Bài 1:
`a)(a+b)³+(a-b)³`
`=(a+b+a-b)[(a+b)²-(a+b)(a-b)+(a-b)²]`
`=2a[a²+2ab+b²-(a²-b²)+a²-2ab+b²]`
`=2a(a²+2ab+b²-a²+b²+a²-2ab+b²)`
`=2a[(a²-a²+a²)+(2ab-2ab)+(b²+b²+b²)]`
`=2a(a²+3b²)`
`b)-x³+9x²-27x+27`
`=27-27x+9x²-x³`
`=3³-3.3².x+3.3.x²-x³`
`=(3-x)³`
Bài 2:
`a)73²-27²`
`=(73+27)(73-27)`
`=100.46`
`=4` `600`
`b)37²-13²`
`=(37+13)(37-13)`
`=50.24`
`=1` `200`
`c)2020²-2²`
`=(2020+2)(2020-2)`
`=2022.2018`
`=4` `080` `396`
Bài 3:
`a)2-25x²=0`
`⇔(`$\sqrt[]{2}$ `)^2-(5x)^2=0`
`⇔(`$\sqrt[]{2}$`+5x)(`$\sqrt[]{2}$`-5x)=0`
`⇔`$\left[\begin{matrix} \sqrt[]{2}+5x=0\\ \sqrt[]{2}-5x=0\end{matrix}\right.$
`⇔`$\left[\begin{matrix} 5x=-\sqrt[]{2}\\ 5x=\sqrt[]{2}\end{matrix}\right.$
`⇔`$\left[\begin{matrix} x=-\dfrac{\sqrt[]{2}}{5}\\ x=\dfrac{\sqrt[]{2}}{5}\end{matrix}\right.$
Vậy `x∈{-`$\dfrac{\sqrt[]{2}}{5}$`;`$\dfrac{\sqrt[]{2}}{5}$`}`
`b)x²-x+1/4=0`
`⇔x²-2.x. 1/2+(1/2)^2=0`
`⇔(x-1/2)^2=0`
`⇔x-1/2=0`
`⇔x=1/2`
Vậy `x=1/2`
