Nhấn vào câu hỏi tương tự
:)))
a, \(11x+11y+x^2+xy=\left(11x+11y\right)+\left(x^2+xy\right)=11\left(x+y\right)+x\left(x+y\right)=\left(x+y\right)\left(x+11\right)\)
b. \(255-4x^2-4xy-y^2=255-\left(4x^2+4xy+y^2\right)=255-\left(2x+y\right)^2=\left(15+2x+y\right)\left(15-2x-y\right)\)
Bài 2:
\(x^2-y^2-4x+4=\left(x^2-4x+4\right)-y^2=\left(x-2\right)^2-y^2=\left(x-2-y\right)\left(x-2+y\right)\)
\(=\left(72-2\right)\left(102-2\right)=70.100=7000\) ( x+y=102, x-y=72 )
Bài 1:
a, 11x+11y+x2+xy=11(x+y)+x(x+y)=(x+y)(11+x)
b,\(b,255-4x^2-4xy-y^2=255-\left(2x+y\right)^2\)
\(=\left(\sqrt{255}-2x-y\right)\left(\sqrt{255}+2x+y\right)\)
Bài 2:
A=x2-y2-4x+4=(x2-4x+4)-y2=(x-2)2-y2=(x-2-y)(x-2+y) (1)
Ta có: x+y=102 và x-y=72
\(\Rightarrow\hept{\begin{cases}x=\frac{102+72}{2}=87\\y=\frac{102-72}{2}=15\end{cases}}\)Thay vào (1) ta có:
A=(87-2-15)(87-2+15)=70.100=7000
Vậy A=7000
