- 5/3+5/3^2+5/3^3+...+5/3^20
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2 × 3 = 6 6 : 3 = 2 6 : 2 = 3 |
3 × 5 = 15 15 : 3 = 5 15 : 5 = 3 |
5 × 4 = 20 20 : 5 = 4 20 : 4 = 5 |
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\(\Leftrightarrow b=\dfrac{5}{3}+\dfrac{5}{3^2}+...+\dfrac{5}{3^{20}}\)
=>\(3b=5+\dfrac{5}{3}+...+\dfrac{5}{3^{19}}\)
=>\(2b=5-\dfrac{5}{3^{20}}=\dfrac{5\cdot3^{20}-5}{3^{20}}\)
=>\(n=\dfrac{5\cdot3^{20}-5}{3^{20}\cdot2}\)
a) 3 x 5 < 3 x 6
3 x 5 > 3x 4
b) 3 x 5 = 5 x 3
4 x 6 = 6 x 4
c) 20 : 4 > 20 : 5
20 : 4 < 20 : 2
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\(3A=5+\dfrac{5}{3}+\dfrac{5}{3^2}+...+\dfrac{5}{3^{19}}\)
\(2A=3A-A=5-\dfrac{5}{3^{20}}\)
\(A=\dfrac{5.3^{20}-5}{2.3^{20}}\)
1) \(2^3\times x-5^2\times x=2\times\left(5^2+2^2\right)-33\)
\(x\times\left(2^3-5^2\right)=2\times\left(25+4\right)-33\)
\(x\times\left(8-25\right)=2\times29-33\)
\(x\times-17=25\)
\(x=-\dfrac{25}{17}\)
2) \(15\div\left(x+2\right)=\left(3^3+3\right)\div1\)
\(15\div\left(x+2\right)=\left(27+3\right)\div1\)
\(15\div\left(x+2\right)=30\div1\)
\(15\div\left(x+2\right)=30\)
\(x+2=\dfrac{1}{2}\)
\(x=-\dfrac{3}{2}\)
3) \(20\div\left(x+1\right)=\left(5^2+1\right)\div13\)
\(20\div\left(x+1\right)=\left(25+1\right)\div13\)
\(20\div\left(x+1\right)=26\div13\)
\(20\div\left(x+1\right)=2\)
\(x+1=20\div2\)
\(x+1=10\)
\(x=9\)
4) \(320\div\left(x-1\right)=\left(5^3-5^2\right)\div4+15\)
\(320\div\left(x-1\right)=\left(125-25\right)\div4+15\)
\(320\div\left(x-1\right)=100\div4+15\)
\(320\div\left(x-1\right)=25+15\)
\(320\div\left(x-1\right)=40\)
\(x-1=8\)
\(x=9\)
5) \(240\div\left(x-5\right)=2^2\times5^2-20\)
\(240\div\left(x-5\right)=4\times25-20\)
\(240\div\left(x-5\right)=100-20\)
\(240\div\left(x-5\right)=80\)
\(x-5=30\)
\(x=35\)
6) \(70\div\left(x-3\right)=\left(3^4-1\right)\div4-10\)
\(70\div\left(x-3\right)=\left(81-1\right)\div4-10\)
\(70\div\left(x-3\right)=80\div4-10\)
\(70\div\left(x-3\right)=20-10\)
\(70\div\left(x-3\right)=10\)
\(x-3=7\)
\(x=10\)
Phương pháp giải:
Nhẩm bảng nhân, chia trong phạm vi đã học rồi điền kết quả vào chỗ trống.
Lời giải chi tiết:
2 × 8 = 16 | 16 : 2 = 8 |
21 : 3 = 7 | 4 × 5 = 20 |
3 × 8 = 24 | 24 : 3 = 8 |
5 × 6 = 30 | 5 × 4 = 20 |
4 × 8 = 32 | 40 : 4 = 10 |
28 : 4 = 7 | 20 : 4 = 5 |
5 × 8 = 40 | 40 : 5 = 8 |
3 × 6 = 18 | 20 : 5 = 4 |
\(2A-A=\left(2^2+2^3+...+2^{21}\right)-\left(2+2^2+...+2^{20}\right)\)
\(A=2^{21}-2\)
B tương tự câu A
\(5C-C=\left(5^2+5^3+...+5^{51}\right)-\left(5+5^2+...+5^{50}\right)\)
\(C=\dfrac{5^{51}-5}{4}\)
\(3D-D=3+3^2+...+3^{101}-\left(1+3+...+3^{100}\right)\)
\(D=\dfrac{3^{101}-1}{2}\)
\(A=2^1+2^2+2^3+...+2^{20}\)
\(2\cdot A=2^2+2^3+2^4+...+2^{21}\)
\(A=2^{21}-2\)
\(B=2^1+2^3+2^5+...+2^{99}\)
\(4\cdot B=2^3+2^5+2^7+...+2^{101}\)
\(B=\)\(\left(2^{101}-2\right):3\)
\(C=5^1+5^2+5^3+...+5^{50}\)
\(5\cdot C=5^2+5^3+5^4+...+5^{51}\)
\(C=(5^{51}-5):4\)
\(D=3^0+3^1+3^2+...+3^{100}\)
\(3\cdot D=3^1+3^2+3^3+...+3^{101}\)
\(D=(3^{101}-1):2\)
53+532+...+5320=5(13+132+...+1320)53+532+...+5320=5(13+132+...+1320)
Gọi A=13+132+...+1320�=13+132+...+1320. Ta có
3A=1+13+...+13193�=1+13+...+1319
3A−A=(1+13+...+1319)−(13+132+...+1320)3�−�=(1+13+...+1319)−(13+132+...+1320)
2A=1−13202�=1−1320
A=1−13202�=1−13202
Suy ra 53+532+...+5320=5(13+132+...+1320)=5⋅1−13202=5−53202