\(\Leftrightarrow Qx^2+Q=10x^2+8x+4\)
\(\Leftrightarrow x^2\left(Q-10\right)-8x+Q-4=0\)(1)
*Neu Q = 10 thi x = ... (ban tu tinh nha)
*Neu Q # 10 thi pt (1) co nghiem khi va chi khi Delta' >
Ta co \(\Delta'\ge0\)
\(\Leftrightarrow16-\left(Q-10\right)\left(Q-4\right)\ge0\)
\(\Leftrightarrow16-Q^2+14Q-40\ge0\)
\(\Leftrightarrow-Q^2+14Q-24\ge0\)
\(\Leftrightarrow2\le Q\le12\)
Ban tu tim dau "=" nha
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