\(P=1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\)

     \(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}\)

      \(=2-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}\)

     \(=2-\frac{1}{n+1}=\frac{2\left(n+1\right)}{n+1}-\frac{1}{n+1}=\frac{2n+2-1}{n+1}=\frac{2n+1}{n+1}\)

\(P=1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{n\left(n+1\right)}=1+1-\frac{1}{2}+\frac{1}{2}-.....-\frac{1}{\left(n+1\right)}\)

\(\Rightarrow P=2-\frac{1}{\left(n+1\right)}=\frac{2n+1}{n+1}\)

vì\(9!-2!=362878\)

Mà \(7!=5040\)

Nên \(9!-2!\ne7!\)

mk cũng bị hết lượt ( T_T )

:))))))))))

Ừ

Ta có \(P=\frac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{a^2-ab+b^2+b^2-bc+c^2+c^2-ac+a^2}\)

\(=\frac{5\left(...\right)}{2\left(...\right)}=\frac{5}{2}\)

Thực hiện phép chia ta có:

Ta có: \(x^3-2x^2+7x-7=\left(x^2+3\right)\left(x-2\right)+4x-1\)

\(x^3-2x^2+7x-7\) chia hết cho \(x^2+3\)

=> \(4x-1⋮x^2+3\) (1)

=> \(4x^2-x=x\left(4x-1\right)⋮x^2+3\)

Mà: \(4x^2+12=4\left(x^2+3\right)⋮x^2+3\)

=> \(\left(4x^2-x\right)-\left(4x^2+12\right)⋮x^2+3\)

=> \(-x-12⋮x^2+3\)

=> \(x+12⋮x^2+3\)

=> \(4x+48⋮x^2+3\) (2)

Từ (1); (2) => \(\left(4x+48\right)-\left(4x-1\right)⋮x^2+3\)

=> \(49⋮x^2+3\)

=> \(x^2+3\in\left\{\pm1;\pm7;\pm49\right\}\) vì \(x^2+3\ge3\) với mọi x

=> \(\begin{cases}x^2+3=7\\x^2+3=49\end{cases}\Rightarrow\orbr{\begin{cases}x^2=4\\x^2=46\left(loại\right)\end{cases}}\)

Với \(x^2=4\Rightarrow x=\pm2\) thử vào bài toán x=-2 loại. x=2 thỏa mãn

Vậy x=2

Em cảm ơn cô