giúp mình với !!!!!!!!!!!!!!!!!!!!!!!!

Câu b, bài b1 chứng minh \(a=2^{2006}-1?\)

1. D

2. Lỗi

3. A

1D

2A

3A

Câu 1: D

Câu 3: A

2:

\(B=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{100^2}-1\right)\)

\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)

\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}+1\right)\)

\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)

\(=-\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{-101}{200}< -\dfrac{100}{200}=-\dfrac{1}{2}\)

Ta có

\(\frac{a+1}{a}=3\Leftrightarrow a+1=3a\Leftrightarrow2a=1\Leftrightarrow a=0,5.\)

Thay a=0,5 vào a^2+1/a^2 ta được

\(a^2+\frac{1}{a^2}=0,5^2+\frac{1}{0,5^2}=4,25\)

Làm tương tự với các câu còn lại

cam on ban

cam on cai gi, k đi

A=1+3+3²+...+3⁷

3A=3+3²+3³+...+3⁸

3A-A=(3+3²+3³+...+3⁸)-(1+3+3²+...+3⁷)

2A=3⁸-1

A=(3⁸-1):2

mà vì chia 2 nên chắc chắn A chia hết cho 2

\(a.A=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}\) 

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)

\(2A-A=1-\frac{1}{2^{99}}\)

\(A=1-\frac{1}{2^{99}}< 1\)

\(b.B=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)

\(3A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)

\(3A-A=\left(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\right)\)

\(2A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(6A=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)

\(6A-2A=\left(3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)

\(4A=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)

\(4A=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)

\(4A=3-\frac{303}{3^{100}}+\frac{100}{3^{100}}\)

\(4A=3-\frac{203}{3^{100}}< 3\)

\(A< \frac{3}{4}\)

Ủng hộ mk nha ^_^

Ta có :

M = 2( a³ + b³ ) - 3( a² + b² ) 

    = 2( a + b ) ( a² - ab + b² ) - 3( a² + b² ) 

    = 2( a² - ab + b² ) - 3 ( a² + b² ) 

   = 2a² - 2ab + 2b² - 3a² - 3b² 

   = -a² - 2ab - b² 

   = - ( a + b )²

   = -1 

M = a³ + b³ + 3ab(a² + b²) + 6a²b²(a + b)

= (a + b)(a² - ab + b²) + 3ab((a + b)² - 2ab) + 6a²b²(a + b)

= (a + b)((a + b)² - 3ab) + 3ab((a + b)² - 2ab) + 6a²b²(a + b)

= 1 - 3ab + 3ab(1 - 2ab) + 6a²b²

= 1 - 3ab + 3ab - 6a²b² + 6a²b² = 1