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

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

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

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)

Câu 41:

a: ĐKXĐ: \(x^2-3\ge0\)

=>\(x^2\ge3\)

=>\(\left[\begin{array}{l}x\ge\sqrt3\\ x\le-\sqrt3\end{array}\right.\)

b: ĐKXĐ: \(x^2+4x-5>0\)

=>(x+5)(x-1)>0

=>x>1 hoặc x<-5

c: ĐKXĐ: \(\begin{cases}2x-1\ge0\\ x-\sqrt{2x-1}>0\end{cases}\Rightarrow\begin{cases}x\ge\frac12\\ \frac{x^2-2x+1}{x+\sqrt{2x-1}}>0\end{cases}\Rightarrow x\ge\frac12\)

d: ĐKXĐ: \(\begin{cases}x^2-3\ge0\\ 1-\sqrt{x^2-3}<>0\end{cases}\Rightarrow\begin{cases}x^2\ge3\\ \sqrt{x^2-3}<>1\end{cases}\Rightarrow\begin{cases}x^2\ge3\\ x^2-3<>1\end{cases}\)

=>\(\begin{cases}x^2\ge3\\ x^2<>4\end{cases}\begin{array}{l}\\ \end{array}\)

=>\(\begin{cases}x\ge\sqrt3\\ x<>2\end{cases}\) hoặc \(\begin{cases}x\le-\sqrt3\\ x<>-2\end{cases}\)

e: ĐKXĐ: \(\begin{cases}x<>0\\ x+\frac{2}{x}<>0\\ -2x\ge0\end{cases}\Rightarrow\begin{cases}x<>0\\ x^2+2<>0\\ x\le0\end{cases}\)

=>x<0

f: ĐKXĐ: \(\begin{cases}3x-1\ge0\\ 5x-3\ge0\\ x^2+x+1\ge0\end{cases}\Rightarrow\begin{cases}3x\ge1\\ 5x\ge3\end{cases}\Rightarrow x\ge\frac35\)

Câu 32:

\(x^2-6x+17\)

\(=x^2-6x+9+8=\left(x-3\right)^2+8\ge8\forall x\)

=>\(A=\frac{1}{x^2-6x+17}\le\frac18\forall x\)

Dấu '=' xảy ra khi x-3=0

=>x=3

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