\(\Leftrightarrow\left(ax+b\right)\left(x-1\right)+c\left(x^2+1\right)=1\)

(a+c)x^2-(a-b)x+(c-b)=1

\(\hept{\begin{cases}a+c=0\\a-b=0\\c-b=1\end{cases}\Leftrightarrow\hept{\begin{cases}c+b=0\\c-b=1\end{cases}\Rightarrow}\hept{\begin{cases}c=\frac{1}{2}\\b=-\frac{1}{2}\\a=-\frac{1}{2}\end{cases}}}\)

=> \(\frac{2}{\left(x^2+1\right)\left(x-1\right)}=\frac{\left(ax+b\right)\left(x-1\right)+c\left(x^2+1\right)}{\left(x^2+1\right)\left(x-1\right)}=\frac{x^2\left(a+c\right)+x\left(b-a\right)+c-b}{\left(x^2+1\right)\left(x-1\right)}\)

=> a+  c = 0  (1)

=> b - a = 0 (2)

=> c - b = 2  (3)

b - a = 0 => a = b Thay (1) ta có :

b + c = 0  Kết hợp với (3) 

=> b + c + c - b = 2 + 0 

=> 2c = 2 

=> c = 1 

=> b = c - 2 = 1 - 2 = -1 = a 

Vậy a = b= -1 ; c = 1

cô Loan ko giải cho thành viên thường đâu

Xét vế phải : \(\frac{a}{x+1}+\frac{b}{x-2}+\frac{c}{\left(x-2\right)^2}=\frac{a\left(x-2\right)^2}{\left(x+1\right)\left(x-2\right)^2}+\frac{b\left(x-2\right)\left(x+1\right)}{\left(x+1\right)\left(x-2\right)^2}+\frac{c\left(x+1\right)}{\left(x+1\right)\left(x-2\right)^2}\)

\(=\frac{a\left(x^2-4x+4\right)+b\left(x^2-x-2\right)+c\left(x+1\right)}{\left(x+1\right)\left(x-2\right)^2}\)

\(=\frac{x^2\left(a+b\right)+x\left(-4a-b+c\right)+\left(4a-2b+c\right)}{\left(x+1\right)\left(x-2\right)^2}\)

So sánh với vế trái, suy ra : 

\(\begin{cases}a+b=2\\-4a-b+c=-1\\4a-2b+c=1\end{cases}\). Giải ra được \(\left(a,b,c\right)=\left(\frac{4}{9};\frac{14}{9};\frac{7}{3}\right)\)

(14,78-a)/(2,87+a)=4/1

14,78+2,87=17,65

Tổng số phần bằng nhau là 4+1=5

Mỗi phần có giá trị bằng 17,65/5=3,53

=>2,87+a=3,53

=>a=0,66.

\(\frac{a}{x}+\frac{b}{x+1}+\frac{c}{x+2}=\frac{a\left(x+1\right)\left(x+2\right)+bx\left(x+2\right)+c\left(x+1\right)x}{x\left(x+1\right)\left(x+2\right)}\)

\(=\frac{a\left(x^2+3x+2\right)+b\left(x^2+2x\right)+c\left(x^2+x\right)}{x\left(x+1\right)\left(x+2\right)}=\frac{ax^2+3ax+2a+bx^2+2bx+cx^2+cx}{x\left(x+1\right)\left(x+2\right)}\)

\(=\frac{x^2\left(a+b+c\right)+x\left(3a+2b+c\right)+2a}{x\left(x+1\right)\left(x+2\right)}=\frac{1}{x\left(x+1\right)\left(x+2\right)}\)

Đồng nhất phân thức ta được : \(\hept{\begin{cases}a+b+c=0\\3a+2b+c=0\\2a=1\end{cases}\Rightarrow\hept{\begin{cases}a=\frac{1}{2}\\b=-1\\c=\frac{1}{2}\end{cases}}}\)

Vậy \(a=\frac{1}{2};b=-1;c=\frac{1}{2}\)

cái trên thì bn dùng BĐT Bunhiakovshi nha

cái dưới hơi rườm tí mik ko bt lm đúng ko

\(f\left(x\right)=x\left(x+1\right)\left(x+2\right)\left(ax+b\right)\)

\(f\left(x-1\right)=\left(x-1\right)x\left(x+1\right)\left(ax-a+b\right)\)

\(\Rightarrow f\left(x\right)-f\left(x-1\right)=x\left(x+1\right)\left(x+2\right)\left(ax+b\right)-\)

\(\left(x-1\right)x\left(x+1\right)\left(ax-a+b\right)\)

\(=x\left(x+1\right)\left[\left(x+2\right)\left(ax+b\right)-\left(x-1\right)\left(ax-a+b\right)\right]\)

\(=x\left(x+1\right)[x\left(ax+b\right)+2\left(ax+b\right)-x\left(ax-a+b\right)\)

\(+\left(ax-a+b\right)]\)

\(=x\left(x+1\right)(ax^2+bx+2ax+2b-ax^2+ax\)

\(-bx+ax-a+b)\)

\(=x\left(x+1\right)\left(4ax-a+3b\right)\)

Mà theo đề \(f\left(x\right)-f\left(x-1\right)=x\left(x+1\right)\left(2x+1\right)\)

Đồng nhất hệ số là ra