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Từ \(4a^2+b^2=5ab\), ta có: \(4a^2-4ab-ab+b^2\)=0

Hay: (a-b) (4a-b)=0

Vì: 2a>b>0 nên 4a-b \(\ne\)0 .

Từ: (.) \(\Rightarrow\)

Từ: a-b=0 . Tức là: a=b

Thay a=b vào C ta được :

C= \(\frac{ab}{4a^2-b^2}=\frac{a^2}{4a^2-a^2}=\frac{1}{3}\)(do a\(\ne\)0)

Sai: do

Sửa: doing

she is tired of doing the same thing

1) a) A= (x+2)(\(x^2-2x+4\)) -(\(x^3-2\))

=(x+2)(\(x^2-2x+2^2\))-(\(x^3-2\))

= \(x^3+2^3\)-\(x^3+2\)

=(\(x^3-x^3\))+(\(2^3+2\))

=10

b) B= (a+2)(a-2)(\(a^2+2a+4\))(\(a^2-2a+4\))

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

=\(a^2-4+a^2+4a^2+16\)

=(\(a^2+a^2+4a^2\))+(-4+16)

=\(6a^2\)+12

Với số tự nhiên n, ta có:

\(\frac{n\left(n+1\right)}{2}+\frac{\left(n+1\right)\left(n+2\right)}{2}=\frac{n\left(n+1\right)}{2}+\frac{n\left(n+1\right)+2\left(n+1\right)}{2}\)

\(=\frac{n\left(n+1\right)}{2}+\frac{n\left(n+1\right)}{2}+n+1\)

\(=n\left(n+1\right)+n+1=\left(n+1\right)\left(n+1\right)=\left(n+1\right)^2\)là số chính phương

Từ gt => ab+bc+ca=0

\(a^2+2bc=a^2+bc+\left(-ab-ac\right)=a\left(a-b\right)-c\left(a-b\right)=\left(a-b\right)\left(a-c\right)\)

Tương tự \(\hept{\begin{cases}b^2+2ac=\left(b-a\right)\left(b-c\right)\\c^2+2ab=\left(c-a\right)\left(c-b\right)\end{cases}}\)

\(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}=\frac{b-c+c-a+a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=0\)

Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow ab+bc+ca=0\)

\(\Rightarrow bc=-ab-ca\)

Vậy thì \(a^2+2bc=a^2+bc-ab-ac=a\left(a-b\right)-c\left(a-b\right)=\left(a-c\right)\left(a-b\right)\)

Tương tự ta cũng có: 

\(b^2+2ac=\left(b-c\right)\left(b-a\right)\)

\(c^2+2ab=\left(c-a\right)\left(c-b\right)\)

Vậy thì \(A=\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\)

\(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\)

\(A=\frac{-b+c+a-c-a+b}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

\(A=0.\)