(/3+2/2)+(/3-2/2)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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) \(3\cdot3^5\cdot3^4\cdot3^2\)
\(=3^{1+5+4+2}\)
\(=3^{12}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2A-A=\left(2^2+2^3+...+2^{21}\right)-\left(2+2^2+...+2^{20}\right)\)
\(A=2^{21}-2\)
B tương tự câu A
\(5C-C=\left(5^2+5^3+...+5^{51}\right)-\left(5+5^2+...+5^{50}\right)\)
\(C=\dfrac{5^{51}-5}{4}\)
\(3D-D=3+3^2+...+3^{101}-\left(1+3+...+3^{100}\right)\)
\(D=\dfrac{3^{101}-1}{2}\)
\(A=2^1+2^2+2^3+...+2^{20}\)
\(2\cdot A=2^2+2^3+2^4+...+2^{21}\)
\(A=2^{21}-2\)
\(B=2^1+2^3+2^5+...+2^{99}\)
\(4\cdot B=2^3+2^5+2^7+...+2^{101}\)
\(B=\)\(\left(2^{101}-2\right):3\)
\(C=5^1+5^2+5^3+...+5^{50}\)
\(5\cdot C=5^2+5^3+5^4+...+5^{51}\)
\(C=(5^{51}-5):4\)
\(D=3^0+3^1+3^2+...+3^{100}\)
\(3\cdot D=3^1+3^2+3^3+...+3^{101}\)
\(D=(3^{101}-1):2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
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![](https://rs.olm.vn/images/avt/0.png?1311)
a: A=3^2(1^2+2^2+...+10^2)
=9*385
=3465
b: B=2^3(1^3+2^3+...+10^3)
=8*3025
=24200
![](https://rs.olm.vn/images/avt/0.png?1311)
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 = \(\frac{\text{3^101 - 3}}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a. M=-1^2+2^2-3^2+4^2-...-99^2+100^2.
M=(2-1)(2+1)+(4-3)(4+3)+...+(100-99)(100+99)
M=3+7+...+199
=>2M=3+7+...+199+3+7+...+199 (198 số)
=(3+199)+(7+195)+...+(199+3) (99 cặp)
=202.99
=19998
=>M=19998:2=9999
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2}{3}\times5+\frac{2}{3}\times7+\frac{2}{3}\times9+\frac{2}{3}\times11+\frac{2}{3}\times13+\frac{2}{3}\times15\)
=\(\frac{2}{3}\)\(\times\)( 5+7+9+11+15)
= \(\frac{2}{3}\)\(\times\)47
= \(\frac{94}{3}\)
Có friend forever II Lê Tiến Đạt giải rồi nhé nên đừng bắt tui giải nữa ( chuồn là thượng sách)
![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(A=\left(\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}\right)\left(\sqrt{3-2\sqrt{2}}+\sqrt{3+2\sqrt{2}}\right)\)
\(=\left(\sqrt{2}+1-\sqrt{2}+1\right)\left(\sqrt{2}-1+\sqrt{2}+1\right)\)
\(=2\cdot2\sqrt{2}=4\sqrt{2}\)