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Đây là câu a/
https://hoc24.vn/hoi-dap/question/693692.html?pos=1903228
Còn câu b thì như sau:
Trước hết, nghi ngờ bạn ghi sai đề ở con này \(\dfrac{1}{a^2+7a+9}\) , số 9 phải là số 12 mới hợp lý. Mình tự sửa lại đề, còn nếu đề đúng như bạn chép thì bạn giữ nguyên nó, phần còn lại rút gọn được còn đâu thì quy đồng giải trâu thôi, chẳng cách nào với đề xấu kiểu ấy cả.
\(B=\dfrac{1}{a\left(a+1\right)}+\dfrac{1}{\left(a+1\right)\left(a+2\right)}+\dfrac{1}{\left(a+2\right)\left(a+3\right)}+\dfrac{1}{\left(a+3\right)\left(a+4\right)}+\dfrac{1}{\left(a+4\right)\left(a+5\right)}\)
\(B=\dfrac{1}{a}-\dfrac{1}{a+1}+\dfrac{1}{a+1}-\dfrac{1}{a+2}+\dfrac{1}{a+2}-\dfrac{1}{a+3}+\dfrac{1}{a+3}-\dfrac{1}{a+4}+\dfrac{1}{a+4}-\dfrac{1}{a+5}\)
\(B=\dfrac{1}{a}-\dfrac{1}{a+5}=\dfrac{5}{a\left(a+5\right)}\)
\(M=\frac{1}{a^2-5a+6}+\frac{1}{a^2-7a+12}+\frac{1}{a^2-9a+20}+\frac{1}{a^2-11a+30}\)
\(M=\frac{1}{\left(a-2\right)\left(a-3\right)}+\frac{1}{\left(a-3\right)\left(a-4\right)}+\frac{1}{\left(a-4\right)\left(a-5\right)}+\frac{1}{\left(a-5\right)\left(a-6\right)}\)
\(M=\frac{1}{a-2}-\frac{1}{a-3}+\frac{1}{a-3}-\frac{1}{a-4}+\frac{1}{a-4}-\frac{1}{a-5}+\frac{1}{a-5}-\frac{1}{a-6}\)
\(M=\frac{1}{a-2}-\frac{1}{a-6}\)
a, ĐK : \(x\ne1;2;3;4;5\)
b, \(\dfrac{1}{x\left(x-1\right)}+\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}\)
\(=\dfrac{1}{x}-\dfrac{1}{x-1}+\dfrac{1}{x-1}-\dfrac{1}{x-2}+\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-4}+\dfrac{1}{x-4}-\dfrac{1}{x-5}\)
\(=\dfrac{1}{x}-\dfrac{1}{x-5}=\dfrac{x-5-x}{x\left(x-5\right)}=\dfrac{-5}{x\left(x-5\right)}\)
a: ĐKXĐ: \(x\notin\left\{0;1;2;3;4;5\right\}\)
b: \(P=\dfrac{1}{\left(x-1\right)\cdot x}+\dfrac{1}{\left(x-2\right)\left(x-1\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}\)
\(=\dfrac{1}{x-1}-\dfrac{1}{x}+\dfrac{1}{x-2}-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-2}+\dfrac{1}{x-4}-\dfrac{1}{x-3}+\dfrac{1}{x-5}-\dfrac{1}{x-4}\)
\(=\dfrac{1}{x-5}-\dfrac{1}{x}=\dfrac{x-x+5}{x\left(x-5\right)}=\dfrac{5}{x\left(x-5\right)}\)
\(\frac{3}{\left(a-2\right)\left(a-3\right)}\). minh khong chac dau nha. neu sai thi thoi.
a: ĐKXĐ: \(x\notin\left\{0;1;2;3;4;5\right\}\)
b: \(P=\dfrac{1}{x^2-x}+\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+6}+\dfrac{1}{x^2-7x+12}+\dfrac{1}{x^2-9x+20}\)
\(=\dfrac{1}{x\left(x-1\right)}+\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}\)
\(=\dfrac{-1}{x}+\dfrac{1}{x-1}-\dfrac{1}{x-1}+\dfrac{1}{x-2}-\dfrac{1}{x-2}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-4}-\dfrac{1}{x-4}+\dfrac{1}{x-5}\)
\(=\dfrac{1}{x-5}-\dfrac{1}{x}\)
\(=\dfrac{x-\left(x-5\right)}{x\left(x-5\right)}=\dfrac{5}{x\left(x-5\right)}\)
c: \(x^3-x^2+2=0\)
=>\(x^3+x^2-2x^2+2=0\)
=>\(x^2\cdot\left(x+1\right)-2\left(x-1\right)\left(x+1\right)=0\)
=>\(\left(x+1\right)\left(x^2-2x+2\right)=0\)
=>x+1=0
=>x=-1
Khi x=-1 thì \(P=\dfrac{5}{\left(-1\right)\left(-1-5\right)}=\dfrac{5}{\left(-1\right)\cdot\left(-6\right)}=\dfrac{5}{6}\)
mẫu thức thứ 2 sai nhé
A=1/a^2-5a+6+1/a^2-7a+12+1/a^2-9a+20
=1/a^2-3a-2a+6+1/a^2-4a-3a+12+1/a^2-5a-4a+20
=1/a(a-3)-2(a-3)+1/a(a-4)-3(a-4)+1/a(a-5)-4(a-5)
=1/(a-2)(a-3)+1/(a-3)(a-4)+1/(a-5)(a-4)
tổng quát: 1/(x-1)x=1/(x-1)-1/x
A=-1/(a-2)+1/(a-3)-1/(a-3)+1/(a-4)-1/(a-4)+1/(a-5)=-1/(a-2)+1/(a-5)=1/(a-5)-1/(a-2)
1,
\(A=\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{4x^2+x-2-\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{4x^2-4}{\left(x-2\right)\left(x+2\right)}\)
\(x=4\Rightarrow A=\dfrac{4.x^2-4}{\left(4-2\right)\left(4+2\right)}=...\)
2.
\(A=\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3-5x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x\left(x+1\right)+3\left(x-1\right)+3-5x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{x+1}\)
3.
Đề lỗi, thiếu dấu trước \(\dfrac{6+5x}{4-x^2}\)
4.
\(A=\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{2x-5\left(x+5\right)-\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4x-20}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4}{x-5}\)
\(x=\dfrac{4}{5}\Rightarrow A=\dfrac{-4}{\dfrac{4}{5}-5}=\dfrac{20}{21}\)
5.
\(M=\dfrac{x^2}{x\left(x+2\right)}+\dfrac{2x}{x\left(x+2\right)}+\dfrac{2\left(x+2\right)}{x\left(x+2\right)}\)
\(=\dfrac{x^2+2x+2\left(x+2\right)}{x\left(x+2\right)}=\dfrac{x^2+4x+4}{x\left(x+2\right)}\)
\(=\dfrac{\left(x+2\right)^2}{x\left(x+2\right)}=\dfrac{x+2}{x}\)
\(x=-\dfrac{3}{2}\Rightarrow M=\dfrac{-\dfrac{3}{2}+2}{-\dfrac{3}{2}}=-\dfrac{1}{3}\)
\(A=\dfrac{1}{\left(a-2\right)\left(a-3\right)}+\dfrac{1}{\left(a-3\right)\left(a-4\right)}+\dfrac{1}{\left(a-4\right)\left(a-5\right)}\)\(A=\dfrac{1}{a-2}-\dfrac{1}{a-3}+\dfrac{1}{a-3}+\dfrac{1}{a-4}-\dfrac{1}{a-4}+\dfrac{1}{a-4}-\dfrac{1}{a-5}\)\(A=\dfrac{1}{a-2}-\dfrac{1}{a-5}=\dfrac{-3}{\left(a-2\right)\left(a-5\right)}\)
\(\dfrac{1}{a^2-5a+6}+\dfrac{1}{a^2-7a+12}+\dfrac{1}{a^2-9a+20}\)
\(\Leftrightarrow\dfrac{1}{\left(a-2\right)\left(a-3\right)}+\dfrac{1}{\left(a-3\right)\left(a-4\right)}+\dfrac{1}{\left(a-4\right)\left(a-5\right)}\)
\(\Leftrightarrow\dfrac{\left(a-4\right)\left(a-5\right)}{\left(a-2\right)\left(a-3\right)\left(a-4\right)\left(a-5\right)}+\dfrac{\left(a-2\right)\left(a-5\right)}{\left(a-2\right)\left(a-3\right)\left(a-4\right)\left(a-5\right)}+\dfrac{\left(a-2\right)\left(a-3\right)}{\left(a-2\right)\left(a-3\right)\left(a-4\right)\left(a-5\right)}\)
\(\Leftrightarrow\dfrac{\left(a-4\right)\left(a-5\right)+\left(a-2\right)\left(a-5\right)+\left(a-2\right)\left(a-3\right)}{\left(a-2\right)\left(a-3\right)\left(a-4\right)\left(a-5\right)}\)
\(\Leftrightarrow\dfrac{\left(a-4\right)\left(a-5\right)+\left(a-2\right)\left[\left(a-3\right)+\left(a-5\right)\right]}{\left(a-2\right)\left(a-3\right)\left(a-4\right)\left(a-5\right)}\)
\(\Leftrightarrow\dfrac{\left(a-4\right)\left(a-5\right)+\left(a-2\right)\left(a-4\right)2}{\left(a-2\right)\left(a-3\right)\left(a-4\right)\left(a-5\right)}\)
\(\Leftrightarrow\dfrac{\left(a-4\right)\left[\left(a-5\right)+2\left(a-2\right)\right]}{\left(a-2\right)\left(a-3\right)\left(a-4\right)\left(a-5\right)}\)
\(\Leftrightarrow\dfrac{3a-9}{\left(x-2\right)\left(x-3\right)\left(x-5\right)}\)
\(\Leftrightarrow\dfrac{3\left(a-3\right)}{\left(a-2\right)\left(a-3\right)\left(a-5\right)}\)
\(\Leftrightarrow\dfrac{3}{\left(a-2\right)\left(a-5\right)}\)