10 cặp tick nhaTrần Thị Hải

A=3x²+2x-5

=3x²+2x+\(\dfrac{1}{3}\) -\(\dfrac{16}{3}\)

=3\(\left(x^2+\dfrac{2}{3}x+\dfrac{1}{9}\right)-\dfrac{16}{3}\)

=3\(\left(x+\dfrac{1}{3}\right)^2-\dfrac{16}{3}\)

do \(\left(x+\dfrac{1}{3}\right)^2\ge0\forall x\)

=>\(3\left(x+\dfrac{1}{3}\right)^2-\dfrac{16}{3}\ge-\dfrac{16}{3}\)

=>GTNNcủa A =-\(\dfrac{16}{3}\) khi

x+\(\dfrac{1}{3}=0\)

=>x=-\(\dfrac{1}{3}\)

vậy ...

a) 3x(x-4)=0

=>\(\left[{}\begin{matrix}3x=0\\x-4=0\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

b) x²-4x+4=0

=>(x-2)²=0

=>x-2=0

=>x=2

3x²+9x-30

=3(x2+3x-10)

=3(x2+5x-2x-10)

=3[(x2+5x)-(2x+10)]

=3[x(x+5)+2(x+5)]

=3(x+5)(x+2)

x²-4x-5

=x²+x-5x-5

=(x²+x)-(5x+5)

=x(x+1)-5(x+1)

=(x+1)(x-5)