ko hỉu

Ko hiểu nốt

do m ;m+k ; m+2k là số nguyên tố > 3

=> m ;m+k ;m+2k lẻ

=> 2m+k chẵn

mà 2m chẵn

=>k ⋮ 2

mặt khác m là số nguyên tố >3 

=> m có dạng 3p+1 và 3p+2 (p∈ N*)

xét m=3p+1

ta lại có k có dạng 3a ;3a+1;3a+2(a∈ N*)

với k =3a+1 ta có 3p+1 + 2(3a+1) = 3(p+1+3a) loại vì m+2k là hợp số 

với k = 3a+2 => m+k = 3(p+a+1) loại

=> k=3a

tương tự với 3p+2

=> k=3a

=> k⋮3

mà (3;2)=1

=> k ⋮ 6

\(y^2+2^x-1=\frac{4^x+y^4}{2}\)

\(\Leftrightarrow4^x+y^4-2y^2-2.2^x+2=0\)

\(\Leftrightarrow\left(y^4-2y^2+1\right)+\left(4^x-2.2^x+1\right)=0\)

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

\(\Leftrightarrow\hept{\begin{cases}y^2=1\Rightarrow\orbr{\begin{cases}y=1\\y=-1\end{cases}}\\2^x=1\Rightarrow x=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}y^2=1\Rightarrow y=\pm1\\2^x=1\Rightarrow x=0\end{cases}}\)

Pt có tập nghiệm \(\left(x;y\right)=\left(0;1\right);\left(0;-1\right)\)

\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=1\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)

\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}=0\Rightarrow\frac{x+y}{xy}+\frac{x+y+z-z}{z\left(x+y+z\right)}=0\)

\(\Rightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\)\(\Rightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right)=0\)

\(\Rightarrow\left(x+y\right)\left(\frac{zx+zy+z^2+xy}{xyz\left(x+y+z\right)}\right)=0\)\(\Rightarrow\left(x+y\right)\left[z\left(x+z\right)+y\left(x+z\right)\right]=0\)

\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)\(\Rightarrow\)\(x=-y\) hoặc \(y=-z\) hoặc \(z=-x\)

\(\Rightarrow A=0\)

Sai đề không

a,\(x^2+2xy+7x+7y+y^2+10=\left(x^2+2xy+y^2\right)+7\left(x+y\right)+10\)

\(=\left(x+y\right)^2+2\left(x+y\right)+5\left(x+y\right)+10\)

\(=\left(x+y\right)\left(x+y+2\right)+5\left(x+y+2\right)\)

\(=\left(x+y+2\right)\left(x+y+5\right)\)

b,\(x^2y+xy^2+x+y=2010\Rightarrow xy\left(x+y\right)+x+y=2010\)

\(\Rightarrow12\left(x+y\right)=2010\Rightarrow x+y=167,5\)

Ta có:\(x^2+y^2=x^2+2xy+y^2-2xy=\left(x+y\right)^2-2xy=\left(167,5\right)^2-2.11=28034,25\)

Ta có \(A=[\frac{2}{\left(x+1\right)^3}\left(\frac{1}{x}+1\right)+\frac{1}{x^2+2x+1}\left(\frac{1}{x^2}+1\right)]:\frac{x-1}{x^3}\)

\(\Leftrightarrow A=\left[\frac{2}{\left(x+1\right)^3}.\frac{x+1}{x}+\frac{1}{\left(x+1\right)^2}.\frac{x^2+1}{x^2}\right].\frac{x^3}{x-1}\)

\(\Leftrightarrow A=\left[\frac{2x+x^2+1}{x^2\left(x+1\right)^2}\right].\frac{x^3}{x+1}=\frac{x}{x+1}\)

Để \(A=\frac{x}{x+1}< 1\Leftrightarrow\frac{1}{x+1}>0\Leftrightarrow x>-1\)

Để \(A=1-\frac{1}{x+1}\text{ nguyên thì }\frac{1}{x+1}\text{ nguyên hay }x\in\left\{-2,0\right\} \)

\(x^2+2y^2=x^2+\frac{4y^2}{2}\ge\frac{\left(x+2y\right)^2}{3}=\frac{1}{3}\)