ko phai bang 4 dau minh thi r ma!

a, a + b + c = 12 + 15 + 9 = 36

b, a  × b  × c = 15  × 0  × 37 = 0

a: \(x^2-8x+5\)

\(=x^2-8x+16-11\)

\(=\left(x-4\right)^2-11\ge-11\forall x\)

Dấu '=' xảy ra khi x-4=0

=>x=4

b: \(a^3+b^3+c^3=3bac\)

=>\(\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)

=>\(\left(a+b+c\right)\left\lbrack\left(a+b\right)^2-c\left(a+b\right)+c^2\right\rbrack-3ab\left(a+b+c\right)=0\)

=>\(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)

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

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

=>\(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)

=>a=b=c

\(N=\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}\)

\(=\frac{a^2+a^2+a^2}{\left(a+a+a\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{3}{3^2}=\frac13\)

giúp mình với

a.=72

b.=9

mình nghĩ là vậy

Th1: a+b+c khác 0

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

\(\Rightarrow2+\frac{a+b-c}{c}=2+\frac{a-b+c}{b}=2+\frac{\left(-a\right)+b+c}{a}\)

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

\(\Rightarrow a=b=c\)

thay a=b=c vào b/t A. ta có:

\(A=\frac{aaa}{\left(a+a\right).\left(a+a\right).\left(a+a\right)}=\frac{aaa}{2a.2a.2a}=\frac{aaa}{8aaa}=\frac{1}{8}\)

th2: a+b+c = 0

=> a+b=-c

b+c=-a

c+a=-b

thay a+b=-c, b+c=-a, c+a=-b vào b/t A ta có:

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

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

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

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

+)Nếu a+b+c=0\(\Rightarrow a+b=-c;b+c=-a;c+a=-b\)

\(\Rightarrow B=\frac{a+b}{a}.\frac{c+a}{c}.\frac{b+c}{b}=\frac{-c}{a}.\frac{-b}{c}.\frac{-a}{b}=\frac{-\left(abc\right)}{abc}=-1\)

Nếu \(a+b+ c\ne0\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có 

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

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

      \(b+ c=2a\)

       \(c+a=2b\)

\(\Rightarrow B=\frac{2c}{a}.\frac{2b}{c}.\frac{2a}{b}=2.2.2=8\)

chumia sư phụ cứu zới !!!