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\(x-y=1\Rightarrow x^2-2xy+y^2=1\Rightarrow x^2+xy+y^2=19\Rightarrow x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)=1.19=19\)
\(2,a^2+b^2+c^2=ab+bc+ca\Leftrightarrow2\left(a^2+b^2+c^2\right)=2ab+2bc+2ca\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)=0\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0ma:\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\Leftrightarrow a=b=c\)
\(a+b+c=0\Leftrightarrow\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ca=0\Leftrightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)\Rightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=4a^2b^2+4b^2c^2+4c^2a^2+4abc\left(a+b+c\right)=4a^2b^2+4c^2a^2+4b^2c^2\Rightarrow a^4+b^4+c^4=2a^2b^2+2b^2c^2+2c^2a^2\Leftrightarrow2\left(a^4+b^4+c^4\right)=a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=\left(a^2+b^2+c^2\right)^2\left(dpcm\right)\)
a/VT=x5+x^4.y+x^3.y^2+x^2.y^4+x.y^4-x^4.y-x^3.y^2-x^2.y^3-x.y^4-y^5
=x^5-y^5=VP
=>dpcm
a) a3+b3+a2c+b2c-abc
= (a+b)(a2-ab+b2)+c(a2+b2)-abc
=(a+b) [ (a+b)2-3ab]+c.[(a+b)2-2ab]-abc
=(a+b)(a+b)2-3ab(a+b)+c(a+b)2-3abc
=(a+b)2(a+b+c)-3ab(a+b+c)
=(a+b)2.0-3ab.0
=0
b) ax+ay+2x+2y+4
=a(x+y)+2(x+y)+4
=(x+y)(a+2)+4
=(a-2)(a+2)+4
=a2-4+4
=a2
c) A=1+x+x2+...+x49=>Ax=x+x2+x3+...+x50
- A=1+x+x2+...+x49
---> Ax-A=x50-1
d)(a+b)(a+c)+(c+a)(c+b)
=a2+ac+ab+bc+c2+bc+ac+ab
=a2+c2+2ac+2ab+2bc
=2b2+2bc+2ac+2ab
=2b(b+c)+2a(b+c)
=2b(b+c)(b+a)
áp dụng BĐT (a - b)² ≥ 0 → a² + b² ≥ 2ab ta có:
x² + y² ≥ 2xy
x² + 1 ≥ 2x
y² + z² ≥ 2yz
y² + 1 ≥ 2y
z² + x² ≥ 2xz
z² + 1 ≥ 2z
Cộng theo vế → 3(x² + y² + z²) + 3 ≥ 2(x + y + z + xy + yz + zx) = 2.6 = 12
→ x² + y² + z² ≥ 9/3 = 3
→ đpcm (dấu = xảy ra khi x = y = z = 1)