A = (2² - 1) (2² +1)(2⁴ +1)...(2⁶⁴ +1) + 1 = (2⁴ - 1)(2⁴ +1)...(2⁶⁴ +1) + 1  = (2⁸ -1)...(2⁶⁴ +1) + 1 = 2¹²⁸ -1 + 1 = 2¹²⁸

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

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

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

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

.......

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

=2^128-1

   3(2²+1)(2⁴+1)(2⁸+1)(2¹⁶+1)(2³²+1)(2⁶⁴+1)

=(2²-1)(2²+1)(2⁴+1)(2⁸+1)(2¹⁶+1)(2³²+1)(2⁶⁴+1)

=(2⁴-1)(2⁴+1)(2⁸+1)(2¹⁶+1)(2³²+1)(2⁶⁴+1)

=(2⁸-1)(2⁸+1)(2¹⁶+1)(2³²+1)(2⁶⁴+1)

=(2¹⁶-1)(2¹⁶+1)(2³²+1)(2⁶⁴+1)

=(2³²-1)(2³²+1)(2⁶⁴+1)

=(2⁶⁴-1)(2⁶⁴+1)

=(2¹²⁸-1)

Nếu thấy đúng thì thích cho mình nha

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. 

câu rút gọn = 7

câu tính nhanh mik chịu thông cảm nha

giải ra luôn đi bạn

(x-1)^2 - 2(x-1)(x+2) + (x+2)^2

= x^2 - 2x + 1 - 2x^2 - 2x + 4 + x^2 + 4x + 4

= x^2 - 2x^2 + x^2 - 2x - 2x + 4x + 1 + 4 + 4 

= 9

kết quả này mới đúng nha

1: Khi x=64 thì \(A=\dfrac{8+2}{8}=\dfrac{10}{8}=\dfrac{5}{4}\)

2: \(B=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

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

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

3: A/B>3/2

=>\(\dfrac{\sqrt{x}+2}{\sqrt{x}}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{3}{2}>0\)

=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{3}{2}>0\)

=>\(\dfrac{2\sqrt{x}+2-3\sqrt{x}}{\sqrt{x}\cdot2}>0\)

=>\(-\sqrt{x}+2>0\)

=>-căn x>-2

=>căn x<2

=>0<x<4

1) Thay x=64 vào A ta có:

\(A=\dfrac{2+\sqrt{64}}{\sqrt{64}}=\dfrac{2+8}{8}=\dfrac{5}{4}\)

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

\(B=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(B=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(B=\dfrac{x-1+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(B=\dfrac{x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(B=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

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

3) Ta có:

\(\dfrac{A}{B}>\dfrac{3}{2}\) khi

\(\dfrac{\sqrt{x}+2}{\sqrt{x}}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}>\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}>\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}}>\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{3}{2}>0\)

\(\Leftrightarrow\dfrac{2\sqrt{x}+2-3\sqrt{x}}{2\sqrt{x}}>0\)

\(\Leftrightarrow\dfrac{2-\sqrt{x}}{2\sqrt{x}}>0\)

Mà: \(2\sqrt{x}\ge0\forall x\)

\(\Leftrightarrow2-\sqrt{x}>0\)

\(\Leftrightarrow\sqrt{x}< 2\)

\(\Leftrightarrow x< 4\)

Kết hợp với đk:

\(0< x< 4\)

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

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