*\(A=\left(x^2-2xy+y^2\right)+\left(3x-3y\right)-10\)
\(A=\left(x-y\right)^2+3\left(x-y\right)-10\)
\(A=\left(x-y\right)\left(x-y+3\right)-10\)
\(A=z\left(z+3\right)-10\) với \(z=x-y\)
\(A=z^2+3z-10\)
\(A=z^2-2z+5z-10\)
\(A=z\left(z-2\right)+5\left(z-2\right)\)
\(A=\left(z+5\right)\left(z-2\right)\)
*\(B=\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(B=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(B=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
\(B=y\left(y+2\right)-24\) với \(y=x^2+7x+10\)
\(B=y^2+2y-24\)
\(B=y^2-4y+6y-24\)
\(B=y\left(y-4\right)+6\left(y-4\right)\)
\(B=\left(y+6\right)\left(y-4\right)\)
*\(C=a^2-2a-4b^2-4b\)
\(C=a^2-2a+1-4b^2-4b-1\)
\(C=\left(a^2-2a+1\right)-\left(4b^2+4b+1\right)\)
\(C=\left(a-1\right)^2-\left(2b+1\right)^2\)
\(C=\left(a-1+2b+1\right)\left(a-1-2b-1\right)\)
\(C=\left(a+2b\right)\left(a-2b-2\right)\)
\(C=a^2-2a-4b^2-4b\)
\(\Rightarrow C=\left(a^2-4b^2\right)-\left(2a+4b\right)\)
\(\Rightarrow C=\left(a-2b\right)\left(a+2b\right)-2\left(a+2b\right)\)
\(\Rightarrow C=\left(a+2b\right)\left(a-2b-2\right)\)
*\(5x^2+10y^2-6xy-4y-2y+9\)
\(=x^2-6xy+9y^2+4x^2+y^2-6y+9\)
\(=\left(x-3y\right)^2+4x^2+\left(y-3\right)^2\)
Vì \(\left(x-3y\right)^2\ge0\), \(4x^2=\left(2x\right)^2\ge0\), \(\left(y-3\right)^2\ge0\)
\(\Rightarrow\left(x-3y\right)^2+4x^2+\left(y-3\right)^2\ge0\)
Hay \(5x^2+10y^2-6xy-4y-2y+9\) sẽ luôn dương
Mình không biết cái đề có sai không mà mình chứng minh là nó luôn dương không phải là luôn âm.