Để mk giải cho

\(\frac{3}{\left(a-2\right)\left(a-3\right)}\). minh khong chac dau nha. neu sai thi thoi.

theo đề ta có :

A= (1 /a² - 2a -3a + 6) + ( 1 /a² -3a -4a + 12 ) + ( 1 /a² - 4a -5a + 20 )

\(\Leftrightarrow\) A = (1 /( a -2 )(a-3 ))+ (1/(a-3)(a-4))+(1/(a-4)(a-5)).

\(\Leftrightarrow\)A = (1/a-2) - (1/ a-3) +(1/ a-3) -(1/ a-4) +(1/ a-4) - (1/ a-5)

\(\Leftrightarrow\)A = (1/a-2) - (1/ a-5)

\(\Leftrightarrow\)A = a-5-a+2 / (a-2)(a-5)

\(\Leftrightarrow\)A= 3 / ( a -2 )(a-5)

chúc bạn học tốt

ĐKXĐ: ...

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

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

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

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

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

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

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: ....

\(P=\frac{1}{a^2-a}+\frac{1}{a^2-3a+2}+\frac{1}{a^2-5a+6}+\frac{1}{a^2-7a+12}+\frac{1}{a^2-9a+20}\)

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

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

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

\(M=\frac{1}{a^2-5a+6}+\frac{1}{a^2-7a+12}+\frac{1}{a^2-9a+20}+\frac{1}{a^2-11a+30}\) 

\(M=\frac{1}{\left(a-2\right)\left(a-3\right)}+\frac{1}{\left(a-3\right)\left(a-4\right)}+\frac{1}{\left(a-4\right)\left(a-5\right)}+\frac{1}{\left(a-5\right)\left(a-6\right)}\)

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

\(M=\frac{1}{a-2}-\frac{1}{a-6}\)

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

Ta có:\(A=\frac{a^2}{a^2-1}-\frac{a^2}{1+a^2}.\frac{a}{a+1}-\frac{a^2}{1+a^2}.\frac{1}{a^2+a}\)

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

a: ĐKXĐ: \(a\notin\left\{0;1;-1\right\}\)

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

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

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

b: Để A=3 thì \(3a^2-3=a\)

\(\Leftrightarrow2a^2=3\)

hay \(a\in\left\{\dfrac{\sqrt{6}}{2};-\dfrac{\sqrt{6}}{2}\right\}\)