Bài 7:
a) Ta có: \(\left(a+b\right)^2=2\left(a^2+b^2\right)\)
\(\Leftrightarrow a^2+2ab+b^2-2a^2-2b^2=0\)
\(\Leftrightarrow a^2-2ab+b^2=0\)
\(\Leftrightarrow\left(a-b\right)^2=0\)
\(\Leftrightarrow a-b=0\)
hay a=b
b) Ta có: \(a^2+b^2+c^2=ab+bc+ca\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\Leftrightarrow a=b=c\)(đpcm)
c) Ta có: \(\left(a+b+c\right)^2=3\left(ab+bc+ca\right)\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac-3ab-3bc-3ac=0\)
\(\Leftrightarrow a^2+b^2+c^2-ac-ab-bc=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\Leftrightarrow a=b=c\)
Bài 7.
a.
$(a+b)^2=2(a^2+b^2)$
$\Leftrightarrow a^2+b^2+2ab=2(a^2+b^2)$
$\Leftrightarrow a^2-2ab+b^2=0$
$\Leftrightarrow (a-b)^2=0$
$\Leftrightarrow a=b$ (đpcm)
b.
$a^2+b^2+c^2=ab+bc+ac$
$\Leftrightarrow 2a^2+2b^2+2c^2=2ab+2bc+2ac$
$\Leftrightarrow 2a^2+2b^2+2c^2-(2ab+2bc+2ac)=0$
$\Leftrightarrow (a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ac+a^2)=0$
$\Leftrightarrow (a-b)^2+(b-c)^2+(c-a)^2=0$
$\Rightarrow a-b=b-c=c-a=0$
$\Leftrightarrow a=b=c$ (đpcm)
c.
$(a+b+c)^2=3(ab+bc+ac)$
$\Leftrightarrow a^2+b^2+c^2+2(ab+bc+ac)=3(ab+bc+ac)$
$\Leftrightarrow a^2+b^2+c^2=ab+bc+ac$
Đến đây tương tự phần b.
Bài 8.
a.
$A=x^2+y^2=(x^2+2xy+y^2)-2xy=(x+y)^2-2xy=a^2-2b$
b.
$B=(x+y)(x^2-xy+y^2)=a(x^2+y^2-xy)=a(a^2-2b-b)=a(a^2-3b)$
Bài 8:
a) Ta có: \(A=x^2+y^2\)
\(=\left(x+y\right)^2-2xy\)
\(=a^2-2b\)
b) Ta có: \(B=x^3+y^3\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)\)
\(=a^3-3ab\)
