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

\(\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{1}{14}}{-1-\dfrac{3}{7}+\dfrac{3}{28}}=\dfrac{\dfrac{2.14+2.6-3}{42}}{\dfrac{-28-12+3}{28}}=\dfrac{37}{42}.\dfrac{-28}{37}=-\dfrac{2}{3}\)

\(\sqrt{9+4\sqrt{2}}=\sqrt{\left(8+4\sqrt{2}+1\right)}=\sqrt{\left(\sqrt{8}+1\right)^2}=\sqrt{8}+1\)

1) \(x^3-8x+7=\left(x-1\right)\left(x^2+x-7\right)\)

2) \(x^3+8x^2-9=\left(x-1\right)\left(x^2+9x+9\right)\)

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

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

5) \(x^4-5x^2+4=\left(x-1\right)\left(x-2\right)\left(x+1\right)\left(x+2\right)\)

\(\sqrt{19+6\sqrt{2}}=\sqrt{\left(\sqrt{18}+1\right)^2}=\sqrt{18}+1\)

Xét ΔOBC và ΔODA có:

OA=OC(gt)

\(\widehat{BOD}\) chung

OB=OD( gt)

=> ΔOBC = ΔODA(c.g.c)

\(\Rightarrow\widehat{OAD}=\widehat{OCB}\) ( 2 góc tương ứng)

Chỗ đó cho mình bổ sung thêm cho dễ hiểu nhé bạn: 

\(\dfrac{-\sqrt{2}-6}{2+\sqrt{2}}=\dfrac{-\left(2+\sqrt{2}\right)}{2+\sqrt{2}}-\dfrac{4}{2+\sqrt{2}}=-1-\dfrac{4\left(2-\sqrt{2}\right)}{\left(2+\sqrt{2}\right)\left(2-\sqrt{2}\right)}=-1-\dfrac{8-4\sqrt{2}}{2^2-\left(\sqrt{2}\right)^2}=-1-\dfrac{8-4\sqrt{2}}{2}=-1-4+2\sqrt{2}=-5+2\sqrt{2}\)

Bài 3 : 

a) \(x^4+2x^2+1=\left(x^2+1\right)^2\)

b) \(4x^2-12xy+9y^2=\left(2x-3y\right)^2\)

c) \(-x^2-2xy-y^2=-\left(x+y\right)^2\)

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

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

g) \(x^3+6x^2+12x+8=\left(x+2\right)^3\)

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

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

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

Để \(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\in Z\) thì \(2⋮\sqrt{x}+3\Rightarrow\sqrt{x}+3\in\) Ư(2)\(=\left\{1;-1;2;-2\right\}\)

Vì \(\sqrt{x}\ge0\Rightarrow x\in\varnothing\)