Câu 1:
\(\text{a) }x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-3xy-3yz-3xz\right)\)
\(\text{b) }x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2=\left(x^3-y^3\right)\left(x^3+y^3\right)=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(\text{c) }xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)+2xyz\\ =x^2y+xy^2+y^2z+yz^2+x^2z+xz^2+xyz+xyz\\ =\left(x^2y+x^2z+xyz\right)+\left(xy^2+y^2z+xyz\right)\\ =x\left(xy+xz+yz\right)+y\left(xy+yz+xz\right)\\ =\left(x+y\right)\left(xy+yz+xz\right)\\ \\ \)
Câu 3:
\(x^2-10x=-25\\ \Leftrightarrow x^2-10x+25=0\\\Leftrightarrow\left(x-5\right)^2=0\\ \Leftrightarrow x-5=0\\ \Leftrightarrow x=5 \)
Vậy \(x=5\)
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Câu 2 bạn thay vào mà làm nhé. Mình ngại lắm