(\(x+1\)) + (\(x-1\))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: 

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

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

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

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

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

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:

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

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

\(a.A=x^2+5x+7\\ =\left[x^2+2\cdot x\cdot\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2\right]+\dfrac{3}{4}\\ =\left(x+\dfrac{5}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)

Dấu "=" xảy ra: `x+5/2=0<=>x=-5/2` 

\(b.B=6x-x^2-5\\ =-\left(x^2-6x+9\right)+4\\ =-\left(x-3\right)^2+4\le4\forall x\)

Dấu "=" xảy ra: `x-3=0<=>x=3`