\(a.25x^2-9=0\\ \Leftrightarrow\left(5x\right)^2-3^2=0\\ \Leftrightarrow\left(5x+3\right)\left(5x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}5x=3\\5x=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\\ b.\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\\ \Leftrightarrow x^2+8x+16-x^2+1=16\\ \Leftrightarrow8x+17=16\\ \Leftrightarrow8x=-1\\ \Leftrightarrow x=-\dfrac{1}{8}\\ c.\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\\ \Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\\ \Leftrightarrow5x^2+2x+10-5x^2+245=0\\ \Leftrightarrow2x+265=0\\ \Leftrightarrow2x=-265\\ \Leftrightarrow x=-\dfrac{265}{2}\)

\(a.A=9x^2+42x+49\\ =\left(3x\right)^2+2\cdot3x\cdot7+7^2\\ =\left(3x+7\right)^2\)

Thay x = 1 vào A ta có:

`A=(3*1+7)^2=10^2=100` 

\(b.B=25x^2-2xy+\dfrac{1}{25}y^2\\ =\left(5x\right)^2-2\cdot5x\cdot\dfrac{1}{5}y+\left(\dfrac{1}{5}y\right)^2\\ =\left(5x-\dfrac{1}{5}y\right)^2\)

Thay x = -1/5 và y = -5 vào B ta có:

\(B=\left(5\cdot\dfrac{-1}{5}-\dfrac{1}{5}\cdot-5\right)^2=\left(-1+1\right)^2=0\)

\(a.\left(x+1\right)^2-\left(x-1\right)^2-3\left(x+1\right)\left(x-1\right)\\ =\left(x^2+2x+1\right)-\left(x^2-2x+1\right)-3\left(x^2-1\right)\\ =x^2+2x+1-x^2+2x-1-3x^2+3\\ =4x-3x^2+3\\b.5\left(x-2\right)\left(x+2\right)-\dfrac{1}{2}\left(6-8x\right)^2+17\\ =5\left(x^2-4\right)-\dfrac{1}{2}\left(36-96x+64x^2\right)+17\\ =5x^2-20-18+48x-32x^2\\ =48x-27x^2-38\)

\(a.\left(x+y+4\right)\left(x+y-4\right)\\ =\left[\left(x+y\right)+4\right]\left[\left(x+y\right)-4\right]\\ =\left(x+y\right)^2-4^2\\ b.\left(x-y+6\right)\left(x+y-6\right)\\ =\left[x-\left(y-6\right)\right]\left[x+\left(y-6\right)\right]\\ =x^2-\left(y-6\right)^2\\ c.\left(y+2z-3\right)\left(y-2z-3\right)\\ =\left[\left(y-3\right)+2z\right]\left[\left(y-3\right)-2z\right]\\ =\left(y-3\right)^2-\left(2z\right)^2\\ d.\left(x+2y+3z\right)\left(2y+3z-x\right)\\ =\left[\left(2y+3z\right)+x\right]\left[\left(2y+3z\right)-x\right]\\ =\left(2y+3z\right)^2-x^2\)

\(\left(x+1\right)^2-\left(x-1\right)^2\\ =\left[\left(x+1\right)-\left(x-1\right)\right]\left[\left(x+1\right)+\left(x-1\right)\right]\\ =\left(x+1-x+1\right)\left(x+1+x-1\right)\\ =2\cdot2x\\ =4x\)

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

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

`= 2x^2 + 2`

`= 2(x^2 +1)`

-----------------------------------

Áp dụng hằng đẳng thức: 

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

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

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

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

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

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

  (\(x+1\)) + (\(x-1\))² 

= \(x\) + 1  + \(x^2\) - 2\(x\) + 1

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

= \(x^2\) - \(x\) + 2

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

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

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