mk chịu bn ơi,nhắn đáp án cho mình nhé

tóc trắng nhỉ

<=>x^2-4x+4=x^2+8x+16

<=>12x+12=0

<=>x=-1

P=\(\left(a^2+b^2+c^2+2ab+2ac+2bc\right)+4\left(ab+bc+ca\right)-\left(a^2+b^2+c^2\right)\)\(+a^3+b^3+c^3-2\left(a^2b+b^2c+c^2a\right)+ab^2+bc^2+ca^2\)\(=1+4\left(ab+bc+ca\right)-\left(a^2+b^2+c^2\right)\left(a+b+c\right)+\left(a^3+b^3+c^3\right)\)\(-2\left(a^2b+b^2c+c^2a\right)+\left(ab^2+bc^2+ca^2\right)\)\(=1+4\left(ab+bc+ca\right)-3\left(a^2b+b^2c+c^2a\right)\)

Mà \(\left(a^2b+b^2c+c^2a\right)\left(b+c+a\right)\ge\left(ab+bc+ca\right)^2\)

=> \(P\le1+4\left(ab+bc+ca\right)-3\left(ab+bc+ca\right)^2\). Đặt \(ab+bc+ca=t\le\frac{1}{3}\)

=> \(P\le-3\left(t^2-\frac{2}{3}t+\frac{1}{9}\right)+2t+\frac{4}{3}\le-3\left(t-\frac{1}{3}\right)^2+\frac{2}{3}+\frac{4}{3}\le2\)

Dấu bằng xảy ra khi \(t=\frac{1}{3}\)<=> \(a=b=c=\frac{1}{3}\)

\(\left(x-\frac{5}{7}\right)-\frac{7}{21}=\frac{5}{3}\)

\(\left(x-\frac{5}{7}\right)-\frac{1}{3}=\frac{5}{3}\)

\(x-\frac{5}{7}=\frac{5}{3}+\frac{1}{3}\)

\(x-\frac{5}{7}=\frac{6}{3}\)

\(x-\frac{5}{7}=2\)

\(x=2+\frac{5}{7}\)

\(x=\frac{19}{7}\)

nhớ tk mk nhé các bn

\(\left(x-\frac{5}{7}\right)-\frac{7}{21}=\frac{5}{3}\)

\(x-\frac{5}{7}=\frac{5}{3}+\frac{7}{21}\)

\(x-\frac{5}{7}=2\)

\(x=2-\frac{5}{7}\)

\(x=\frac{9}{7}\)

Ta có: 2005 đồng dư với 1 theo mod 2004 (hay 2005 chia 2004 dư 1)

=> 2005³ đồng dư với 1³ theo mod 2004 

<=> 2005³ đồng dư với 1 theo mod 2004 

=> 2005³-1 chia hết cho 2004

cảm ơn đạm thị trà hương