Ta có: \(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x.\left(x+1\right)}=1\frac{1993}{1995}\)   ( ĐK: \(x\ne0,\)\(x\ne-1\))

    \(\Leftrightarrow2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}\right)=1\frac{1993}{1995}\)

    \(\Leftrightarrow2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{3988}{1995}\)

    \(\Leftrightarrow1-\frac{1}{x+1}=\frac{1994}{1995}\)

    \(\Leftrightarrow\frac{1}{x+1}=\frac{1}{1995}\)

    \(\Leftrightarrow x+1=1995\)

    \(\Leftrightarrow x=1994\)\(\left(TM\right)\)

Vậy..........

Ta có: \(\left(x^2+5x\right)^2-2\left(x^2+5x\right)=24\)

\(\Leftrightarrow\left[\left(x^2+5x\right)^2+4\left(x^2+5x\right)\right]-\left[6\left(x^2+5x\right)+24\right]=0\)

\(\Leftrightarrow\left(x^2+5x\right)\left(x^2+5x+4\right)-6\left(x^2+5x+4\right)=0\)

\(\Leftrightarrow\left(x^2+5x-6\right)\left(x^2+5x+4\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+6\right)\left(x+1\right)\left(x+4\right)=0\)

\(\Rightarrow x\in\left\{-6;-4;-1;1\right\}\)

\(\left(x^2+5x\right)^2-2\left(x^2+5x\right)=24\)

\(\Leftrightarrow\left(x^2+5x\right)^2-2\left(x^2+5x\right)-24=0\)

Đặt: \(x^2+5x=t\)

\(\Rightarrow t^2-2t-24=0\)

\(\Leftrightarrow t^2+4t-6t-24=0\)

\(\Leftrightarrow t\left(t+4\right)-6\left(t+4\right)=0\)

\(\Leftrightarrow\left(t+4\right)\left(t-6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t+4=0\\t-6=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}t=-4\\t=6\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x^2+5x=-4\\x^2+5x=6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2+5x+4=0\\x^2+5x-6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1;x=-4\\x=1;x=-6\end{cases}}}\)

Vậy: \(S=\left\{-1;-4;1;-6\right\}\)

Ta có: \(2a^2+a=3b^2+b\)

\(\Leftrightarrow\left(2a^2-2b^2\right)+\left(a-b\right)=b^2\)

\(\Leftrightarrow\left(2a+2b\right)\left(a-b\right)+\left(a-b\right)=b^2\)

\(\Leftrightarrow\left(2a+2b+1\right)\left(a-b\right)=b^2\)

*CM 2a+2b+1 và a-b nguyên tố cùng nhau

=> 2a+2b+1 cũng là 1 SCP

Ta có: 

\(2a^2+a=3b^2+b\)

\(\Leftrightarrow2a^2-2b^2+a-b=b^2\)

\(\Leftrightarrow\left(a-b\right)\left(2a+2b+1\right)=b^2\)

Ta có: 

Đặt \(d=\left(a-b,2a+2b+1\right)\).

\(\Rightarrow\hept{\begin{cases}a-b⋮d\\2a+2b+1⋮d\end{cases}}\Rightarrow\left(a-b\right)\left(2a+2b+1\right)=b^2⋮d^2\Rightarrow b⋮d\)

\(\Rightarrow\left(a-b\right)+b=a⋮d\)

\(\Rightarrow\left(2a+2b+1\right)-2a-2b=1⋮d\Rightarrow d=1\).

Do đó \(a-b,2a+2b+1\)là hai số chính phương. 

\(A=n^3+n^2-n+2=\left(n+2\right)\left(n^2-n+1\right)\)là số nguyên tố suy ra 

\(\orbr{\begin{cases}n+2=1\\n^2-n+1=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}n=-1\\n=1;n=0\end{cases}}\)

Thử lại đều thỏa mãn. 

kết quả là 13,25 nhé

chúc bn học tốt

13,25 bạn nhé

jhyfhregrjhesdftruiejxfhrjehxgmjfd;j03169543256545449526u4tnkuyfnikuyf42b 4r 6e524brd62v4utq7w8e9r96f5d4s1d323g5t5esd232df2f5e2s2sd

từ phương trình thứ nhất ta có :

\(y=-x+3m+2\) thế xuống phương trình dười : \(3x+2x-6m-4=11-m\Leftrightarrow x=3+m\Rightarrow y=2m-1\)

b. ta có \(x^2-y^2=\left(m+3\right)^2-\left(2m-1\right)^2=-3m^2+10m+8=-3\left(m-\frac{5}{3}\right)^2+\frac{49}{3}\le\frac{49}{3}\)

Dấu bằng xảy ra khi m=5/3