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a)
\(A=\frac{6^3+3.6^3+3^3}{-13}=\frac{3^3.2^3+3^3.2^2+3^3}{-13}=\frac{3^3\left(8+4+1\right)}{-13}=\frac{27.13}{-13}=-27\)
b)
A=1+5+52+53+...+550
5A=5+52+53+...551
5A-A=(5+52+53+...+551)-(1+5+52+...+550)
4A=551-1
A=\(\frac{5^{51}-1}{4}\)
c)
A=2100-299+298-...+22-2
2A=2101-2100+299-...+23-22
2A+A=(2101-2100+...+23-22)+(2100-299+...+22-2)
3A=2101-2
A=\(\frac{2^{101}-2}{3}\)
b.
\(A=1+5+5^2+5^3+...+5^{49}+5^{50}\)
\(5A=5+5^2+5^3+...+5^{50}+5^{51}\)
\(5A-A=\left(5+5^2+5^3+...+5^{50}+5^{51}\right)-\left(1+5+5^2+..+5^{50}\right)\)
\(4A=5^{51}-1\)
\(A=\frac{5^{51}-1}{4}\)
1 cách khác
M = 299 + 2 . 298 + 3 . 297 + 4 . 296 + ... + 98 . 22 + 99 . 2 + 100 . 20
M = 299 + 2 . ( 299 - 298 ) + 3 . ( 298 - 297 ) + 4 . ( 297 - 296 ) + ... + 99 . ( 22 - 2 ) + 100 . ( 2 - 1 )
M = 299 + 2100 - 2 . 298 + 3 . 298 - 3 . 297 + 4 . 297 - 4. 296 + ... + 99 . 22 - 99 . 2 + 100 . 2 - 100
M = 2100 + 299 +298 + 297 + 296 + ... + 2 - 100
M = 2101 - 102
a)Ta có:
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)....\left(\frac{1}{98^2}-1\right)\left(\frac{1}{99^2}-1\right)\)
\(=\left(\frac{1}{2.2}-1\right)\left(\frac{1}{3.3}-1\right)\left(\frac{1}{4.4}-1\right)....\left(\frac{1}{98.98}-1\right)\left(\frac{1}{99.99}-1\right)\)
\(=\left(-\frac{3}{2.2}\right).\left(-\frac{8}{3.3}\right).\left(-\frac{15}{4.4}\right)...\left(-\frac{9603}{98.98}\right).\left(-\frac{9800}{99.99}\right)\)
\(=\left[\left(-1\right).\left(-1\right).\left(-1\right)...\left(-1\right)\right].\frac{3}{2.2}.\frac{8}{3.3}.\frac{15}{4.4}...\frac{9603}{98.98}.\frac{9800}{99.99}\)
|------------------------98 số -1--------------------|
\(=\left(-1\right)^{98}.\frac{1.3}{2.3}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{95.97}{98.98}.\frac{98.100}{99.99}\)
\(=\frac{1.3}{2.3}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{95.97}{98.98}.\frac{98.100}{99.99}\)
\(=\frac{1.3.2.4.3.5...95.97.98.100}{2.2.3.3.4.4...98.98.99.99}\)
Ta sẽ rút gọn các thừa số chung ở tử và mẫu
\(=\frac{1.100}{2.99.99}\)
\(=\frac{50}{9801}\)
Vậy \(A=\frac{50}{9801}\)
cho mik hỏi bước 3 chỗ \(\frac{3}{2.2}\)sai o duoi lai la\(\frac{3}{2.3}\)vay
Đặt A=\(\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{99}}\)
3A=\(\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+....+\frac{1}{3^{98}}\)
2A = 3A - A = \(\frac{1}{3}-\frac{1}{3^{98}}\)<\(\frac{1}{2}\)
=> A = \(\frac{\frac{1}{3}-\frac{1}{3^{98}}}{2}<\frac{1}{2}\)(đpcm)
Đặt B=2100-299+298-297+.....+22-2
2A=2101-2100+299-298+...+23-22
3A=2A+A=2101-2
=> A=\(\frac{2^{101}-2}{3}\)
a, A = 2 + 22 + 23 + 24 +....+ 260
A = (2 + 22) + ( 23 + 24) +...+ (259 + 260)
A = 2.(1 + 2) + 23.(1 + 2) +...+ 259.(1 + 2)
A = 2.3 + 23.3 +...+ 259.3
A = 3.( 2 + 23+...+ 259) vì 3 ⋮ 3 ⇒ A = 3.(2 + 23 +...+ 259) ⋮ 3 (đpcm)
A = 2 + 22 + 23+ 24+...+ 260
A = ( 2 + 22 + 23) + ( 24 + 25 + 26) +...+ (258 + 259 + 260)
A = 2.( 1 + 2 + 4) + 24.(1 + 2 + 4)+...+ 258.(1 + 2+4)
A = 2.7 + 24.7 +...+258.7
A = 7.(2 + 24 + ...+ 258) vì 7 ⋮ 7 ⇒ A = 7.(2 + 24+...+ 258)⋮ 7(đpcm)
A = 2 + 22 + 23 + 24 +...+ 260
A = (2 + 22 + 23 + 24) +...+( 257 + 258 + 259+ 260)
A = 2.(1 + 2 + 22 + 23) +...+ 257.(1 + 2 + 22+23)
A = 2.30 + ...+ 257. 30
A = 30.( 2 +...+ 257) vì 30 ⋮ 15 ⇒ 30.( 2 + ...+ 257) ⋮ 15 (đpcm)
\(a)2x^2-98=0\)
\(2x^2=0+98\)
\(2x^2=98\)
\(x^2=98:2\)
\(x^2=49\)
\(\rightarrow x^2=7^2\)
\(\rightarrow x=7\)
Vậy x = 7
a, 64/125