Ta có: a² + b² + c² = ab + bc + ca

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

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

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

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

\(\Rightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Rightarrow a=b=c\)(đpcm)

đây nha

  PS : nhớ k :33

                                                                                                                                                      # Aeri # 

A = (2x - 1)x + (2x - 1)2 = 2x + 4x -2 = 6x - 2 (x là chữ x). Sai thì thông cảm nha. Chúc bn học tốt!

A = 26² - 24² = (26 - 24)(26 + 24) = 2.50 = 100

B = 27² - 25² = (27 - 25)(27 + 25) = 2.52 = 104

\(A=26^2-24^2\)

\(A=\left(26-24\right)\left(26+24\right)\)

\(A=2.50\)

\(B=27^2-25^2\)

\(B=\left(27-25\right)\left(27+25\right)\)

\(B=2.52\)

\(2.50< 2.52\)

\(< =>A< B\)

x( x² + 1 ) = 10( x² + 1 )

<=> x( x² + 1 ) - 10( x² + 1 ) = 0

<=> ( x² + 1 )( x - 10 ) = 0

<=> x - 10 = 0 <=> x = 10 ( vì x² + 1 > 0 ) 

\(ĐK:x\ne\pm3\)

\(P=\left[\frac{\left(2x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3-10x}{\left(x-3\right)\left(x+3\right)}\right]\cdot\frac{x-3}{x+2}\)

\(=\frac{2x^2-7x+3+x^2+3x-3+10x}{\left(x-3\right)\left(x+3\right)}\cdot\frac{x-3}{x+2}\)

\(=\frac{3x^2+6x}{x+3}\cdot\frac{1}{x+2}=\frac{3x\left(x+2\right)}{\left(x+3\right)\left(x+2\right)}=\frac{3x}{x+3}\)

Ta có : \(D=3x^2+2x+1=3\left(x^2+\frac{2}{3}x+\frac{1}{3}\right)=3\left(x^2+\frac{2}{3}x+\frac{1}{9}+\frac{2}{9}\right)=3\left(x+\frac{1}{3}\right)^2+\frac{2}{3}\ge\frac{2}{3}\)

\(\Rightarrow\)Min D = 2/3

Dấu "=" xảy ra khi x + 1/3 = 0 

\(\Rightarrow x=-\frac{1}{3}\)

Vậy Min D = 2/3 khi x = -1/3 

D = 3x² + 2x + 1 = 3( x² + 2/3x + 1/9 ) + 2/3 = 3( x + 1/3 )² + 2/3 ≥ 2/3 ∀ x

Dấu "=" xảy ra <=> x = -1/3 . Vậy MinD = 2/3

\(D=3x^2+2x+1\)

\(D=\left(3x^2+2x+\frac{\sqrt{3}}{3}^2\right)+\frac{2}{3}\)

\(D=\left(\sqrt{3}x+\frac{\sqrt{3}}{3}\right)^2+\frac{2}{3}\ge\frac{2}{3}\)

dấu "=" xảy ra khi và chỉ khi

\(x=\frac{1}{3}\)

\(< =>MIN:D=\frac{2}{3}\)

Khó nhìn quá em ạ

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