\(A=\sqrt{3-\sqrt{5}}-\sqrt{4-\sqrt{15}}+\sqrt{6-3\sqrt{3}}\)

\(=\dfrac{1}{\sqrt{2}}\cdot\left(\sqrt{6-2\sqrt{5}}-\sqrt{8-2\sqrt{15}}+\sqrt{12-6\sqrt{3}}\right)\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{5}-1-\sqrt{5}+\sqrt{3}+3-\sqrt{3}\right)\)

=2/căn 2=căn 2

\(B=\sqrt{4-\sqrt{7}}-\sqrt{14-5\sqrt{3}}-\sqrt{5+\sqrt{21}}\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{8-2\sqrt{7}}-\sqrt{28-10\sqrt{3}}-\sqrt{10+2\sqrt{21}}\right)\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{7}-1-5+\sqrt{3}-\sqrt{7}-\sqrt{3}\right)\)

=-6/căn 2=-3căn2

\(C=\sqrt{11-6\sqrt{2}}-\sqrt{6-4\sqrt{2}}+\sqrt{7-2\sqrt{6}}\)

=3-căn 2-2+căn 2+căn 6-1

=căn 6

\(D=\sqrt{6-\sqrt{11}}-\sqrt{10+3\sqrt{11}}+2\sqrt{2}-1\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{12-2\sqrt{11}}-\sqrt{20+6\sqrt{11}}\right)+2\sqrt{2}-1\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{11}-1-\sqrt{11}-3\right)+2\sqrt{2}-1\)

=-1

\(F=\sqrt{6+3\sqrt{3}}-\sqrt{2+\sqrt{3}}+\sqrt{6-4\sqrt{2}}\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{12+6\sqrt{3}}-\sqrt{4+2\sqrt{3}}\right)+2-\sqrt{2}\)

=1/căn 2(3+căn 3-căn 3-1)+2-căn 2

=căn 2+2-căn 2

=2

Gấp quá em nhỉ! 

 A         =  2 + 2² + 2³ +2⁴ + 2⁵ +....+2²⁰⁰

2A       =         2²  + 2³ + 2⁴ + 2⁵+....+2²⁰⁰ + 2²⁰¹

2A - A  =         2²⁰¹ - 2

A         =         2²⁰¹ - 2

A         =        (2⁴)⁵⁰.2 - 2

A         =       (\(\overline{...6}\))⁵⁰.2 - 2

A         =        \(\overline{...6}\) . 2 - 2

A         =         \(\overline{...2}\) - 2

A        =           \(\overline{...0}\) \(\Rightarrow\) A = C . 10 ( C \(\in\) N*)

Thay A = C .10  vào biểu thức B = -1,9 . A  ta có :

B =    -1,9 . C .10

B = -(1,9 .10) .C

B = - 19.C vì C \(\in\) N \(\Rightarrow\) B \(\in\) Z (đpcm)

thank you

  chọn D

a: \(f\left(x\right)=-x^4-8x^3+5x^2+6x-7\)

\(g\left(x\right)=x^4+8x^3-5x^2+5\)

b: \(f\left(x\right)+g\left(x\right)=6x-2\)

\(f\left(x\right)-g\left(x\right)=-2x^2-16x^3+10x^2+6x-12\)

c: |x|=1 thì x=-1 hoặc x=1

h(-1)=6x(-1)-2=-8

h(1)=6x1-2=4

a/ với f(x)

có : \(-x^4-8x^3+5x^2+6x-7\)

với g(x)

có :\(x^4+8x^3-5x^2+5\)

b,     f(x)  \(-x^4-8x^3+5x^2+6x-7\)

        g(x)   \(x^4+8x^3-5x^2\)            + 5

f(x)+g(x) = 6x-2

1: Để C là số nguyên thì 2n+2-3 chia hết cho n+1

=>\(n+1\in\left\{1;-1;3;-3\right\}\)

=>\(n\in\left\{0;-2;2;-4\right\}\)