bn áp dụng HĐT sẽ ra

=(x+3)^3

=x³ + 3x².3 + 3x.3² + 3³

Câu hỏi của bggvf - Toán lớp 8 - Học toán với OnlineMath

Em tham khảo tại link bên trên nhé.

Giải:
Chú ý sử dụng hằng đẳng thức : 
(m+n+p)² = m² + n² + p² + 2mn + 2mp + 2np
Áp dụng hằng đẳng thức trên, ta có:
ax+by+cz = 0 ⇒ (ax+by+cz)² = 0

⇒a²x²+b²y²+c²z²+2ax.by+2ax.cz+2by.cz=0

⇒a²x²+b²y²+c²z²= − (2abxy+2aczx+2bcyz)

Ta lại có:
bc.(y−z)²+ac.(x−z)²+ab.(x−y)²

=bc(y²−2yz+z²)+ac(x²−2xz+z²)+ab(x²−2xy+y²)

=bcy²+bcz²−2bcyz+acx²+acz²−2acxz+abx²+aby²−2abxy

=(bcy²+bcz²+acx²+acz²+abx²+aby²)−(2abxy+2aczx+2bcyz)

=bcy²+bcz²+acx²+acz²+abx²+aby²+a²x2+b²y²+c²z²

=x²(ac+ab+a²)+y²(bc+ab+b²)+z²(bc+ac+c²)

=ax²(a+b+c)+by²(a+b+c)+cz²(a+b+c)

=(a+b+c)(a.x²+b.y²+c.z²)

Vậy:
A=a.x²+b.y²+c.z2bc.(y−z)²+ac.(x−z)²+ab.(x−y)²=ax²+by²+cz²(a+b+c)(a.x²+b.y²+c.z²)

A=1a+b+cA=1a+b+c

\(\frac{x+1}{2015}+\frac{x+2}{2014}=\frac{x+3}{2013}+\frac{x+4}{2012}\)

\(\Leftrightarrow\frac{x+1}{2015}+1+\frac{x+2}{2014}+1=\frac{x+3}{2013}+1+\frac{x+4}{2012}+1\)

\(\Leftrightarrow\frac{x+2016}{2015}+\frac{x+2016}{2014}=\frac{x+2016}{2013}+\frac{x+2016}{2012}\)

\(\Leftrightarrow\frac{x+2016}{2015}+\frac{x+2016}{2014}-\frac{x+2016}{2013}-\frac{x+2016}{2012}=0\)

\(\Leftrightarrow\left(x+2016\right)\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)

\(\Leftrightarrow x+2016=0\).Do \(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\ne0\)

\(\Leftrightarrow x=-2016\)

\(\left\{-2016\right\}\)