Bài 1:
a) Sửa đề : \(A=3x^2-11x+10\)
\(A=3x^2-6x-5x+10\)
\(A=3x\left(x-2\right)-5\left(x-2\right)\)
\(A=\left(x-2\right)\left(3x-5\right)\)
b) \(B=5a^3+b^3-27+12ab\)( chịu )
Bài 2 :
\(A=-5x^2+21x+2019\)
\(A=-5\left(x^2-\frac{21}{5}x-\frac{2019}{5}\right)\)
\(A=-5\left[\left(x-\frac{21}{10}\right)^2-\frac{40821}{100}\right]\)
\(A=\frac{40821}{20}-5\left(x-\frac{21}{10}\right)^2\le\frac{40821}{20}\forall x\)
Dấu "=" \(\Leftrightarrow x=\frac{21}{10}\)
Bài 3 :
Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)\rightarrow\left(x;y;z\right)\)
Ta có : \(\left\{{}\begin{matrix}x+y+z=2\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{xyz}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y+z=2\\\frac{xy+yz+xz}{xyz}=\frac{1}{xyz}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y+z=2\\xy+yz+xz=1\end{matrix}\right.\)
\(A=x^3+y^3+z^3-3xyz\)
\(A=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)
\(A=\left(x+y+z\right)\left[\left(x+y+z\right)^2-3\left(xy+yz+xz\right)\right]\)
\(A=2\cdot\left(2^2-3\cdot1\right)\)
\(A=2\)