nói thật nhá hóa ko có công thức đâu, học thuộc là chính

\(A=2x^2+10x-1\)

\(=2\left(x^2+5x-\frac{1}{2}\right)\)

\(=2\left(x^2+2x.\frac{5}{2}+\frac{25}{4}-\frac{27}{4}\right)\)

\(=2\left[\left(x^2+\frac{5}{2}\right)^2-\frac{27}{4}\right]\)

\(=2\left(x+\frac{5}{2}\right)^2-\frac{27}{2}\ge\frac{-27}{2}\)(Vì \(\left(x+\frac{5}{2}\right)^2\ge0\))

Dấy " = " xảy ra khi :

\(x+\frac{5}{2}=0\)

\(\Leftrightarrow x=\frac{-5}{2}\)

Vậy GTNN của A là \(\frac{-27}{2}\)khi \(x=\frac{-5}{2}\)

          Hk tốt ~

\(a)xy+3x-2y=11\)

\(\Leftrightarrow xy+3x-2y-6=5\)

\(\Leftrightarrow x\left(y+3\right)-2\left(y+3\right)=5\)

\(\Leftrightarrow\left(y+3\right)\left(x-2\right)=5\)

\(\Leftrightarrow\hept{\begin{cases}y+3=-1\\x-2=-5\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-4\\x=-3\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y+3=1\\x-2=5\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-2\\x=7\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y+3=-5\\x-2=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-8\\x=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y+3=5\\x-2=1\end{cases}}\Leftrightarrow\hept{\begin{cases}y=2\\x=3\end{cases}}\)

\(b)2x^2-2xy+x-y=12\)

\(\Leftrightarrow2x\left(x-y\right)+\left(x-y\right)=12\)

\(\Leftrightarrow\left(x-y\right)\left(2x+1\right)=12\)

\(\Rightarrow\left(x-y\right);\left(2x+1\right)\inƯ\left(12\right)\)

\(\RightarrowƯ\left(12\right)\in\left\{-1;1;-2;2;-3;3;-4;4;-6;6;-12;12\right\}\)

Vì 2x+1 luôn lẻ

\(\Rightarrow2x+1\in\left\{-1;1;-3;3\right\}\)

\(\Leftrightarrow\hept{\begin{cases}2x+1=-1\\x-y=-12\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\y=11\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}2x+1=1\\x-y=12\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\y=-12\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}2x+1=-3\\x-y=-4\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-2\\y=2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}2x+1=3\\x-y=4\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-3\end{cases}}\)

a, Đặt : \(C=1+5+5^2+5^3+...+5^9\)

\(\Leftrightarrow5C=5+5^2+5^3+5^4+...+5^{10}\)

\(\Leftrightarrow5C-C=5^{10}-1\)

\(\Leftrightarrow4C=5^{10}-1\)

\(\Leftrightarrow C=\frac{5^{10}-1}{4}\)

Ta có mẫu là : 

\(\frac{5^9-1}{4}\)

Đặt vào A ta đc 

\(A=\frac{1+5+5^2+5^3+...+5^9}{1+5+5^2+5^3+...+5^8}\)

\(\Leftrightarrow A=\frac{\frac{5^{10}-1}{4}}{\frac{5^9-1}{4}}\)

\(\Leftrightarrow\frac{5^{10}-1}{5^9-1}\)

Vậy ...

 Tương tự a , ta có 

\(B=\frac{\frac{3^{10}-1}{2}}{\frac{3^9-1}{2}}\)

\(\Leftrightarrow B=\frac{3^{10}-1}{3^9-1}\)

Vậy ...

\(\Rightarrowđpcm\)

a) \(x^3-6x^2+9x\)

\(=x\left(x^2-6x+9\right)\)

\(=x\left(x-3\right)^2\)

b) \(x^2-2xy+3x-6y\)

\(=x\left(x-2y\right)+3\left(x-2y\right)\)

\(=\left(x+3\right)\left(x-2y\right)\)

c) \(x^2-8x+7\)

\(=x^2-7x-x+7\)

\(=x\left(x-7\right)-\left(x-7\right)\)

\(=\left(x-1\right)\left(x-7\right)\)

Câu tính chia mk lm đc nhg ko cs phần mềm trình bày