Câu hỏi của Kfkfj

Question

c/m đẳng thức:

a. $\left(a-b+c\right)^2-\left(a+b+c\right)^2=-4ab-4bc$

b. $6,3-5x+x^2>0\forall x$

giúp mk với cảm ơn các bạn nhiều ạ!

Trang Thùy · Accepted Answer

ta có: $\left(a-b+c\right)^2-\left(a+b+c\right)^2$

VT   $=\left(a-b+c\right)\left(a-b+c\right)-\left(a+b+c\right)\left(a+b+c\right)$

$=a^2-ab+ac-ab+b^2-bc+ac-bc+c^2-a^2-ab-ac-ab-b^2-bc-ac-c-c^2$

= $-4ab-4bc=VT\left(đpcm\right)$Câu trả lời: ta có: $\left(a-b+c\right)^2-\left(a+b+c\right)^2$

VT   $=\left(a-b+c\right)\left(a-b+c\right)-\left(a+b+c\right)\left(a+b+c\right)$

$=a^2-ab+ac-ab+b^2-bc+ac-bc+c^2-a^2-ab-ac-ab-b^2-bc-ac-c-c^2$

= $-4ab-4bc=VT\left(đpcm\right)$

Khôi Bùi · Answer

a ) $\left(a-b+c\right)^2-\left(a+b+c\right)^2$

$=\left(a-b+c-a-b-c\right)\left(a-b+c+a+b+c\right)$

$=-2b\left(2a+2c\right)$

$=-4ab-4bc\left(đpcm\right)$

b ) $6,3-5x+x^2$

$=x^2-5x+\dfrac{63}{10}$

$=x^2-5x+\dfrac{25}{4}+\dfrac{1}{20}$

$=\left(x-\dfrac{5}{2}\right)^2+\dfrac{1}{20}\ge\dfrac{1}{20}>0\forall x\left(đpcm\right)$

:DCâu trả lời: a ) $\left(a-b+c\right)^2-\left(a+b+c\right)^2$