Vì $abc\ne0$ nên từ $a+b+c=\dfrac1a+\dfrac1b+\dfrac1c$

suy ra $abc(a+b+c)=ab+bc+ca$.

Xét hiệu hai vế cần chứng minh:

$b(a^2-bc)(1-ac)-a(1-bc)(b^2-ac)$

$=-(a-b)\left[abc(a+b+c)-(ab+bc+ca)\right]$

$=-(a-b)\cdot0$

$=0$

Suy ra $\boxed{b(a^2-bc)(1-ac)=a(1-bc)(b^2-ac)}$

Ta có $(a+b+c)^2=a^2+b^2+c^2$

$\Rightarrow 2(ab+bc+ca)=0$

$\Rightarrow ab+bc+ca=0$

Ta có $\dfrac1{a^3}+\dfrac1{b^3}+\dfrac1{c^3}
=\dfrac{a^3b^3+b^3c^3+c^3a^3}{a^3b^3c^3}$

Mà $a^3b^3+b^3c^3+c^3a^3-3a^2b^2c^2$

$=(ab+bc+ca)(a^2b^2+b^2c^2+c^2a^2-abc(a+b+c))$

$=0$

Do đó $a^3b^3+b^3c^3+c^3a^3=3a^2b^2c^2$

Suy ra $\dfrac1{a^3}+\dfrac1{b^3}+\dfrac1{c^3}$

$=\dfrac{3a^2b^2c^2}{a^3b^3c^3}$

$=\boxed{\dfrac3{abc}}$

Ta có A=\(\left(ab+bc+ca\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-abc\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)

=\(2\left(a+b+c\right)+\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}-\frac{ab}{c}-\frac{bc}{a}-\frac{ca}{b}=2\left(a+b+c\right)\)

\(A=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2=a^2-ab+b^2+3ab\left(1-2ab\right)+6a^2b^2\)

=\(\left(a+b\right)^2-3ab+3ab-6a^2b^2+6a^2b^2=1\)

2) Ta có \(A=\left(a-1\right)\left(b-1\right)\left(c-1\right)=abc-ab-bc-ca+a+b+c-1=0\)

Ta có  \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{3}{4}\)

\(\Leftrightarrow\frac{b^2c^2+c^2a^2+a^2b^2}{\left(abc\right)^2}=\frac{3}{4}\)

\(\Leftrightarrow\frac{b^2c^2+c^2a^2+a^2b^2}{64}=\frac{3}{4}\)

\(\Leftrightarrow b^2c^2+c^2a^2+a^2b^2=\frac{3.64}{4}=48\)

Do đó  \(T=\frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}=\frac{b^2c^2+c^2a^2+a^2b^2}{abc}=\frac{48}{8}=6\)

\(\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)abc=\frac{3}{4}8\Rightarrow\frac{abc}{a^2}+\frac{abc}{b^2}+\frac{abc}{c^2}=\frac{3.8}{4}\Leftrightarrow\)\(\frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}=6\)

lm đc trước rồi nhưng cũng k cho

Ai giúp mik vs, mik cam on !!!! :)

Lời giải:

Vì $a+b+c=1$ nên:

\(a^2+b^2+abc-1=(a+b)^2-2ab+abc-1\)

\(=(a+b)^2-1+ab(c-2)=(1-c)^2-1+ab(c-2)\)

\(=-c(2-c)+ab(c-2)=c(c-2)+ab(c-2)=(c+ab)(c-2)\)

Do đó:

\(\frac{c+ab}{a^2+b^2+abc-1}=\frac{c+ab}{(c+ab)(c-2)}=\frac{1}{c-2}\)

Hoàn toàn tương tự với các phân thức còn lại, suy ra:

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

\(=\frac{ab+bc+ac-4(a+b+c)+12}{(a-2)(b-2)(c-2)}=\frac{ab+bc+ac+8}{(a-2)(b-2)(c-2)}\)

Ta có đpcm.

Akai Haruma