(a+b+c)²=3(ab+bc+ca)

<=> a²+b²+c²+2ab+2ac+2bc=3ab+3bc+3ca

<=> a²+b²+c²+2ab+2ac+2bc-3ab-3bc-3ca=0

<=> a²+b²+c²-ab-bc-ca=0

<=> 2a²+2b²+2c²-2ab-2bc-2ca=0

<=> (a²-2ab+b²)+(b²-2bc+c²)+(c²-2ca+a²)=0

<=> (a-b)²+(b-c)²+(c-a)²=0

Vì \(\hept{\begin{cases}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{cases}}\)\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Rightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Rightarrow}a=b=c}\) (đpcm)

\(\left(a+b\right)^2=2\left(a^2+b^2\right)\Rightarrow a^2+2ab+b^2=2a^2+2b^2\Rightarrow a^2-2ab+b^2=0\Rightarrow\left(a-b\right)^2=0\Rightarrow a-b=0\Rightarrow a=b\left(đpcm\right)\)

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

Ta có: \(2\left(a^2+b^2\right)=\left(a+b\right)^2\)

\(\Leftrightarrow2a^2+2b^2-a^2-b^2-2ab=0\)

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

\(\Leftrightarrow a-b=0\)

hay a=b

ta có \(a^2+b^2+c^2=ab+bc+ca\Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)

\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

\(\Rightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2=0\)\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

mà \(\left(a-b\right)^2\ge0;\left(b-c\right)^2\ge0;\left(c-a\right)^2\ge0\)

=> \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)

dấu = xảy ra <=> \(\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Leftrightarrow a=b=c}\) (ĐPCM)

\(a^2+b^2+c^2=ab+bc+ca\)

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

<=> \(2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)

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

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

Vì \(\left(a-b\right)^2\ge0;\left(b-c\right)^2\ge0;\left(c-a\right)^2\ge0\)

=> a-b=0 ; b-c =0 ; a-c=0

=> a=b ; b=c ; c=a

=> a=b=c

\(a^2+b^2+c^2=ab+bc+ca\)

\(\Rightarrow2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ca\right)\)

\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)

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

Vì \(\hept{\begin{cases}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{cases}\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0}\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Leftrightarrow}\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Leftrightarrow}a=b=c}\) (đpcm)

TA có a²+b²+c²-ab-bc-ac=0

    <=>a(a-b)+b(b-c)+c(c-a)=0

=>\(\orbr{\begin{cases}a=b=c=0\\\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\end{cases}}\)

=>a=b=c=0(đpcm)

z 2  = ( a + b i ) 2  = a 2  − b 2  + 2abi

( z ) 2  = ( a - b i ) 2  =  a 2  −  b 2  − 2abi

z.z− = (a + bi)(a − bi) =  a 2  +  b 2

Từ đó suy ra các kết quả.

Biến đổi vế trái ta có:

VT = (a + b)( a 2  – ab +  b 2 ) + (a – b)( a 2  + ab +  b 2 )

=  a 3  +  b 3  +  a 3  –  b 3  = 2 a 3  = VP

Vế trái bằng vế phải nên đẳng thức được chứng minh.

Ta có: \(2\left(a^2+b^2\right)=\left(a+b\right)^2\)

\(\Leftrightarrow2a^2+2b^2-a^2-2ab-b^2=0\)

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

\(\Leftrightarrow\left(a-b\right)^2=0\)

\(\Leftrightarrow a-b=0\)

hay a=b

dê mà, thôi mik giải cho k mik vs nha

A = 5 + 5^2 + 5^3 + .......... + 5^8

5A = 5^2 + 5^3 + 5^4 + .................. + 5^9 

5A - A =  5^2 + 5^3 + 5^4 + .................. + 5^9 - 5 - 5^2 - 5^3 -  .......... - 5^8

4A = 5^9 - 5

Suy ra A = ( 5^9 - 5 ) : 4 = 488280 chia hết cho 30

đừng quên k nha

dễ thế cũng hỏi,ko bít động não hả

kb voi tui di roi tui noi cach lam