\(\left(x+y+t\right)^3-x^3-y^3-t^3\\ =\left(x+y\right)^3+3\left(x+y\right)^2t+3\left(x+y\right)t^2+t^3-x^3-y^3-t^3\\ =\left(x+y\right)^3+3\left(x+y\right)^2t+3\left(x+y\right)t^2-\left(x^3+y^3\right)\\ =\left(x+y\right)\left[\left(x+y\right)^2+3\left(x+y\right)t+3t^2\right]-\left(x+y\right)\left(x^2-xy+y^2\right)\\ =\left(x+y\right)\left[x^2+2xy+y^2+3\left(x+y\right)t+3t^2-x^2+xy-y^2\right]\\ =\left(x+y\right)\left[3\left(x+y\right)t+3xy+3t^2\right]\\ =3\left(x+y\right)\left(xt+yt+xy+t^2\right)\\ =3\left(x+y\right)\left[t\left(x+t\right)+y\left(x+t\right)\right]\\ =3\left(x+y\right)\left[\left(x+y\right)t+xy+t^2\right]\\ =3\left(x+y\right)\left(y+t\right)\left(t+x\right)\\ \Rightarrow2011=3\left(x+y\right)\left(y+t\right)\left(t+x\right)\\ \Rightarrow D=\dfrac{3\left(x+y\right)\left(y+t\right)\left(t+x\right)}{\left(x+y\right)\left(y+t\right)\left(t+x\right)}=3\)

Ta có: n+2 = (n-3)+2 

Vì n-3 chia hết cho n-3 => 2 chia hết cho n-3

=> n thuộc Ư(2) ={1;2;-1;-2}

\(P\left(x\right)=2x-7+\left(x-14\right)=0\\ \Leftrightarrow2x-7+x-14=0\\ \Leftrightarrow3x=21\\ \Leftrightarrow x=7\\ Q\left(x\right)=x^2-64=0\\ \Leftrightarrow\left(x+8\right)\left(x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+8=0\\x-8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-8\\x=8\end{matrix}\right.\)

\(\dfrac{5}{n-3}\in Z\Rightarrow n-3\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)

Vậy \(n\in\left\{-2;2;4;8\right\}\)

\(x-\left(2x+3\right)=4-3x\\ \Leftrightarrow x-2x-3=4-3x\\ \Leftrightarrow x-2x+3x=4+3\\ \Leftrightarrow2x=7\\ \Leftrightarrow x=\dfrac{7}{2}\)

\(x^{2016}\cdot x^{2017}=0\\ \Leftrightarrow x^{4033}=0\\ \Leftrightarrow x=0\)

\(1^2+2^2+3^2+...+99^2\\ =1\cdot\left(0+1\right)+2\cdot\left(1+1\right)+3\cdot\left(2+1\right)+...+99\cdot\left(98+1\right)\\ =1+1\cdot2+2+2\cdot3+3+...+98\cdot99+99\\ =\left(1\cdot2+2\cdot3+...+98\cdot99\right)+\left(1+2+3+...+99\right)\\ 1+2+3+...+99=\dfrac{99\cdot100}{2}=4950\\ 1\cdot2+2\cdot3+...+98\cdot99\\ =\dfrac{3\left(1\cdot2+2\cdot3+...+98\cdot99\right)}{3}\\ =\dfrac{1\cdot2\cdot3+2\cdot3\cdot3+...+98\cdot99\cdot3}{3}\\ =\dfrac{1\cdot2\cdot\left(3-0\right)+2\cdot3\cdot\left(4-1\right)+...+98\cdot99\cdot\left(100-97\right)}{3}\\ =\dfrac{1\cdot2\cdot3-0+2\cdot3\cdot4-1\cdot2\cdot3+...+98\cdot99\cdot100-97\cdot98\cdot99}{3}\\ =\dfrac{98\cdot99\cdot100}{3}\\ =323400\\ 1^2+2^2+3^2+...+99^2\\=4950+323400\\=328350\)

\(b-\left(b+a-c\right)???\\ =b-b-a+c\\ =c-a\)

\(VT=\left|x-1\right|+\left|x+3\right|=\left|1-x\right|+\left|x+3\right|\ge\left|1-x+x+3\right|=4=VP\\ \Rightarrow\text{Dấu "=" phải xảy ra}\Leftrightarrow\left(1-x\right)\left(x+3\right)\ge0\Leftrightarrow-3\le x\le1\)

417

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tick nhé