a+b\right)}\)

= \(\frac{-b\left(a+b\right)\left(a+3b\right)}{b\left(a+b\right)\left(a-b\right)}\)

= \(\frac{-a-3b}{a-b}\)

b/ đk: a\(\ne0,a\ne\pm3\)

= \(\left[\frac{3a+1}{a\left(a-3\right)}+\frac{3a-1}{a\left(a+3\right)}\right].\frac{\left(a-3\right)\left(a+3\right)}{a^2+1}\)

= \(\frac{\left(3a+1\right)\left(a+3\right)+\left(3a-1\right)\left(a-3\right)}{a\left(a-3\right)\left(a+3\right)}.\frac{\left(a-3\right)\left(a+3\right)}{a^2+1}\)

= \(\frac{6a^2+6}{a\left(a-3\right)\left(a+3\right)}.\frac{\left(a-3\right)\left(a+3\right)}{a^2+1}\)

= \(\frac{6\left(a^2+1\right)\left(a-3\right)\left(a+3\right)}{a\left(a^2+1\right)\left(a-3\right)\left(a+3\right)}\)

= \(\frac{6}{a}\)