chứng minh phương trình luôn có nghiệm

C

conga222222

$\eqalign{
& \left( {{x^2} + ax + b} \right)\left( {{x^2} + bx + a} \right) = 0 \cr
& \leftrightarrow \left[ \matrix{
{x^2} + ax + b = 0\;\;(1) \cr
{x^2} + bx + a = 0\;\;(2) \cr} \right. \cr
& {\Delta _1} = {a^2} - 4b \cr
& {\Delta _2} = {b^2} - 4a \cr
& neu\;a < 0 \to {\Delta _2} > 0 \to co\;nghiem \cr
& neu\;b < 0 \to {\Delta _1} > 0 \to \;co\;nghiem \cr
& neu\left\{ \matrix{
a > 0 \cr
b > 0 \cr} \right. \cr
& co{1 \over a} + {1 \over b} = {1 \over 2} \leftrightarrow 2a + 2b = ab\;\left( {do\;dk\;la\;a,b \ne 0\;roi} \right) \cr
& \to {\Delta _1} + {\Delta _1} = {a^2} + {b^2} - 4\left( {a + b} \right) \ge 2ab - 4\left( {a + b} \right) = 0 \cr
& \to \left[ \matrix{
{\Delta _1} \ge 0 \cr
{\Delta _2} \ge 0 \cr} \right. \to co\;nghiem \cr
& \to xong \cr} $
 
You must log in or register to reply here.
Top Bottom