\(x=\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}\)

Cộng 1 vào mỗi tỉ số ta được :

\(x+1=\frac{a+b}{c}+1=\frac{b+c}{a}+1=\frac{c+a}{b}+1\)

\(x+1=\frac{a+b+c}{c}=\frac{a+b+c}{a}=\frac{a+b+c}{b}\)

Xét a + b + c = 0 

\(\Rightarrow\)a + b = -c ; b + c = -a ; a + c = -b

\(\Rightarrow\)\(x=\frac{-c}{c}=\frac{-a}{a}=\frac{-b}{b}=-1\)

\(\Rightarrow\)( x² + x + 1 )¹⁰⁰ = [ ( -1 )² + ( -1 ) + 1 ]¹⁰⁰ = 1¹⁰⁰ = 1

Xét a + b + c \(\ne\)0  thì a = b = c

\(\Rightarrow x=\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=2\)

\(\Rightarrow\)( x² + x + 1 )¹⁰⁰ = ( 2² + 2 + 1 )¹⁰⁰ = 7¹⁰⁰ 

thanks nhìu 

Lời giải:

$B=|\frac{(a-c)(b-a)(b+c)}{abc}|$

Do $a-b-c=0$ nên: $b-a=-c; a-c=b; b+c=a$

$\Rightarrow (a-c)(b-a)(b+c)=b(-c)a=-abc$

$\Rightarrow B=|\frac{-abc}{abc}|=|-1|=1$

Cho abc(a+b+c) khác 0. Giải phương trình ẩn x:

(x-a)/bc+(x-b)/ac+(x-c)/ab=1/2(1/a+1/b+1/c)

.

bf

\(\dfrac{a}{b}=\dfrac{c-a}{b-c}\Rightarrow ab-ac=bc-ab\\ \Rightarrow ac+bc=2ab\\ \dfrac{1}{c}=x\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=\dfrac{x}{a}+\dfrac{x}{b}=\dfrac{ax+bx}{ab}\\ \Rightarrow ac.x+bc.x=ab\\ \Rightarrow x\left(ac+bc\right)=ab\\ \Rightarrow2x\left(ac+bc\right)=2ab\\ \Rightarrow2x.2ab=2ab\\ \Rightarrow2x=1\Rightarrow x=\dfrac{1}{2}\)

thank bạn

..............c

........1...........1

a.....................

0.....................

a x (b+c)= a x b + a x c

a x 1= 1x a = a

a : 1 = a

a : a=1

0 : a = 0

( a khác 0)

a x (b+c)=a x b + a x c

a x 1= 1x a=a

a : 1 =a

a : a = 1

0 : a = 0

Bài 1 rõ thế còn gì.