Theo bài ra:

\(\dfrac{a}{b+c}=\dfrac{b}{a+c}=\dfrac{c}{a+b};a\ne b\ne c;a,b,c\ne0\)

\(P=\dfrac{b+c}{a}+\dfrac{a+c}{b}+\dfrac{a+b}{c}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\dfrac{a}{b+c}=\dfrac{b}{a+c}=\dfrac{c}{a+b}=\dfrac{a+b+c}{b+c+a+c+a+b}=\dfrac{a+b+c}{2a+2b+2c}=\dfrac{a+b+c}{2\left(a+b+c\right)}=\dfrac{1}{2}\)

\(hay:\dfrac{a}{b+c}=\dfrac{1}{2}\Rightarrow a=\dfrac{b+c}{2}\)

Thay \(a=\dfrac{b+c}{2}\) vào \(P\), ta có:

\(P=\dfrac{b+c}{\dfrac{b+c}{2}}+\dfrac{b+c+c}{b}+\dfrac{b+c+b}{c}\\ P=\dfrac{2\left(b+c\right)}{b+c}+\dfrac{2c+b}{b}+\dfrac{2b+c}{c}\\ P=2+\dfrac{2c}{b}+\dfrac{b}{b}+\dfrac{2b}{c}+\dfrac{c}{c}\\ P=2+\dfrac{2c}{b}+1+\dfrac{2b}{c}+1\\ P=\left(2+1+1\right)+\dfrac{2c}{b}+\dfrac{2b}{c}\\ P=4+\dfrac{2c}{b}+\dfrac{2b}{c}\\ P=4+\dfrac{2c+2b}{b+c}\\ P=4+\dfrac{2\left(b+c\right)}{b+c}\\ P=4+2\\ P=6\)

Vậy: \(P=6\)

thiếu đè

Thiếu nhé:

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

\(\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{bk}{b\left(k-1\right)}=\dfrac{k}{k-1}\)

\(\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{dk}{d\left(k-1\right)}=\dfrac{k}{k-1}\)

Ta có điều phải chứng minh

Có \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=2^2\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ac}=4\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{c+a+b}{abc}\right)=4\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2=4\) (do \(a+b+c=abc\))
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\). (đpcm).

vi a mu 2=bc nen

Chỗ giả thiết vế phải có đúng ko vậy

Bài 1:

\(a^2+b^2+c^2=16\Rightarrow\left(a+b+c\right)^2-2ab-2bc-2ac=16\)\(\Leftrightarrow-2\left(ab+bc+ac\right)=16\Rightarrow ab+bc+ac=-8\)\(\Rightarrow\left(ab+bc+ac\right)^2=64\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2=64\)\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=64\)

\(\Rightarrow a^2b^2+b^2c^2+a^2c^2=64\)

Ta có:

\(a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2a^2b^2-2b^2c^2-2a^2c^2\)\(=16^2-2\left(a^2b^2+b^2c^2+a^2c^2\right)=256-2.64=128\)

1. Ta có $a + b + c = 0$

\(\Rightarrow\) $( a + b + c)^2 = 0$

\(\Leftrightarrow\) $a^2+b^2 +c^2 +2ab+2bc+2ac = 0
\(\Leftrightarrow\) $a^2 + b^2 + c^2 = -2(ab+bc+ac)$

Thay $a^2 + b^2 + c^2 = 2$

\(\Rightarrow\)$2 = -2(ab+bc+ac)$ \(\Rightarrow\) $ab + bc +ac = -1 $
Ta có: $(a^2+b^2+c^2) = 2$

\(\Leftrightarrow\) $(a^2+b^2+c^2)^2 = 4$

\(\Leftrightarrow\)$a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2 = 4$

\(\Leftrightarrow\) $a^4+b^4+c^4 + 2(a^b^2+b^2c^2+a^2c^2) = 4$ (1)
Do $2(ab+bc+ac)^2 = 2(a^2b^2+b^2c^2+a^2c^2 + 2a^2bc+2ab^2c+2abc^2)$ (2)
Từ (1)(2) => $a^4+b^4+c^4+2(ab+bc+ac)^2 - 4abc(a+b+c) = 4$(3)
Thay $(ab+bc+ac) = -1$ và $a+b+c = 0$ (4)
Từ (3)(4) => $a^4 + b^4 + c^4 +2(-1)^2 -4abc.(0) = 4 $
<=> $a^4 + b^4 + c^4 + 2 = 4 => a^4 + b^4 + c^4 = 2 $