x^4+2008x^2+2007x+2008

=x^4+2008x^2+2008x+2008-x

=x(x^3-1)+2008(x^2+x+1)

=x(x-1)(x^2+x+1)+2008(x^2+x+1)

=(x^2+x+1)(x^2-x+2008)

x⁴+2008x²+2007x+2008

=(x⁴+x²+1)+(2007x²+2007x+2007)

=(x⁴+2x²+1-x²)+2007(x²+x+1)

=(x²+1)²-x²+2007(x²+x+1)

=(x²+1-x)(x²+1+x)+2007(x²+x+1)

=(x²+x+1)(x²+1-x+2007)=(x²+x+1)(x²-x+2008)

1, không hiêu dề

2, 

\(\frac{a+b-c}{c}+2=\frac{b+c-a}{a}+2=\frac{c+a-b}{b}+2\)

\(\Leftrightarrow\frac{a+b+c}{a}=\frac{a+b+c}{b}=\frac{a+b+c}{c}\)

\(\Leftrightarrow a=b=c\)

Thay vào B dễ tính được B=2.2.2=8

199 chắc chắn

x²+x+1=x²+2x+1-x-1+1=(x+1)²-(x+1)+1

x²+2x+1=(x+1)²

đặt x+1=y , A=(y²-y+1)/y²=1-1/y+1/y²

tiếp tục đặt 1/y=z ta đc A=z²-z+1=(z-1/2)²+3/4 >= 3/4

minA=3/4 dấu "=" khi z=1/2<=>y=2<=>x=1

\(64x^4+y^4=64x^4+16x^2y^2-16x^2y^2+y^4\)

\(=\left(64x^4+16x^2y^2+y^4\right)-16x^2y^2\)

\(=\left(8x^2+y^2\right)^2-\left(4xy\right)^2\)

\(=\left(8x^2+y^2-4xy\right)\left(8x^2+y^2+4xy\right)\)

a)Từ \(2\left(a^2+b^2\right)=5ab\)\(\Rightarrow2a^2+2b^2-5ab=0\)

\(\Rightarrow2a^2-4ab-ab+2b^2=0\)

\(\Rightarrow2a\left(a-2b\right)-b\left(a-2b\right)=0\)

\(\Rightarrow\left(2a-b\right)\left(a-2b\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2a-b=0\\a-2b=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}2a=b\\a=2b\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}a=\frac{b}{2}\\a=2b\end{cases}}\)

Thay vào tính được P

b)sai đề