1,

\(x^2-2x+\left(x+2\right)x-x\left(2x+1\right)+5=x^2-2x+x^2+2x-2x^2-x+5=-x+5\)

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

N²⁺⁴ chia het cho n+ 2 khi nao ???

\(x^2\left(y-z\right)+y^2\left(z+x\right)+z^2\left(x-y\right)\)

\(\Leftrightarrow x^2\left(y-z\right)+y^2\left(z-x \right)+z^2\left[-\left(y-z\right)-\left(z-x\right)\right]\)\(\Leftrightarrow x^2\left(y-z\right)+y^2\left(z-x\right)-z^2\left(y-z\right)-z^2\left(z-x\right)\)\(\Leftrightarrow\left(y-z\right)\left(x-z\right)\left(x+z\right)+\left(z-x\right)\left(y-z\right)\left(y+z\right)\)\(\Leftrightarrow\left(y-z\right)\left(x-z\right)\left(x+z\right)-\left(y-z\right)\left(x-z\right)\left(z+y\right)\)\(\Leftrightarrow\left(y-z\right)\left(x-z\right)\left(x+z-y-z\right)=\left(y-z\right)\left(x-z\right)\left(x-y\right)\)

Gọi kim loại cần tìm là R

\(\Rightarrow\) Vì R có hóa trị II nên CTHH của oxit đó là RO

Số mol : \(n_{HCl}=\dfrac{2,19}{36,5}=0,06mol\)

PTHH:

RO + 2 HCl ---> \(RCl_2\) + \(H_2O\)

0,03mol 0,06mol

\(\Rightarrow M_{RO}=\dfrac{2,4}{0,03}=80\)

Ta có :

\(M_{RO}=M_R+M_O\Rightarrow M_R=M_{RO}-M_O=80-16=64\)Vậy R là kim loại Cu

Công thức oxit CuO

323

112

358

425

\(2x^3+5x^2=7x\)

\(\Leftrightarrow2x^3-2x^2+7x^2-7x=0\)

\(\Leftrightarrow2x^2\left(x-1\right)+7x\left(x-1\right)=0\)

\(\Leftrightarrow\left(2x^2+7x\right)\left(x-1\right)=0\)

\(\Leftrightarrow x\left(2x+7\right)\left(x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\2x+7=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\2x=-7\\x=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-2}{7}\\x=1\end{matrix}\right.\)

\(x^4+5x^3-8x-40=0\)

\(\Leftrightarrow x^3\left(x+5\right)-8\left(x+5\right)=0\)

\(\Leftrightarrow\left(x^3-8\right)\left(x+5\right)=0\)

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

Ta có : \(x^2+2x+4=x^2+2x+1+3=\left(x+1\right)^2+3\ge3\)\(\Rightarrow\left[{}\begin{matrix}x+5=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

\(a_1=1\)

\(a_2=3=1+2\)

\(a_3=6=1+2+3\)

\(a_4=1+2+3+4\)

\(a_5=15=1+2+3+4+5\)

.............................................

\(a_{100}=1+2+3+...+100=\dfrac{100\left(100+1\right)}{2}=5050\)\(a_n=1+2+3+...+n=\dfrac{n\left(n+1\right)}{2}\)