\(\left(7x+1\right)^2-\left(x+7\right)^2=\left(7x+1\right).\left(7x+1\right)-\left(x+7\right)\left(x+7\right)=\left(49x^2+7x+7x+1\right)-\left(x^2+7x+7x+49\right)\)\(=49x^2+14x+1-x^2-7x-7x-49=\left(49x^2-x^2\right)+\left(14x-7x-7x\right)-\left(49-1\right)=48x^2-48=48.\left(x^2-1\right)\)

a: \(\dfrac{2x^4-x^3-x^2+7x-4}{x^2+x-1}\)

\(=\dfrac{2x^4+2x^3-2x^2-3x^3-3x^2+3x+4x^2+4x-4}{x^2+x-1}\)

=2x^2-3x+4

b: \(=\dfrac{y}{x\left(2x-y\right)}+\dfrac{4x}{y\left(y-2x\right)}\)

\(=\dfrac{y^2-4x^2}{xy\left(2x-y\right)}=\dfrac{-\left(2x-y\right)\left(2x+y\right)}{xy\left(2x-y\right)}=\dfrac{-2x-y}{xy}\)

c: \(=\dfrac{6\left(x+8\right)}{7\left(x-1\right)}\cdot\dfrac{\left(x-1\right)^2}{\left(x-8\right)\left(x+8\right)}=\dfrac{6\left(x-1\right)}{7\left(x-8\right)}\)

BÀI 1:

Ta có:   \(VT=\left(7x+1\right)^2-\left(x+7\right)^2\)

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

                    \(=\left(8x+8\right)\left(6x-6\right)\)

                   \(=8\left(x+1\right).6\left(x-1\right)\)

                  \(=48\left(x^2-1\right)=VP\)  (đpcm)

Bài 2:

         \(16x^2-\left(4x-5\right)^2=15\)

\(\Leftrightarrow\)\(16x^2-16x^2+40x-25=15\)

\(\Leftrightarrow\)\(40x=40\)

\(\Leftrightarrow\)\(x=1\)

Vậy...

Bài 3:

\(A=x^2+2x+3=\left(x+1\right)^2+2\ge2\)

Vậy MIN A = 2  khi  x = -1

a)4x²+4x+1-x²-10x-25=0

`<=>(2x+1)^2-(x+5)^2=0`

`<=>(2x+1-x-5)(2x+1+x+5)=0`

`<=>(x-4)(3x+6)=0`

`<=>(x-4)(x+2)=0`

`<=>` \(\left[ \begin{array}{l}x=2\\x=-2\end{array} \right.\) 
b)(x^2+x+7)(x^2+x-7)=(x²+x)²-7x

`<=>(x^2+x)^2-7^2=(x^2+x)^2-7x`

`<=>-7^2=-7x`

`<=>-49=-7x`

`<=>x=7`

Vậy x=7

Xét \(\left(7x+1\right)^2-\left(x+7\right)^2-48\left(x^2-1\right)\)

\(=49x^2+14x+1-x^2-14x-49-48x^2+48\)

\(=0\)

Vậy \(\left(7x+1\right)^2-\left(x+7\right)^2=48\left(x^2-1\right)\)

a)

 ⇔ \(x^2-16=9\)

⇔ \(x^2=25\)

⇔ \(x=\pm5\)

b)

 ⇔ \(x^2-4x+4-25x^2+20x-4=0\)

⇔ \(16x-24x^2=0\)

⇔ \(8x\left(2-3x\right)=0\)

⇒ \(\left[{}\begin{matrix}x=0\\2-3x=0\end{matrix}\right.\)   ⇔   \(\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy \(x=0\) hoặc \(x=\dfrac{2}{3}\)

c)  

⇔ \(3x^2-10x-20=0\)

⇔ \(x^2-2.x.\dfrac{5}{3}+\dfrac{25}{9}-\dfrac{205}{9}=0\)

⇔ \(\left(x-\dfrac{5}{3}\right)^2=\dfrac{205}{9}\)

⇒ \(\left[{}\begin{matrix}x-\dfrac{5}{3}=\sqrt{\dfrac{205}{9}}\\x-\dfrac{5}{3}=-\sqrt{\dfrac{205}{9}}\end{matrix}\right.\)  ⇔ \(\left[{}\begin{matrix}x=\dfrac{\sqrt{\text{205}}}{\text{3}}+\dfrac{5}{3}\\x=-\dfrac{\sqrt{\text{205}}}{\text{3}}+\dfrac{5}{3}\end{matrix}\right.\)  ⇔ \(\left[{}\begin{matrix}x=\dfrac{15+\text{9}\sqrt{\text{205}}}{\text{9}}\\\text{x}=-\dfrac{15+\text{9}\sqrt{\text{205}}}{\text{9}}\end{matrix}\right.\)

Vậy... 

d) 

⇔ \(\left(x^2+x\right)^2-49=\left(x^2+x\right)^2-7x\)

⇔ 7x = 49

⇔ x=7

Vậy...

Theo đầu bài ta có:
\(\left(7x+1\right)^2-\left(x+7\right)^2=48\left(x^2-1\right)\)
\(\Rightarrow\left[\left(7x+1\right)+\left(x+7\right)\right]\left[\left(7x+1\right)-\left(x+7\right)\right]=\left(7^2-1^2\right)\left(x^2-1^2\right)\)
\(\Rightarrow\left(8x+8\right)\left(6x-6\right)=\left[\left(7+1\right)\left(7-1\right)\right]\left[\left(x+1\right)\left(x-1\right)\right]\)
\(\Rightarrow8\left(x+1\right)\cdot6\left(x-1\right)=8\left(x+1\right)\cdot6\left(x-1\right)\)( đpcm )