\(A=\frac{3}{1\times3}+\frac{3}{3\times5}+...+\frac{3}{44\times51}\)
Giúp mình với, mk cần gấp.
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Giúp mình với!!!!!
Tính:
G=\(\frac{1}{1\times3\times5}+\frac{1}{5\times7\times9}+\frac{1}{9\times11\times13}+...+\frac{1}{49\times51\times53}\)
\(\frac{1}{1x3x5}+\frac{1}{5x7x9}+\frac{1}{9x11x13}+.....+\frac{1}{49x51x53}=\)
\(1-\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}-\frac{1}{9}+.....+\frac{1}{49}-\frac{1}{51}-\frac{1}{53}=\)
\(1-\frac{1}{3}-\frac{1}{7}-....-\frac{1}{51}-\frac{1}{53}=\)
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Tính nhanh : \(\frac{1}{1\times3\times5}+\frac{1}{3\times5\times7}+\frac{1}{5\times7\times9}+.....+\frac{1}{45\times47\times49}\) Help me . Mình cần gấp , HELP
Đặt \(A=\frac{1}{1.2.3}+\frac{1}{3.5.7}+...+\frac{1}{45.47.49}\)
\(\Rightarrow4A=\frac{4}{1.3.5}+\frac{4}{3.5.7}+...+\frac{4}{45.47.49}\)
\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{45.47}-\frac{1}{47.49}\)
\(=\frac{1}{3}-\frac{1}{47.49}\)
\(\Rightarrow A=\frac{\frac{1}{3}-\frac{1}{47.49}}{4}=\frac{575}{6909}\)
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a.\(\frac{7^3\times5^8}{49\times25^4}\)
b.\(\frac{3^9\times25\times5^3}{15\times625\times3^8}\)
c.\(\frac{2^{50}\times3^{61}+2^{90}\times3^{16}}{2^{51}\times3^{61}+2^{91}\times3^{16}}\)
d.\((\frac{2}{5}-\frac{1}{2})^2+(\frac{1}{2}+\frac{3}{5})^2\)
a) \(\frac{7^3.5^8}{49.25^4}=\frac{7^3.5^8}{7^2.5^8}=7\)
b) \(\frac{3^9.25.5^3}{15.625.3^8}=\frac{3^9.5^2.5^3}{3.5.5^4.3^8}=\frac{3^9.5^5}{3^9.5^5}=1\)
c) \(\frac{2^{50}.3^{61}+2^{90}.3^{16}}{2^{51}.3^{61}+2^{91}.3^{16}}=\frac{2^{50}.3^{16}\left(3^{45}+2^{40}\right)}{2^{51}.3^{16}\left(3^{45}+2^{40}\right)}=\frac{1}{2}\)
d) \(\left(\frac{2}{5}-\frac{1}{2}\right)^2+\left(\frac{1}{2}+\frac{3}{5}\right)^2\)
\(=\left(\frac{-1}{10}\right)^2+\left(\frac{11}{10}\right)^2\)
\(=\frac{1}{100}+\frac{121}{100}=\frac{122}{100}=\frac{61}{50}\)
a) \(\frac{7^3.5^8}{49.25^4}=\frac{7^3.5^8}{7^2.\left(5^2\right)^4}=\frac{7^3.5^8}{7^2.5^8}=7\)
b) \(\frac{3^9.25.5^3}{15.625.3^8}=\frac{3^9.5^2.5^3}{5.3.5^4.3^8}=\frac{3^9.5^5}{5^5.3^9}=1\)
c) \(\frac{2^{50}.3^{61}+2^{90}.3^{16}}{2^{51}.3^{61}+2^{91}.3^{16}}=\frac{2^{50}.3^{16}\left(3^{45}+2^{40}\right)}{2^{51}.3^{16}\left(3^{45}+2^{40}\right)}=\frac{1}{2}\)
d) \(\left(\frac{2}{5}-\frac{1}{2}\right)^2+\left(\frac{1}{2}+\frac{3}{5}\right)^2=\left(-\frac{1}{10}\right)^2+\left(\frac{11}{10}\right)^2=\frac{1}{100}+\frac{121}{100}=\frac{122}{100}=1,22\)
Xem thêm câu trả lời
a,tìm x :( 1/2- 1/3 - 1/6 ) x (1/2 + 2/3 + 3/4 +4/5 + .......+ 2019 /2020 ) + 3/4 : x = 9/10
b,\(X+\left(\frac{3}{1\times3}+\frac{3}{3\times5}+\frac{3}{5\times7}+....+\frac{3}{13\times15}\right)=2\frac{1}{5}\)\(2\frac{1}{5}\)
MK ĐANG CẦN RẤT GẤP BẠN NÀO NHANH MK TICK CHO NHỚ LÀ PHẢI CÓ CÁCH GIẢI NHÉ!
Câu b
Ta có :x + 3 /1.3 +3/3.5 + 3/5.7+...+3/13.15=2 1/5
X + 2/3.(1-1/3+1/3-1/5+1/5-1/7+...+1/13-1/15)1=11/5
X+2/3.(1-1/15)=11/5
X+ 2/3.14/15=11/5
X + 28/45=11/5
X = 11/5 -28/45
X=71/45
Câu a gợi ý
1/2-1/3/1/6=0
1/2- 1/3 - 1/6 ) x (1/2 + 2/3 + 3/4 +4/5 + .......+ 2019 /2020 ) =0
3/4:x=9/10
X = 3/4:9/10
X = 5/6
Câu b
Ta có : x + 3 /1.3 +3/3.5 + 3/5.7+...+3/13.15 = 2 1/5
X + 2/3.(1-1/3+1/3-1/5+1/5-1/7+...+1/13-1/15)1 = 11/5
X + 2/3.(1-1/15) = 11/5
X + 2/3.14/15 = 11/5
X + 28/45 = 11/5
X = 11/5 - 28/45
X = 71/45
đúng 100% nhé
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A =\(\frac{1}{1\times2\times3}\)+\(\frac{1}{2\times3\times4}\)+\(\frac{1}{3\times4\times5}\)+........+\(\frac{1}{36\times37\times38}\)+\(\frac{1}{37\times38\times39}\) GIÚP MK VỚI MK CẦN GẤP
Vũ Thị Yến Đan
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tham khảo bài của bạn Hoàng Xuân Ngân nhé
Tính N =\(\frac{2^2}{1\times3}\times\frac{3^2}{2\times4}\times...\times\frac{50^2}{49\times51}\)
Các bạn giúp mk, mk cần gấp!
\(\frac{4}{1\times3}+\frac{4}{3\times5}+\frac{4}{5\times7}+......+\frac{4}{99\times101}\)
\(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{99.101}\)
\(=2.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=2.\left(1-\frac{1}{101}\right)\)
\(=2.\frac{100}{101}=\frac{200}{101}\)
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Đặt \(A=\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+..+\frac{4}{99.101}\)
\(A=2.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(A=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(A=2.\left(1-\frac{1}{101}\right)\)
\(A=\frac{2.100}{101}=\frac{200}{101}\)
Ủng hộ mk nha !!! ^_^
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Ta gọi tổng trên là A
Ta có: A = 4/1x3+4/3x5 + 4/5x7+...+4/99x101
=>A x 2/4 = 2/1x3+2/3x5+2/5x7+...+2/99x101
=1-1/2+1/2-1/3+1/3-1/4+....+1/99+1/101
=1-1/101=100/101
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\(\frac{1\times2}{2\times3}+\frac{2\times3}{3\times4}+\frac{3\times4}{4\times5}+...+\frac{98\times99}{99\times100}\)
\(=\frac{1.2}{99.100}\)
\(=\frac{2}{9900}=\frac{1}{4950}\)
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Tính:
\(A=\frac{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+.....\frac{1}{97}+\frac{1}{99}}{\frac{1}{1\times99}+\frac{1}{3\times97}+\frac{1}{5\times95}+\frac{1}{97\times3}+\frac{1}{99\times1}}\)
Có khó không giúp mình với chiều nay cần gấp
Ta xét riêng tử số:
\(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+......+\frac{1}{97}+\frac{1}{99}\)
\(=\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+\left(\frac{1}{5}+\frac{1}{95}\right)+......+\left(\frac{1}{49}+\frac{1}{51}\right)\)
\(=\frac{100}{1\times99}+\frac{100}{3\times97}+\frac{100}{5\times95}+......+\frac{100}{49\times51}\)
\(=100\times\left(\frac{1}{1\times99}+\frac{1}{3\times97}+\frac{1}{5\times95}+......+\frac{1}{49\times51}\right)\)
Bây giờ xét đến mẫu số:
\(\frac{1}{1\times99}+\frac{1}{3\times97}+\frac{1}{5\times95}+......+\frac{1}{97\times3}+\frac{1}{99\times1}\)
\(=\frac{2}{1\times99}+\frac{2}{3\times97}+\frac{2}{5\times95}+......+\frac{2}{49\times51}\)
\(=2\times\left(\frac{1}{1\times99}+\frac{1}{3\times97}+\frac{1}{5\times95}+......+\frac{1}{49\times51}\right)\)
Vậy giá trị của biểu thức là: \(\frac{100}{2}=50\)
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