a, \(x^2+20x+100\)

\(=x^2+10x+10x+100\)

\(=x.\left(x+10\right)+10.\left(x+10\right)=\left(x+10\right)^2\)

b, \(x^2-20xy+100y^2\)

\(=x^2-10xy-10xy+100y^2\)

\(=x.\left(x-10y\right)-10y.\left(x-10y\right)=\left(x-10y\right)^2\)

c, \(x^2-20x+100\)

\(=x^2-10x-10x+100\)

\(=x.\left(x-10\right)-10.\left(x-10\right)=\left(x-10\right)^2\)

Chúc bạn học tốt!!!

a) $x^2+20x+100=x^2=2.10x+10^2=(x+10)^2$

b) $x^2-20xy+100y^2=x^2-2.x.10y+(10y)^2=(x+10y)^2$

c) $x^2-20x+100=x^2-2.10x+10^2=(x-10)^2$

Ta có thể viết thành: (x - 5)3/10 = x/5 + 5

Nhân hai vế với 10, ta có : 3(x - 5) = (x/5 + 5) * 10

                                           3x - 15 = 2x + 50

                                           3x = 2x + 50 = 2x + 65

                                           3x - 2x = 65

                                           x = 65

Chúc bạn ht ~~

\(x=7\Rightarrow\left\{{}\begin{matrix}4=x-3\\20=3x-1\end{matrix}\right.\)\(\Rightarrow P\left(7\right)=x^{100}-4x^{99}-20x^{98}-4x^{97}-...-20x^2-4x\\ =x^{100}-\left(x-3\right)x^{99}-\left(3x-1\right)x^{98}-\left(x-3\right)x^{97}-...-\left(3x-1\right)x^2-\left(x-3\right)x\\ =x^{100}-x^{100}+3x^{99}-3x^{99}+x^{98}-x^{98}+3x^{97}-...-3x^3+x^2-x^2+3x\\ =3x\\ =21\)

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