Ta có: \(ac=b^2\)

=>\(\frac{a}{b}=\frac{b}{c}\)

Ta có: \(ab=c^2\)

=>\(\frac{b}{c}=\frac{c}{a}\)

=>\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\)

Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=k\)

=>\(\begin{cases}c=ak\\ b=ck=ak\cdot k=ak^2\\ a=bk=ak^2\cdot k=ak^3\end{cases}\Rightarrow\begin{cases}c=ak\\ b=ak^2\\ a\left(k^3-1\right)=0\end{cases}\)

=>\(\begin{cases}c=ak\\ b=ak^2\\ k^3-1=0\end{cases}\Rightarrow\begin{cases}k=1\\ c=a\cdot1=a\\ b=a\cdot1^2=a\end{cases}\)

=>a=b=c

\(P=\frac{a^{555}}{b^{222}\cdot c^{333}}+\frac{b^{555}}{c^{222}\cdot a^{333}}+\frac{c^{555}}{a^{222}\cdot b^{333}}\)

\(=\frac{a^{555}}{a^{222}\cdot a^{333}}+\frac{a^{555}}{a^{222}\cdot a^{333}}+\frac{a^{555}}{a^{222}\cdot a^{333}}\)

=1+1+1

=3

Bài 3: 

a: \(\left(a-b\right)^2=\left(a+b\right)^2-4ab=7^2-4\cdot12=1\)

b: \(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\)

\(=7^3-3\cdot12\cdot7\)

\(=343-252=91\)

Tham khảo:

CHÚC EM HỌC TỐT NHÁ 

a = 2;b= (-2);c= 3

Thay : a+b+c=2+(-2)+3

                 .     =[2+(-2)]+3

                       =0+3=3

B)vì a và b là 2 số đối nhau nên ta có :

a =2;b= (-2) và là 2số đối nhau vì

|-2|=2

Áp dụng cô si ,ta có

\(a^2+b^2\ge2ab\)

\(c^2+b^2\ge2bc\)

\(a^2+c^2\ge2ac\)

\(\Rightarrow2a^2+2b^2+2c^2\ge2ab+2ac+2bc\)

\(a^2+b^2+c^2\ge ab+ac+bc\)

\(\Rightarrow a^2+b^2+c^2+2ab+2ac+2bc\ge3ab+3ac+3bc\)

\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ac\right)\)

\(\Rightarrow200^2\ge3\left(ab+bc+ac\right)\)

\(\Rightarrow\frac{40000}{3}\ge ab+bc+ac\)

Dấu = xảy ra khi a=b=c=200/3

Chọn A

Ta có: 

Do đó, . Vậy