a) \(2^5\cdot2^7\)

\(=2^{5+7}\)

\(=2^{12}\)

b) \(2^3\cdot2^2\)

\(=2^{3+2}\)

\(=2^5\)

c) \(2^4\cdot2^3\cdot2^5\)

\(=2^{4+3+5}\)

\(=2^{12}\)

d) \(2^2\cdot2^4\cdot2^6\cdot2\)

\(=2^{2+4+6+1}\)

\(=2^{13}\)

e) \(2\cdot2^3\cdot2^7\cdot2^4\)

\(=2^{1+3+7+4}\)

\(=2^{15}\)

f) \(3^8\cdot3^7\)

\(=3^{8+7}\)

\(=3^{15}\)

g) \(3^2\cdot3\)

\(=3^{2+1}\)

\(=3^3\)

h) \(3^4\cdot3^2\cdot3\)

\(=3^{4+2+1}\)

\(=3^7\)

I) \(3\cdot3^5\cdot3^4\cdot3^2\)

\(=3^{1+5+4+2}\)

\(=3^{12}\)

Lời giải chi tiết

1²  1                                     1³ 1² – 0²        (0 + 1)²  0²  +1²

2² 1 + 3                                2³ 3² – 1²         (1 + 2)² 1² + 2²

3² 1 + 3 + 5                           3³ 6² – 3²      (2 + 3)²  2² +  3²

                                                     4³ 10² – 6²

= , = , = , = , = , > nhé

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\(2A-A=\left(2^2+2^3+...+2^{21}\right)-\left(2+2^2+...+2^{20}\right)\)

\(A=2^{21}-2\)

B tương tự câu A

\(5C-C=\left(5^2+5^3+...+5^{51}\right)-\left(5+5^2+...+5^{50}\right)\)

\(C=\dfrac{5^{51}-5}{4}\)

\(3D-D=3+3^2+...+3^{101}-\left(1+3+...+3^{100}\right)\)

\(D=\dfrac{3^{101}-1}{2}\)

\(A=2^1+2^2+2^3+...+2^{20}\)

\(2\cdot A=2^2+2^3+2^4+...+2^{21}\)

\(A=2^{21}-2\)

\(B=2^1+2^3+2^5+...+2^{99}\)

\(4\cdot B=2^3+2^5+2^7+...+2^{101}\)

\(B=\)\(\left(2^{101}-2\right):3\)

\(C=5^1+5^2+5^3+...+5^{50}\)

\(5\cdot C=5^2+5^3+5^4+...+5^{51}\)

\(C=(5^{51}-5):4\)

\(D=3^0+3^1+3^2+...+3^{100}\)

\(3\cdot D=3^1+3^2+3^3+...+3^{101}\)

\(D=(3^{101}-1):2\)

giải chi tiết ý b ddc ko

a) 2A = 2 + 2^2 + 2^3 +...+ 2^11 

2A-A = (2 + 2^2 + 2^3 +...+ 2^11) - (1 + 2 + 2^2 +...+ 2^10)

      A = 2^11 - 1

b) 3B = 3 + 3^2 + 3^3 +...+ 3^101 

    3B-B = (3 + 3^2 + 3^3 +...+ 3^101) - (1 + 3 + 3^2 +...+ 3^100)

        2B = 3^101 - 3

          B = \(\frac{\text{3^101 - 3}}{2}\)

a, Ta có : \(\left\{{}\begin{matrix}\sqrt{3+2\sqrt{2}}=\sqrt{2+2\sqrt{2}+1}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\\\sqrt{3-2\sqrt{2}}=\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}-1\end{matrix}\right.\)

- Thay lần lượt vào A ta được :

\(A=\left(\sqrt{2}+1-\sqrt{2}+1\right)\left(\sqrt{2}-1+\sqrt{2}+1\right)=2.2\sqrt{2}=4\sqrt{2}\)

b, \(B=\sqrt{2+\sqrt{3}}\sqrt{2^2-\left(\sqrt{2+\sqrt{3}}\right)^2}=\sqrt{2+\sqrt{3}}\sqrt{4-2-\sqrt{3}}\)

\(=\sqrt{2-\sqrt{3}}\sqrt{2+\sqrt{3}}=\sqrt{4-3}=\sqrt{1}=1\)

c, \(C=\dfrac{\left(2+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{2-\sqrt{3}}\right)+\left(2-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{2+\sqrt{3}}\right)}{\left(\sqrt{2}+\sqrt{2+\sqrt{3}}\right)\left(\sqrt{2}-\sqrt{2-\sqrt{3}}\right)}\)

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

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

a) Ta có: \(A=\left(\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}\right)\left(\sqrt{3-2\sqrt{2}}+\sqrt{3+2\sqrt{2}}\right)\)

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

\(=2\cdot2\sqrt{2}=4\sqrt{2}\)

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