Bài 2:

\(a.6x\left(2x-3y\right)+12xy^2\left(2x-3y\right)\\ =\left(2x-3y\right)\left(6x+12xy^2\right)\\ =6x\left(2x-3y\right)\left(2y^2+1\right)\\ b.14x^2y\left(6x+1\right)-21xy^2\left(6x+1\right)\\ =\left(6x+1\right)\left(14x^2y-21xy^2\right)\\ =7xy\left(6x+1\right)\left(2x-3y\right)\\ c.-3a\left(x-3\right)-a^2\left(3-x\right)\\ =-3a\left(x-3\right)+a^2\left(x-3\right)\\ =\left(x-3\right)\left(a^2-3a\right)\\ =a\left(x-3\right)\left(a-3\right)\\ d.4x^2y\left(7-2y\right)-24x^3y^2\left(2y-7\right)\\ =4x^2y\left(7-2y\right)+24x^3y^2\left(7-2y\right)\\ =\left(7-2y\right)\left(4x^2y+24x^3y^2\right)\\ =4x^2y\left(7-2y\right)\left(1+6xy\right)\\ e.4ab^2\left(x+2y\right)-16a^3y\left(-x-2y\right)\\ =4ab^2\left(x+2y\right)+16a^3y\left(x+2y\right)\\ =\left(x+2y\right)\left(4ab^2+16a^3y\right)\\ =4a\left(x+2y\right)\left(b^2+4a^2y\right)\)

Bài 3:

\(a.4x^2-12x=0\\ \Leftrightarrow4x\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}4x=0\\x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\\ b.x^3-25x=0\\ \Leftrightarrow x\left(x^2-25\right)=0\\ \Leftrightarrow x\left(x-5\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\\ c.\left(2x-1\right)^2-3x\left(x+2\right)=1\\ \Leftrightarrow4x^2-4x+1-3x^2-6x=1\\ \Leftrightarrow x^2-10x+1=1\\ \Leftrightarrow x^2-10x=0\\ \Leftrightarrow x\left(x-10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-10=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\\ d.7x\left(x-18\right)-x+18=0\\ \Leftrightarrow7x\left(x-18\right)-\left(x-18\right)=0\\ \Leftrightarrow\left(x-18\right)\left(7x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-18=0\\7x-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=18\\7x=1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=18\\x=\dfrac{1}{7}\end{matrix}\right.\)

ΔABC vuông tại A nên ta có:

\(sinB=\dfrac{AC}{BC}\\ =>AC=BC\cdot sinB=8\cdot sin60^o=4\sqrt{3}\left(cm\right)\)

Áp dụng định lý Pythagore cho tam giác ABC ta có:

\(BC^2=AC^2+AB^2\\ =>AB=\sqrt{BC^2-AC^2}\\ =>AB=\sqrt{8^2-\left(4\sqrt{3}\right)^2}=4\left(cm\right)\) 

 3x²-75=0                                                                                                 3x²=0+75                                                                                       3x²=75                                                                                                x²=75:3                                                                                                x²=25                                                                                              x²=5²                                                                                                    x=5                                                                                                          Vậy x=5

          \(3x^2-75=0\)

     \(3\left(x^2-25\right)=0\)

        \(x^2-25^2=0\)

\(\left(x-5\right)\left(x+5\right)=0\)

\(\left[{}\begin{matrix}x-5=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)

\(a.x^2+4x+4=\left(x+2\right)^2\\ b.x^2-5=\left(x+\sqrt{5}\right)\left(x-\sqrt{5}\right)\\ c.9x^2+6x+1=\left(3x+1\right)^2\\ d.64x^3-27y^3=\left(4x\right)^3-\left(3y\right)^3=\left(4x-3y\right)\left(16x^2+12xy+9y^2\right)\\ e.\left(x+1\right)^2-4y^2=\left(x+1\right)-\left(2y\right)^2=\left(x-2y+1\right)\left(x+2y+1\right)\\ f.8x^3+12x^2+6x+1=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2+1^3=\left(2x+1\right)^3\)

a, bn xem lại nhé

b, \(x^2-5=\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)\)

c, \(9x^2+6x+1=\left(3x\right)^2+2.3x+1=\left(3x+1\right)^2\)

d, \(64x^3-27y^3=\left(4x\right)^3-\left(3y\right)^3=\left(4x-3y\right)\left(16x^2+12xy+9y^2\right)\)

e, \(\left(x+1\right)^2-4y^2=\left(x+1-2y\right)\left(x+1+2y\right)\)

f, \(8x^3+12x^2+6x+1=\left(2x\right)^3+3.\left(2x\right)^2+3.2x.1^2+1=\left(2x+1\right)^3\)

g, \(6x^2-24y^2=\left(\sqrt{6}x\right)^2-\left(2\sqrt{6}y\right)^2=\left(\sqrt{6}x-2\sqrt{6}y\right)\left(\sqrt{6}x+2\sqrt{6}y\right)\)

h, \(\left(x+y\right)^3+8y^3=\left(x+y+2y\right)\left[\left(x+y\right)^2-2y\left(x+y\right)+4y^2\right]\)

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

k, \(1975x^4-1975x^2=1975x^2\left(x^2-1\right)=1975x^2\left(x-1\right)\left(x+1\right)\)

i, \(x^3-4x=x\left(x^2-4\right)=x\left(x-2\right)\left(x+2\right)\)

m, \(x^4-2x^3+x^2=x^2\left(x^2-2x+1\right)=x^2\left(x-1\right)^2\)

  (\(x+1\)) + (\(x-1\))² 

= \(x\) + 1  + \(x^2\) - 2\(x\) + 1

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

= \(x^2\) - \(x\) + 2

\(\left(x+1\right)+\left(x-1\right)^2\\ =\left(x+1\right)+\left(x^2-2x+1\right)\\ =x+1+x^2-2x+1\\ =x^2+\left(x-2x\right)+\left(1+1\right)\\ =x^2-x+2\)

Đặt \(x^2-x+1=a;x+1=b\)

Phương trình sẽ trở thành: \(3a^2-2b^2=5ab\)

=>\(3a^2-5ab-2b^2=0\)

=>\(3a^2-6ab+ab-2b^2=0\)

=>3a(a-2b)+b(a-2b)=0

=>(a-2b)(3a+b)=0

=>\(\left[{}\begin{matrix}a-2b=0\\3a+b=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2-x+1-2\left(x+1\right)=0\\3\left(x^2-x+1\right)+x+1=0\end{matrix}\right.\)

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

=>\(x^2-3x-1=0\)

=>\(x=\dfrac{3\pm\sqrt{13}}{2}\)

`(x+1)^2 + (x-1)^2`

`= x^2 + 2x + 1 + x^2 - 2x + 1`

`= 2x^2 + 2`

`= 2(x^2 +1)`

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Áp dụng hằng đẳng thức: 

\(\left(a\pm b\right)^2=a^2\pm2ab+b^2\)