Ta có: a^2 + b^2 + c^2 = ab + bc + ca 
<=> 2.a^2 + 2.b^2 + 2.c^2 = 2.ab + 2.bc + 2.ca 
<=> ( a^2 - 2ab + b^2 ) + ( b^2 - 2bc +c^2 ) + ( c^2 - 2ac + a^2 ) =0 
<=> (a-b)^2 + (b-c)^2 + (c -a)^2 =0 (1) 
Vì (a-b)^2 ; (b-c)^2 ; (c -a)^2 ≧ 0 với mọi a,b,c. 
=> (a-b)^2 + (b-c)^2 + (c -a)^2 ≧ 0 (2) 
Từ (1) và (2) khẳng định dấu "=" khi: 
a - b = 0; b - c = 0 ; c - a = 0 => a=b=c 
Vậy a=b=c.
=> đpcm

Ta có 2(a²+b²+c²)=2(ab+bc+ca)

VT-VP= (a-b)²+(b-c)²+(c-a)²=0

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

\(gt\Rightarrow\left(2z-1\right)^3=\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)^3\) \(=6+3\left(\sqrt[3]{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}\right).\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)\)

\(\Leftrightarrow\left(2z-1\right)^3=6+3.\sqrt[3]{9-8}\left(2z-1\right)=6z+3\)

\(\Leftrightarrow\left(2z-1\right)^3-6z-3=0\Leftrightarrow8z^3-12z^2-4=0\)

=> Đa thức bậc 3 có 1 nghiệm là z là:\(8x^3-12x^2-4=0\Leftrightarrow2x^3-3x^2-1=0\)

#)Giải :

(Hình bạn tự vẽ nhé ^^)

Ta có \(\widehat{xOy}+\widehat{yOx}+\widehat{zOt}+\widehat{xOz}=360^o\)

\(\Rightarrow\widehat{xOy}+90^o+\widehat{zOt}+90^o=360^o\)

\(\Rightarrow\widehat{xOy}+\widehat{zOt}=180^o\left(đpcm\right)\)

Ta có xOy + zOt = ( xOz + yOz)+(yOt+yOz)

=> 90 + yOz + 90 + yOz 

Mà yOz = yOz

=> xOy + zOt = 90 + 90 

=> XOy + zOt = 180 độ(dpcm)

12334556757566 + 457905

=12334557215471

12334557215571 bấm !

#)Giải :

\(\left(\frac{1}{2}\right)^{15}\div\left(\frac{1}{4}\right)^{20}=\left(\frac{1}{2}\right)^{15}\div\left(\frac{1}{2}\right)^{40}=\left(\frac{1}{2}\right)^{-25}\)

Cảm ơn bạn nha

\(\frac{1}{2}-\left[\frac{1}{3}+\frac{3}{4}\right]< x< \frac{8}{3}-\left[\frac{1}{5}+\frac{3}{4}\right]\)

\(\Leftrightarrow\frac{1}{2}-\left[\frac{4}{12}+\frac{9}{12}\right]< x< \frac{8}{3}-\left[\frac{4}{20}+\frac{15}{20}\right]\)

\(\Leftrightarrow\frac{1}{2}-\frac{13}{12}< x< \frac{8}{3}-\frac{19}{20}\)

\(\Leftrightarrow\frac{6}{12}-\frac{13}{12}< x< \frac{160}{60}-\frac{57}{60}\)

\(\Leftrightarrow-\frac{7}{12}< x< \frac{103}{60}\)

Đến đây ez rùi

-35/60<x<103/60