B

Chọn B

Giúp tui với mấy bạn ơi

A = 3(2²+1)(2^4 + 1)....(2^64 + 1) + 1 

= (2²-1)(2²+1)(2^4 + 1)....(2^64 + 1) + 1 

= (2^4 - 1)(2^4 + 1)....(2^64 + 1) + 1 

= (2^8 - 1).(2^8 + 1)(2^16 + 1)(2^32 + 1)(2^64 + 1) + 1 

= (2^16 - 1)(2^16 + 1)(2^32 + 1)(2^64 + 1) + 1 

= (2^32 - 1)(2^32 + 1)(2^64 + 1) + 1 

= (2^64 - 1)(2^64 + 1) + 1 = 2^128 - 1 + 1 = 2^128. 

B = 3(2²+1)(2^4 + 1)....(2^64 + 1) + 1 

= (2²-1)(2²+1)(2^4 + 1)....(2^64 + 1) + 1 

= (2^4 - 1)(2^4 + 1)....(2^64 + 1) + 1 

= (2^8 - 1).(2^8 + 1)(2^16 + 1)(2^32 + 1)(2^64 + 1) + 1 

= (2^16 - 1)(2^16 + 1)(2^32 + 1)(2^64 + 1) + 1 

= (2^32 - 1)(2^32 + 1)(2^64 + 1) + 1 

= (2^64 - 1)(2^64 + 1) + 1 = 2^128 - 1 + 1 = 2^128. 

Rút gọn:

a: Khi x=64 thì \(A=\dfrac{2}{8-2}=\dfrac{2}{6}=\dfrac{1}{3}\)

b: \(P=B:A\)

\(=\dfrac{3\sqrt{x}+\sqrt{x}-2-2\left(\sqrt{x}+2\right)}{x-4}:\dfrac{2}{\sqrt{x}-2}\)

\(=\dfrac{4\sqrt{x}-2-2\sqrt{x}-4}{x-4}\cdot\dfrac{\sqrt{x}-2}{2}\)

\(=\dfrac{2\sqrt{x}-6}{2\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\)

c: P<0

=>căn x-3<0

=>0<=x<9

mà x nguyên và x<>4

nên \(x\in\left\{0;1;2;3;5;6;7;8\right\}\)

a: \(A=4\cdot\dfrac{5}{2}\sqrt{x}-\dfrac{8}{3}\cdot\dfrac{3}{2}\sqrt{x}-\dfrac{4}{3x}\cdot\dfrac{3x}{8}\cdot\sqrt{x}\)

\(=10\sqrt{x}-4\sqrt{x}-\dfrac{1}{2}\sqrt{x}\)

\(=\dfrac{11}{2}\sqrt{x}\)

b: \(B=\dfrac{y}{2}+\dfrac{3}{4}\cdot\left|2y-1\right|-\dfrac{3}{2}\)

\(=\dfrac{y}{2}+\dfrac{3}{4}\left(1-2y\right)-\dfrac{3}{2}\)

=1/2y+3/4-3/2y-3/2

=-y-3/4

Bài làm:

Ta có: \(A=64-\left(x-4\right)\left(x^2+4x+16\right)\)

\(A=64-x^3+64\)

\(A=128-x^3\)

Tại \(x=-\frac{1}{2}\) ta được:

\(A=128-\left(-\frac{1}{2}\right)^3=\frac{1025}{8}\)

A = 64 - ( x - 4 )( x² + 4x + 16 )

A = 64 - ( x³ + 4x² + 16x - 4x² - 16x - 64 )

A = 64 - ( x³ - 64 )

A = 64 - x³ + 64

A = -x³ + 128

Thế x = -1/2 vào A ta được :

A = -(-1/2)³ + 128 = 1/8 + 128 = 1025/8

\(A=64-\left(x-4\right)\left(x^2+4x+16\right)\)

\(=64-\left(x-4\right)\left(x^2-4x+4^2\right)\)

\(=64-\left(x^3-4^3\right)=64-x^3+4^3\)

\(=64+64-x^3=128-x^3\)

Với \(x=-\frac{1}{2}\)thì \(128-x^3=128-\left(-\frac{1}{2}\right)^3\)

\(=128-\left(-\frac{1}{8}\right)=128+\frac{1}{8}=\frac{1025}{8}\)

a: Khi x=64 thì \(A=\dfrac{3\cdot8+1}{8+2}=\dfrac{25}{10}=\dfrac{5}{2}\)

b: \(B=\dfrac{2\sqrt{x}-4-\sqrt{x}+5}{x-4}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}+2}\)

`a)|x-2|=2<=>[(x=4(ko t//m)),(x=0(t//m)):}`

Thay `x=0` vào `A` có: `A=[2\sqrt{0}-3]/[\sqrt{0}-2]=3/2`

`b)` Với `x >= 0,x ne 4` có:

`B=[2(\sqrt{x}-3)+\sqrt{x}(\sqrt{x}+3)-4\sqrt{x}]/[(\sqrt{x}+3)(\sqrt{x}-3)]`

`B=[2\sqrt{x}-6+x+3\sqrt{x}-4\sqrt{x}]/[(\sqrt{x}+3)(\sqrt{x}-3)]`

`B=[x+\sqrt{x}-6]/[(\sqrt{x}+3)(\sqrt{x}-3)]`

`B=[(\sqrt{x}+3)(\sqrt{x}-2)]/[(\sqrt{x}+3)(\sqrt{x}-3)]`

`B=[\sqrt{x}-2]/[\sqrt{x}-3]`

`c)` Với `x >= 0,x ne 4` có:

`C=A.B=[2\sqrt{x}-3]/[\sqrt{x}-2].[\sqrt{x}-2]/[\sqrt{x}-3]=[2\sqrt{x}-3]/[\sqrt{x}-3]`

Có: `C >= 1`

`<=>[2\sqrt{x}-3]/[\sqrt{x}-3] >= 1`

`<=>[2\sqrt{x}-3-\sqrt{x}+3]/[\sqrt{x}-3] >= 0`

`<=>[\sqrt{x}]/[\sqrt{x}-3] >= 0`

  Vì `x >= 0=>\sqrt{x} >= 0`

  `=>\sqrt{x}-3 > 0`

`<=>x > 9` (t/m đk)

Ta có

A   =   5 ( x   +   4 ) 2   +   4 ( x   –   5 ) 2   –   9 ( 4   +   x ) ( x   –   4 )     =   5 ( x 2   +   2 . x . 4   +   16 )   +   4 ( x 2   –   2 . x . 5   +   5 2 )   –   9 ( x 2   –   4 2 )     =   5 ( x 2   +   8 x   +   16 )   +   4 ( x 2   –   10 x   +   25 )   –   9 ( x 2   –   4 2 )     =   5 x 2   +   40 x   +   80   +   4 x 2   –   40 x   +   100   –   9 x 2   +   144     = ( 5 x 2   +   4 x 2   –   9 x 2 )   +   ( 40 x   –   40 x )   +   ( 80   + 100   +   144 )

= 324

Đáp án cần chọn là: C