\(4,=\dfrac{6\left(\sqrt{2}-\sqrt{3}-3\right)}{5-2\sqrt{6}-9}=\dfrac{6\left(\sqrt{2}-\sqrt{3}-3\right)}{-4-2\sqrt{6}}\\ =\dfrac{3\left(3-\sqrt{2}-\sqrt{3}\right)}{2+\sqrt{6}}=\dfrac{\left(9-3\sqrt{2}-3\sqrt{3}\right)\left(\sqrt{6}-2\right)}{2}\\ =\dfrac{9\sqrt{6}-18-6\sqrt{3}+6\sqrt{2}-9\sqrt{2}+6\sqrt{3}}{2}\\ =\dfrac{9\sqrt{6}-3\sqrt{2}-18}{2}\)

\(7,=\dfrac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-2-\sqrt{3}\\ =\sqrt{3}+2+\sqrt{2}+1-2-\sqrt{3}=1+\sqrt{2}\)

\(10,\dfrac{1}{\sqrt{a}+\sqrt{a+2}}=\dfrac{\sqrt{a}-\sqrt{a+2}}{a-a-2}=\dfrac{\sqrt{a-2}-\sqrt{a}}{2}\)

Do đó \(\dfrac{1}{\sqrt{1}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{5}}+...+\dfrac{1}{\sqrt{47}+\sqrt{49}}\)

\(=\dfrac{\sqrt{3}-\sqrt{1}+\sqrt{5}-\sqrt{3}+...+\sqrt{49}-\sqrt{47}}{2}=\dfrac{-1+\sqrt{49}}{2}=\dfrac{7-1}{2}=3\)

10, \(\dfrac{1}{\sqrt{1}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{5}}+...+\dfrac{1}{\sqrt{17}+\sqrt{19}}=\dfrac{\sqrt{1}-\sqrt{3}}{\left(\sqrt{1}+\sqrt{3}\right)\left(\sqrt{1}-\sqrt{3}\right)}+\dfrac{\sqrt{3}-\sqrt{5}}{\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{3}-\sqrt{5}\right)}+...+\dfrac{\sqrt{17}-\sqrt{19}}{\left(\sqrt{17}+\sqrt{19}\right)\left(\sqrt{17}-\sqrt{19}\right)}=\dfrac{1-\sqrt{3}+\sqrt{3}-\sqrt{5}+...+\sqrt{17}-\sqrt{19}}{-2}=-\dfrac{1-\sqrt{19}}{2}\)

= y.2 + (5,3 +4,7) =15

   y .2 +10             =15

    y .2                  =15-10 =5

       y                    =5 :2

         y =   2.5

(Y+5,3)+(Y+4,7)=15

=>2Y+10=15

=>2Y=15-10=5

=>Y=5/2=2,5

(y+5,3)+(y+4,7)=15

y+5,3+y+4,7=15

2y+5,3+4,7=15

2y+10=15

2y=5 => y=2,5

Như đã hứa, kkkkkkk cho mk nha

Sửa đề: -cy đổi thành -xy

P=-x^3y+4x^3y+x^2-xy+2xy+1

=3x^3y+x^2+xy+1

\(\left(5x+3y\right)\left(25x^2-15xy+9y^2\right)\)

\(=\left(5x+3y\right)\left[\left(5x\right)^2-5x.3y+\left(3y\right)^2\right]\)

\(=\left(5x\right)^3+\left(3y\right)^3=125x^3-27y^3\)