Vì 

\(x^2\ge0\) \(\forall x\)

\(\Rightarrow x^2+13\ge13\) \(\forall x\)

\(\Rightarrow C=\left(x^2+13\right)^2\ge13^2\) \(\forall x\)

\(\Rightarrow C=\left(x^2+13\right)^2\ge169\) \(\forall x\)

Dấu "=" xảy ra <=> \(x=0\)

Vậy \(C_{min}=169\) tại \(x=0\)

x=8 do dua nao cay ong va vo mom

Ta có : \(\left(x-y\right)^2=x^2-2xy+y^2=x^2-2.2+y^2\)

\(\Rightarrow x^2+y^2=4\)

\(\Rightarrow x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=\left(x+y\right)\left[\left(x^2+y^2\right)-xy\right]\)

\(=4\left(4-2\right)=8\)

\(B=\frac{2012}{1}+\frac{2011}{2}+\frac{2010}{3}+....+\frac{1}{2012}\)

\(=1+\left(\frac{2011}{2}+1\right)+\left(\frac{2010}{3}+1\right)+....+\left(\frac{1}{2012}+1\right)\)

\(=\frac{2013}{2}+\frac{2013}{3}+.....+\frac{2013}{2012}+\frac{2013}{2013}\)

\(=2013\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}\right)\)

\(\Rightarrow\frac{B}{A}=\frac{2013\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2013}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2013}}=2013\)

có vẻ đề bài bn sai

Ta có : 

\(\frac{a}{k}=\frac{x}{a}\Rightarrow a^2=kx\) (1)

\(\frac{b}{k}=\frac{y}{b}\Rightarrow b^2=ky\) (2)

Chia hai vế của (1) cho (2) ta được :

\(\frac{a^2}{b^2}=\frac{kx}{ky}\)\(\Rightarrow\frac{a^2}{b^2}=\frac{x}{y}\) (đpcm)

\(A=\left[-a^5.\left(-a^5\right)\right]^2+\left[-a^2.\left(-a^2\right)\right]^5=0\)O

=>\(\left(-a^{10}\right)^2+\left(-a^4\right)^5=a^{20}-a^{20}=0\)

\(B;\left(-1\right)^n.a^{a+k}=\left(-a\right)^n.a^k\)

\(=\left(-1\right)^n.a^n.a^k=\left(-1.a\right)^n.a^k\)

=\(\left(-a^n\right).a^k\)

chào bạn

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