\(\Rightarrow\left(x-1\right)^2-\left(2x-3\right)^2=0\\ \Rightarrow\left(x-1-2x+3\right)\left(x-1+2x-3\right)=0\\ \Rightarrow\left(2-x\right)\left(3x-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{4}{3}\end{matrix}\right.\)

Nguyễn Hoàng Minh  thank you

hơi mờ

nhỏ quá

đề?

c1 x²-3x-1

c2 3x²-5x-2

1:Ta có: \(x^2-3x-1\)

\(=x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{13}{4}\)

\(=\left(x-\dfrac{3}{2}\right)^2-\dfrac{13}{4}\ge-\dfrac{13}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{3}{2}\)

cần gấp xin cảm ơn ạ

\(\left\{{}\begin{matrix}4x^2-12x+9\\x^2+10x+25\end{matrix}\right.\)

=2x² - 2.2x.3 + 3²

=4x²-12x+ 9

b) =x²+ 2.x.5+5²

=x²+ 10x+25

tính cái gì bạn

\(2^x+2^{x-1}+2^{x-2}=2^{x-2}.2^2+2^{x-2}.2+2^{x-2}\)\(=2^{x-2}\left(4+2+1\right)=2^{x-2}.7\)

thank you