Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
tham khảo thôi nhé ko giống y sì đâu
https://olm.vn/hoi-dap/detail/213882782299.html
\(f\left(x\right)=ax^2+bx+c\)
\(\Rightarrow f\left(\frac{1}{2}\right)=\frac{1}{4}a+\frac{1}{2}b+c\)
\(\Rightarrow f\left(-2\right)=4a-2b+c\)
\(\Rightarrow f\left(\frac{1}{2}\right)+f\left(-2\right)=\frac{17}{4}a-\frac{3}{2}b+2c\)
\(\Rightarrow4\left[f\left(\frac{1}{2}\right)+f\left(-2\right)\right]=17a-6b+8c=0\)( vì 17a-6b+8c=0)
\(\Rightarrow f\left(\frac{1}{2}\right)+f\left(-2\right)=0\)
\(\Rightarrow f\left(\frac{1}{2}\right)=-f\left(-2\right)\)
\(\Rightarrow f\left(\frac{1}{2}\right).f\left(-2\right)=-\left[f\left(-2\right)\right]^2\le0\left(đpcm\right)\)
13a+b+2c=0
=>b=-13a-2c
f(-2)=4a-2b+c=4a+c+26a+4c=30a+5c
f(3)=9a+3b+c=9a+c-39a-6c=-30a-5c
=>f(-2)*f(3)<=0
\(f\left(-2\right)=4a-2b+c\)
\(f\left(3\right)=9a+3b+c\)
\(f\left(-2\right)+f\left(3\right)=13a+b+2c=0\)
\(\Rightarrow f\left(-2\right)=-f\left(3\right)\Rightarrow f\left(-2\right).f\left(3\right)=-f\left(-2\right)^2\le0\)
p/s: nhớ t nữa ko :>
\(f\left(x\right)=ax^2+bx+c\)
\(f\left(-2\right)=a.\left(-2\right)^2+\left(-2\right).b+c=4a-2b+c\)
\(f\left(3\right)=a.3^2+3.b+c=9a+3b+c\)
\(f\left(3\right)+f\left(-2\right)=4a-2b+c+9a+3b+c=13a+b+2c=0\)
\(\Rightarrow f\left(3\right)=-f\left(-2\right)\Rightarrow f\left(3\right)f\left(-2\right)=-\left[f\left(3\right)\right]^2\le0\left(đpcm\right)\)
\(f\left(3\right).f\left(-2\right)=\left(9a+3b+c\right)\left(4a-2b+c\right)\)
\(=\left[3\left(a+b\right)+6a+c\right]\left[-2\left(a+b\right)+6a+c\right]\)
\(=\left(6a+c\right)\left(6a+c\right)=\left(6a+c\right)^2\ge0\) (đpcm)
