\(\Leftrightarrow a^3+6a^2b+12ab^2+8b^3=6a^2b+12ab^2-6ab+1\)

\(\Leftrightarrow\left(a+2b\right)^3=6ab\left(a+2b-1\right)+1\)

\(\Leftrightarrow\left(a+2b\right)^3-1-6ab\left(a+2b-1\right)=0\)

\(\Leftrightarrow\left(a+2b-1\right)\left(\left(a+2b\right)^2+a+2b+1\right)-6ab\left(a+2b-1\right)=0\)

\(\Leftrightarrow\left(a+2b-1\right)\left(a^2+4ab+4b^2+a+2b+1-6ab\right)=0\)

\(\Leftrightarrow\left(a+2b-1\right)\left(a^2-2ab+4b^2+a+2b+1\right)=0\)

TH1: Nếu \(a+2b-1=0\)

\(\Leftrightarrow a+2b-1=0\)

\(\Rightarrow a+2b=1\)

TH2: \(a^2-2ab+4b^2+a+2b+1=0\)

\(\Leftrightarrow a^2-2ab+4b^2+a+2b+1=0\)

\(\Leftrightarrow\left(a-b+\frac{1}{2}\right)^2+3\left(b+\frac{1}{2}\right)^2=0\)

\(\Rightarrow\left\{{}\begin{matrix}a-b+\frac{1}{2}=0\\b+\frac{1}{2}=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-1\\b=-\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow a+2b=-2\)

a + b + c = 0
(a + b + c)² = 0
a² + b² + c² + 2(ab + bc + ca) = 0
a² + b² + c² = -2(ab + bc + ca)

\(\Rightarrow\left(a^2+b^2+c^2\right)^2=\left[-2\left(ab+bc+ca\right)\right]^2\)

\(a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2=4\left(a^2b^2+a^2c^2+b^2c^2+2ab^2c+2a^2bc+2abc^2\right)\)

=>\(a^4+b^4+c^4=2a^2b^2+2a^2c^2+2b^2c^2+4ab^2c+4abc^2+4a^2bc\)

=>\(a^4+b^4+c^4=2\left(ab+bc+ca\right)^2\left(đpcm\right)\)

Sakura Harunoo bạn Nguyễn Thị Hồng Nhung bạn đó copp cho a+b+c=0. CMR:a 4 +b 4 +c 4 = 2(a 2 b - Online Math

Sakura Harunoo nhớ nhìn kĩ nhé

\(M=\left(-a+b\right)-\left(b+c-a\right)+\left(c-a\right)\)

\(M=-a+b-b-c+a+c-a\)

\(M=\left(-a+a\right)+\left(-b+b\right)+\left(-c+c\right)-a\)

\(M=-a\)

Vì \(a< 0\Rightarrow-a=-\left(-a\right)>0\)

\(\Rightarrow M>0\)

có ai giúp mk không vậy ?

a/ 1/2 quá dễ

b/ Kẻ DO//FC có d là tđ BC suy ra O là tđ BF\(\Rightarrow\frac{OF}{BF}=\frac{OF}{AF}=\frac{1}{2}\)

FG//OD suy ra \(\frac{\Rightarrow OF}{AF}=\frac{GD}{AG}=\frac{1}{2}\) vậy AG/GD=2

câu c tương tự b nhé, ko bk thì hỏi

DO//FC

\(\Rightarrow\left\{{}\begin{matrix}2DO=FC\\\frac{AF}{AO}=\frac{FG}{OD}=\frac{2}{3}\Rightarrow\frac{3}{2}DO=FG\end{matrix}\right.\)

Vậy FG/FC=3/4

a) a+|a|

=-3+|-3|

=-3+3=0

b) a-|a|

=-3-|-3|

=-3-3=-6

c) a.|a|

=-3.|-3|

=-3.3=-9

a ) a + / a /

= -3 + 3

= 0

b ) a - / a /

= -3 - 3

= -6

c ) a . / a /

= -3 . 3

= -9

đúng k pn ? nếu sai cho mk xl

\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a} = \frac{a+b+c}{b+c+a}=1\) (tính chất của dãy tỉ số bằng nhau)

=> a=b=c

chúc bn học giỏi

ta có 

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

\(\Rightarrow a=b=c\)