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\(A=2\left(1.30+2.29+3.28+...+15.16\right)=\)
\(=2.\left[1.30+2.\left(30-1\right)+3.\left(30-2\right)+...+15.\left(30-14\right)\right]=\)
\(=2.\left(1.30-1.2+2.30+3.30-2.3+4.30-3.4+...+15.30-14.15\right)=\)
\(=2.\left[30.\left(1+2+3+...+15\right)-\left(1.2+2.3+3.4+...+14.15\right)\right]=\)
Đặt \(B=1+2+3+...+15=\frac{15.\left(1+15\right)}{2}=120\)
Đặt \(C=1.2+2.3+3.4+...+14.15\)
\(\Rightarrow3C=1.2.3+2.3.3+3.4.3+...+14.15.3=\)
\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+14.15.\left(16-13\right)=\)
\(=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...-13.14.15+14.15.16=14.15.16\)
\(\Rightarrow C=\frac{14.15.16}{3}=5.14.16=1120\)
\(\Rightarrow A=2.\left(30.B-C\right)=2\left(30.120-1120\right)=4960\)
Ta có:
A=\(\frac{1}{1.101}+\frac{1}{2.102}+...+\frac{1}{25.125}\)
=\(\frac{1}{100}\left(\frac{100}{1.101}+\frac{100}{2.102}+...+\frac{100}{25.125}\right)\)
=\(\frac{1}{100}\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{25}-\frac{1}{125}\right)\)
=\(\frac{1}{100}\left[\left(1+\frac{1}{2}+...+\frac{1}{25}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{125}\right)\right]\)
B=\(\frac{1}{1.26}+\frac{1}{2.27}+...+\frac{1}{100.125}\)
=\(\frac{1}{25}\left(\frac{25}{1.26}+\frac{25}{2.27}+...+\frac{25}{100.125}\right)\)
=\(\frac{1}{25}\left(1-\frac{1}{26}+\frac{1}{2}-\frac{1}{27}+...+\frac{1}{100}-\frac{1}{125}\right)\)
=\(\frac{1}{25}\left[\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{125}\right)\right]\)
=\(\frac{1}{25}\left[\left(1+\frac{1}{2}+...+\frac{1}{25}\right)+\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{100}\right)-\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{100}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{125}\right)\right]\)
= \(\frac{1}{25}\left[\left(1+\frac{1}{2}+...+\frac{1}{25}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{125}\right)\right]\)
=> \(\frac{A}{B}\)=\(\frac{\frac{1}{100}\left[\left(1+\frac{1}{2}+...+\frac{1}{25}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{125}\right)\right]}{\frac{1}{25}\left[\left(1+\frac{1}{2}+...+\frac{1}{25}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{125}\right)\right]}\)=\(\frac{1}{\frac{100}{\frac{1}{25}}}\)=\(\frac{1}{100}\cdot25=\frac{25}{100}=\frac{1}{4}\)
ko bít minh làm đúng ko, nhưng nếu sai đừng trách mình nhá
ta có:
1/20=1/4-1/5, 1/30=1/5-1/6,1/42=1/6-1/7,1/54=1/7-1/8,1/72=1/8-1/9,1/90=1/9-1/10
1/4-1/5x1/5-1/6x1/6-1/7x1/7-1/8x1/8-1/9x1/9-1/10
gạch đi mấy phân số giống nhau, ta còn:1/4x1/10=1/40
Sửa đề: 8666
8666*15+170*4333
=4333*(2*15+170)
=200*4333
=866600
A = 44 . 82 -22 + 18 . 44
= 44 . 82 - 4 + 18 . 44
= 44. ( 82 + 18 ) - 4
= 44 . 100 - 4
= 4400 - 4
= 4396
Ta có: \(A=44\cdot82-2^2+18\cdot44\)
\(=44\cdot\left(82+18\right)-4\)
\(=44\cdot100-4\)
\(=4400-4=4396\)
13: \(\left(-15\right)+8+7\)
\(=\left(-15\right)+\left(8+7\right)\)
=-15+15
=0
14: \(\left(-8\right)+2+6\)
\(=\left(-8\right)+\left(2+6\right)\)
=-8+8
=0
15: \(\left(-1\right)+3-2\)
\(=\left(-1-2\right)+3\)
=-3+3
=0
16: \(25-8-7\)
\(=25-\left(8+7\right)\)
=25-15
=10
17: \(8-2-6\)
\(=8-\left(2+6\right)\)
=8-8
=0
18: \(\left(-12\right)-3+15\)
\(=\left(-12-3\right)+15\)
=-15+15
=0