a)\(\left(x+y\right)^2+\left(x-y\right)^2=x^2+2xy+y^2+x^2-2xy+y^2=2\left(x^2+y^2\right)\)

b)\(m^3+n^3+p^3-3mnp=\left(m+n\right)^3-3mn\left(m+n\right)+p^3-3mnp=\left(m+n+p\right)^3-3\left(m+n\right)p\left(m+n+p\right)-3mn\left(m+n+p\right)=\left(m+n+p\right)\left(m^2+n^2+p^2+2mn+2np+2mp-3mp-3np-3mn\right)=\left(m+n+p\right)\left(m^2+n^2+p^2-mn-np-mp\right)\)

33+35+37+39+41+43+45+47=(33+47)+(35+45)+(37+43)+(39+41)=80+80+80+80=320

14+15+16+17+18+19+20+21+22+23+24+25+26=(14+26)+(15+25)+(16+24)+(17+23)+(18+22)+(19+21)+20=40+40+40+40+40+40+20=40x6+20=240+20=260

\(\Delta ANM=\Delta DKE;\Delta MAN=\Delta EDK;\Delta MNA=\Delta EKD;\Delta NAM=\Delta KED;\Delta NMA=\Delta KDE\)

a)\(x\left(6-x\right)+\left(x-2\right)\left(x-4\right)=6x-x^2+x^2-6x+8=8\)

b)

\(\left(12x^3-7x^2-14x+5\right):\left(4x-5\right)\\ =\left[\left(12x^3-15x^2\right)+\left(8x^2-10x\right)-\left(4x-5\right)\right]:\left(4x-5\right)\\ =\left[3x^2\left(4x-5\right)+2x\left(4x-5\right)-\left(4x-5\right)\right]:\left(4x-5\right)\\ =\left(3x^2+2x-1\right)\left(4x-5\right):\left(4x-5\right)\\ =3x^2+2x-1\\ =\left(x-1\right)\left(x+3\right)\)

\(\dfrac{-9}{7}\cdot x=-1\dfrac{1}{2}\\ \Leftrightarrow x=\dfrac{-3}{2}\cdot\dfrac{-7}{9}\\ \Leftrightarrow x=\dfrac{7}{6}\)

Vậy x = 7/6

\(x^2-4x+4=\left(x-2\right)^2\)

a) Ta có:

\(n^2\left(n+1\right)-n\left(n+1\right)=n\left(n-1\right)\left(n+1\right)\)

Vì trong 3 số nguyên liên tiếp, có ít nhất 1 số chia hết cho 3 và 1 số chia hết cho 2 nên tích n(n-1)(n+1) chia hết cho 6 hay \(n^2\left(n+1\right)-n\left(n+1\right)\) chia hết cho 6(đpcm).

b) Ta có:

\(20^{n+1}-20^n=20^n\cdot19\)

Vì \(20^n\) là số nguyên nên \(20^n\cdot19⋮19\). Hay \(20^{n+1}-20^n⋮19\left(đpcm\right)\)

\(y^2+2xy-12x+4\left(x+y\right)+2x^2+40=0\\ \Leftrightarrow\left[\left(x^2+2xy+y^2\right)+4\left(x+y\right)+4\right]+\left(x^2-12x+36\right)=0\\ \Leftrightarrow\left(x+y+2\right)^2+\left(x-6\right)^2=0\)

Vì \(\left\{{}\begin{matrix}\left(x+y+2\right)^2\ge0\forall x,y\\\left(x-6\right)^2\ge0\forall x\end{matrix}\right.\) 

Nên \(\left(x+y+2\right)^2+\left(x-6\right)^2\ge0\forall x,y\)

Dấu"=" xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}x+y+2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-8\\x=6\end{matrix}\right.\)

Vậy x = 6 và y = -8

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