Lời giải:

Theo đề bài ta có:

\(\frac{2ab+1}{2b}=\frac{2bc+1}{c}=\frac{ac+1}{a}\Leftrightarrow a+\frac{1}{2b}=2b+\frac{1}{c}=c+\frac{1}{a}\)

\(\Rightarrow \left\{\begin{matrix} a-2b=\frac{1}{c}-\frac{1}{2b}=\frac{2b-c}{2bc}\\ a-c=\frac{1}{a}-\frac{1}{2b}=\frac{2b-a}{2ab}\\ 2b-c=\frac{1}{a}-\frac{1}{c}=\frac{c-a}{ac}\end{matrix}\right.\)

Nhân theo vế:
\((a-2b)(a-c)(2b-c)=\frac{(2b-c)(2b-a)(c-a)}{4a^2b^2c^2}=\frac{(2b-c)(a-2b)(a-c)}{4a^2b^2c^2}\)

\(\Leftrightarrow (a-2b)(a-c)(2b-c)\left[1-\frac{1}{4a^2b^2c^2}\right]=0\)

$\Rightarrow (a-2b)(a-c)(2b-c)=0$ hoặc $1-\frac{1}{4a^2b^2c^2}=0$

TH1: $(a-2b)(a-c)(2b-c)=0$\(\Rightarrow \left\{\begin{matrix} a=2b\\ a=c\\ 2b=c\end{matrix}\right.\)

+Nếu $a=2b$ thì $\frac{2b-c}{2bc}=a-2b=0\Rightarrow 2b-c=0\Rightarrow 2b=c$

$\Rightarrow a=2b=c$

+ Nếu $a=c, 2b=c$: hoàn toàn tương tự suy ra $a=2b=c$

TH2: $1-\frac{1}{4a^2b^2c^2}=0\Rightarrow 4a^2b^2c^2=1$

Vậy ta có đpcm.

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)

=> bc+ac+ab=0

ta có

\(bc+ac=-ab\)

<=> \(\left(bc+ac\right)^2=a^2b^2\)

<=> \(b^2c^2+a^2c^2+2abc^2=a^2b^2\)

<=> \(b^2c^2+a^2c^2-a^2b^2=-2abc^2\)

tương tự

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

\(c^2a^2+a^2b^2-b^2c^2=-2a^2bc\)

thay vào E ta đc

\(E=\dfrac{-a^2b^2c^2}{2ab^2c}-\dfrac{a^2b^2c^2}{2abc^2}-\dfrac{a^2b^2c^2}{2a^2bc}\)

=\(-\dfrac{ac}{2}-\dfrac{ab}{2}-\dfrac{bc}{2}=\dfrac{-\left(ac+ab+bc\right)}{2}=0\) (vì ac+bc+ab=0 cmt)

Hình như sai đề :

Ta có : \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)

\(\Leftrightarrow\dfrac{bc}{abc}+\dfrac{ac}{abc}+\dfrac{ab}{abc}=0\)

\(\Leftrightarrow\dfrac{ab+ac+bc}{abc}=0\)

\(\Leftrightarrow ab+ac+bc=0\) ( do \(a;b;c\ne0\) ) ( 1 )

Từ ( 1 ) \(\Rightarrow ab+bc=-ac\)

\(\Rightarrow\left(ab+bc\right)^2=\left[-\left(ac\right)\right]^2\)

\(\Rightarrow a^2b^2+b^2c^2+2ab^2c=a^2c^2\) ( * )

CMTT , ta được : \(\left\{{}\begin{matrix}b^2c^2+c^2a^2+2bc^2a=a^2b^2\\c^2a^2+a^2b^2+2a^2cb=b^2c^2\end{matrix}\right.\) ( *' )

Thay ( * ) và ( * ') vào E , ta được :

\(E=\dfrac{a^2b^2c^2}{a^2b^2+b^2c^2-\left(a^2b^2+b^2c^2+2b^2ac\right)}+\dfrac{a^2b^2c^2}{b^2c^2+c^2a^2-\left(b^2c^2+c^2a^2+2bc^2a\right)}\)

\(+\dfrac{a^2b^2c^2}{c^2a^2+a^2b^2-\left(c^2a^2+a^2b^2+2a^2cb\right)}\)

\(=\dfrac{a^2b^2c^2}{-2b^2ac}+\dfrac{a^2b^2c^2}{-2c^2ab}+\dfrac{a^2b^2c^2}{-2a^2cb}\)

\(=\dfrac{-ac}{2}+\dfrac{-ab}{2}+\dfrac{-bc}{2}\)

\(=\dfrac{-\left(ac+ab+bc\right)}{2}\)

\(=\dfrac{0}{2}=0\)

Vậy \(E=0\)

Lời giải:

Ta có: \(a^2b+b^2c+c^2a\geq \frac{9a^2b^2c^2}{1+2a^2b^2c^2}\)

\(\Leftrightarrow (a^2b+b^2c+c^2a)(1+2a^2b^2c^2)\geq 9a^2b^2c^2\)

\(\Leftrightarrow a^2b+b^2c+c^2a+2a^4b^3c^2+2a^2b^4c^3+2a^3b^2c^4\geq 3a^2b^2c^2(a+b+c)(*)\)

--------------------------

Áp dụng BĐT AM-GM ta có:

\(a^2b+a^4b^3c^2+a^3b^2c^4\geq 3\sqrt[3]{a^9b^6c^6}=3a^3b^2c^2\)

\(b^2c+a^2b^4c^3+a^4b^3c^2\geq 3a^2b^3c^2\)

\(c^2a+a^3b^2c^4+a^2b^4c^3\geq 3a^2b^2c^3\)

Cộng theo vế:

\(\Rightarrow a^2b+b^2c+c^2a+2a^4b^3c^2+2a^2b^4c^3+2a^3b^2c^4\geq 3a^2b^2c^2(a+b+c)\)

Vậy $(*)$ đúng

Do đó ta có đpcm

Dấu bằng xảy ra khi $a=b=c=1$

Lời giải:
\((2a+2b-c)^2+(2b+2c-a)^2+(2c+2a-b)^2\)

\(=(2a+2b)^2-2c(2a+2b)+c^2+(2b+2c)^2-2a(2b+2c)+a^2+(2c+2a)^2-2b(2c+2a)+b^2\)

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

\(=4(a^2+2ab+b^2)+4(b^2+2bc+c^2)+4(c^2+2ac+a^2)+(c^2+a^2+b^2)-8(ab+bc+ac)\)

\(=9(a^2+b^2+c^2)=9.9=81\)

Bài 1 : 

\(a)\)Ta có : 

\(A=\frac{2.6^9-4^5.9^4}{20.6^8+2^{10}.3^8}\)

\(A=\frac{2.\left(2.3\right)^9-\left(2^2\right)^5.\left(3^2\right)^4}{\left(2^2.5\right).\left(2.3\right)^8+2^{10}.3^8}\)

\(A=\frac{2.2^9.3^9-2^{10}.3^8}{2^2.5.2^8.3^8+2^{10}.3^8}\)

\(A=\frac{2^{10}.3^9-2^{10}.3^8}{2^{10}.3^8.5+2^{10}.3^8}\)

\(A=\frac{2^{10}.3^8\left(3-1\right)}{2^{10}.3^8\left(5+1\right)}\)

\(A=\frac{2}{6}\)

\(A=\frac{1}{3}\)

Vậy \(A=\frac{1}{3}\)

Năm mới zui zẻ nhé ^^

thanks

\(B=x^3+3x^2-4\)

\(B=x^3-x^2+4x^2-4\)

\(B=x^2\left(x-1\right)+4\left(x^2-1\right)\)

\(B=x^2\left(x-1\right)+4\left(x+1\right)\left(x-1\right)\)

\(B=\left(x-1\right)\left(x^2+4x+4\right)\)

\(B=\left(x-1\right)\left(x+2\right)^2\)