\(P=2.\left(x^3-y^3\right)+3.\left(x^2+y^2\right)\)

\(=2.\left(x-y\right).\left(x^2+xy+y^2\right)+3.\left(x^2+y^2\right)\)

Thay vào ta được

\(P=2.\left(-1\right).[\left(x^2-2xy+y^2\right)+3xy]+3.[\left(x^2-2xy+y^2\right)+2xy]\)

\(=-2.[\left(x-y\right)^2+3xy]+3.[\left(x-y\right)^2+2xy]\)

Thay vảo ta được

\(P=-2.[\left(-1\right)^2+3xy]+3.[\left(-1\right)^2+2xy]\)

\(=-2.\left(1+3xy\right)+3.\left(1+2xy\right)\)

\(=-2-6xy+3+6xy\)

\(=1\)

\(A=-x^2+2x-5\)

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

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

\(=-\left(x-1\right)^2-4\)

Mà: \(\left(x-1\right)^2\ge0\forall x\Rightarrow-\left(x-1\right)^2\le0\Rightarrow A< 0\)

\(B=-x^2+x-1\)

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

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

\(=-\left(x-\frac{1}{2}\right)^2-\frac{3}{4}\)

Mà: \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\Rightarrow-\left(x-\frac{1}{2}\right)^2\le0\Rightarrow B< 0\)

Bạn xem lại đề phần \(C\)nhé.

a/\(x^2+8x+17=\left(x^2+8x+16\right)+1=\left(x+4\right)^2+1\)

\(\left(x+4\right)^2\ge0\Rightarrow\left(x+4\right)^2+1\ge1>0\)

b/\(x^2-10x+29=\left(x^2-10x+25\right)+4=\left(x-5\right)^2+4\)

\(\left(x-5\right)^2\ge0\Rightarrow\left(x-5\right)^2+4\ge4>0\)

c/

Ta có 

\(\left(2m-a\right)^2+\left(3m-b\right)^2+\left(3m-c\right)^2=\)

\(=4m^2-4ma+a^2+9m^2-6mb+b^2+9m^2-6mc+c^2=\)

\(=22m^2-2m\left(2a+3b+3c\right)+a^2+b^2+c^2=\)

\(=22m^2-2m.11m+a^2+b^2+c^2=a^2+b^2+c^2\)

đúng nha bạn

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

\(\Leftrightarrow x^2+2xy+y^2=x^2+y^2\)

\(\Leftrightarrow2xy=0\)

Vậy điều trên chỉ đúng khi một trong hai số bằng 0

\(35^2-15^2\)

\(=\left(35+15\right)\left(35-15\right)\)

\(=50.20\)

\(=1000\)

\(x+5x^2=0\)

\(\Leftrightarrow x\left(5x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{-1}{5}\end{cases}}\)