1) Cho $(a+b+c)^2 = 3(ab+bc+ca)$ Chứng minh $a=b=c$ 2) Cho $x^2 – y^2 – z^2 = 0$ Chứng minh $(5x – 3y + 4z)(5x – 3y-4z)=(3x-5y)^2$

Đáp án:

Giải thích các bước giải:

 1)

Ta có:

$(a+b+c)^2 = 3(ab+bc+ca)$

$⇒ a^2+b^2+c^2+ab+bc+ca – 3ab – 3bc – 3ca = 0$
$⇒ a^2 + b^2 +c^2 -2ab – 2bc – 2ca = 0$
$⇒ 2(a^2+b^2+c^2-2ab-2bc-2ca)=2.0$
$⇒2a^2+2b^2+2c^2-4ab-4bc-4ca = 0$
$⇒(a^2-4ab+b^2)+(b^2-4bc+c^2)+(a^2-4ac+c^2)=0$
$⇒(a-b)^2 + (b-c)^2 + (a-c)^2 = 0$
⇒\(\left[ \begin{array}{l}(a-b)^2=0\\(b-c)^2=0\\(a-c)^2=0\end{array} \right.\)

⇒\(\left[ \begin{array}{l}a=b\\b=c\\a=c\end{array} \right.\) 

$⇒a = b =c$

2) 

Xét: $(5x – 3y+4z)(5x – 3y – 4z) – (3x – 5y)^2$
$= (5x – 3y)^2 – 16z^2 – (3x – 5y)^2$
$= (5x-3y +3x – 5y)(5x-3y-3x+5y) – 16z^2$
$= (8x-8y)(2x+2y) – 16z^2$
$= 8(x-y).2(x+y) – 16z^2$
$= 16(x^2 – y^2) – 16z^2$
$= 16(x^2-y^2-z^2)$

Thay $x^2-y^2-z^2 = 0$
$16 . 0$
$= 0$
$⇒(5x – 3y+4z)(5x – 3y – 4z) – (3x – 5y)^2 = 0$

$⇒ (5x – 3y+4z)(5x – 3y – 4z) = (3x – 5y)^2(Đpcm)$