Tính nhanh:
B=\(2\dfrac{1}{3154}\cdot\dfrac{3}{6517}-\dfrac{3}{3154}\cdot\dfrac{6516}{6517}+\dfrac{6}{1577}-\dfrac{6}{3154\cdot6517}\)
Những câu hỏi liên quan
Tinhs nhanh B=2\(\frac{1}{3154}\).\(\frac{3}{6517}-\frac{3}{3154}.3\frac{6516}{6517}+\frac{6}{1577}-\frac{6}{3154.6517}\)
Bài 1.tính giá trị của các biểu thức sau bằng cách thay số bởi chữ hợp lí
A2dfrac{1}{3179}-dfrac{3}{1111}-dfrac{3178}{3179}*dfrac{1}{1111}-dfrac{4}{3179cdot1111}
B2dfrac{1}{3150}*dfrac{3}{6517}-dfrac{1}{1050}*3dfrac{6516}{6517}+dfrac{4}{1050}-dfrac{6}{3150cdot3897}
Cdfrac{1}{4587}*7dfrac{1}{3897}-3dfrac{4586}{4587}*dfrac{2}{3897}-dfrac{7}{4587}-dfrac{3}{4587cdot3897}
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Bài 1.tính giá trị của các biểu thức sau bằng cách thay số bởi chữ hợp lí
A=\(2\dfrac{1}{3179}\)-\(\dfrac{3}{1111}\)-\(\dfrac{3178}{3179}\)*\(\dfrac{1}{1111}\)-\(\dfrac{4}{3179\cdot1111}\)
B=\(2\dfrac{1}{3150}\)*\(\dfrac{3}{6517}\)-\(\dfrac{1}{1050}\)*\(3\dfrac{6516}{6517}\)+\(\dfrac{4}{1050}\)-\(\dfrac{6}{3150\cdot3897}\)
C=\(\dfrac{1}{4587}\)*\(7\dfrac{1}{3897}\)-\(3\dfrac{4586}{4587}\)*\(\dfrac{2}{3897}\)-\(\dfrac{7}{4587}\)-\(\dfrac{3}{4587\cdot3897}\)
Tính hợp lý:
\(a.\dfrac{3}{17}+\dfrac{-5}{13}+\dfrac{-18}{35}+\dfrac{14}{17}+\dfrac{17}{-35}+\dfrac{-8}{13}\)
\(b.\dfrac{-3}{8}\cdot\dfrac{1}{6}+\dfrac{3}{-8}\cdot\dfrac{5}{6}+\dfrac{-10}{16}\)
\(c.\dfrac{-4}{11}\cdot\dfrac{5}{15}\cdot\dfrac{11}{-4}\)
a: \(=\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{-5}{13}-\dfrac{8}{13}\right)+\left(\dfrac{-18}{35}-\dfrac{17}{35}\right)\)
=1-1-1
=-1
b: \(=\dfrac{-3}{8}\left(\dfrac{1}{6}+\dfrac{5}{6}\right)+\dfrac{-5}{8}=\dfrac{-3}{8}-\dfrac{5}{8}=-1\)
c: \(=\dfrac{4}{4}\cdot\dfrac{5}{15}\cdot\dfrac{11}{11}=\dfrac{1}{3}\)
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a) \(=\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(-\dfrac{5}{13}+-\dfrac{8}{13}\right)+\left(-\dfrac{18}{35}+-\dfrac{17}{35}\right)=1+-1+-1=-1\)
b) \(=-\dfrac{3}{8}\cdot\left(\dfrac{1}{6}+\dfrac{5}{6}\right)-\dfrac{10}{16}=-\dfrac{3}{8}-\dfrac{10}{16}=-1\)
c) \(=\left(-\dfrac{4}{11}\cdot-\dfrac{11}{4}\right)\cdot\dfrac{5}{15}=1\cdot\dfrac{1}{3}=\dfrac{1}{3}\)
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a)\(=\left(-\dfrac{5}{13}+\dfrac{-8}{13}\right)+\left(-\dfrac{18}{35}-\dfrac{17}{35}\right)+\left(\dfrac{3}{14}+\dfrac{14}{17}\right)=-1-1+1=-1\)
b)\(=\dfrac{-3}{8}.\left(\dfrac{1}{6}+\dfrac{5}{6}\right)-\dfrac{10}{16}=-\dfrac{3}{8}.1-\dfrac{10}{16}=-\dfrac{6}{16}-\dfrac{10}{16}=-\dfrac{16}{16}=-1\)
c)\(\dfrac{-4.5.11}{11.5.3.-4}=\dfrac{1}{3}\)
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Tính giá trị của các biểu thức sau 1) A1+2+2^2+...+2^{2015} 2) Bleft(dfrac{1}{4}-1right)cdotleft(dfrac{1}{9}-1right)cdotleft(dfrac{1}{16}-1right)cdotcdotcdotcdotcdotleft(dfrac{1}{400}-1right) 3) Cleft(dfrac{1}{4cdot9}+dfrac{1}{9cdot14}+dfrac{1}{14cdot19}+...+dfrac{1}{44cdot49}right)cdotdfrac{1-3-5-7-...-49}{89} 4) Ddfrac{2^{12}cdot3^5-4^6cdot9^2}{left(2^2cdot3right)^6+8^4cdot3^5}-dfrac{5^{10}cdot7^3-25^5cdot49^2}{left(125cdot7right)^3+5^9cdot14^3} 5) Edfrac{dfrac{1}{2003}+dfrac{1}{2004}-dfrac{1}...
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Tính giá trị của các biểu thức sau 1) \(A=1+2+2^2+...+2^{2015}\) 2) \(B=\left(\dfrac{1}{4}-1\right)\cdot\left(\dfrac{1}{9}-1\right)\cdot\left(\dfrac{1}{16}-1\right)\cdot\cdot\cdot\cdot\cdot\left(\dfrac{1}{400}-1\right)\) 3) \(C=\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\) 4) \(D=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\) 5) \(E=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{3}{2004}}\) 6) Cho 13+23+...+103=3025 Tính S= 23+43+63+...+203
Tìm số thích hợp cho ?:
a) \(\dfrac{-2}{3}\cdot\dfrac{?}{4}=\dfrac{1}{2};\)
b) \(\dfrac{?}{3}\cdot\dfrac{5}{8}=\dfrac{-5}{12};\)
c) \(\dfrac{5}{6}\cdot\dfrac{3}{?}=\dfrac{1}{4}.\)
\(a.\)
\(-\dfrac{2}{3}\cdot\dfrac{?}{4}=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{?}{4}=\dfrac{1}{2}:-\dfrac{2}{3}=\dfrac{1}{2}\cdot-\dfrac{3}{2}=-\dfrac{3}{4}\)
\(\Leftrightarrow?=-3\)
\(b.\)
\(\dfrac{?}{3}\cdot\dfrac{5}{8}=-\dfrac{5}{12}\)
\(\Leftrightarrow\dfrac{?}{3}=\dfrac{-5}{12}:\dfrac{5}{8}=\dfrac{-5}{12}\cdot\dfrac{8}{5}=-\dfrac{2}{3}\)
\(\Leftrightarrow?=-2\)
\(c.\)
\(\dfrac{5}{6}\cdot\dfrac{3}{?}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{3}{?}=\dfrac{1}{4}:\dfrac{5}{6}=\dfrac{1}{4}\cdot\dfrac{6}{5}=\dfrac{3}{10}\)
\(\Leftrightarrow?=10\)
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Mk gọi ? = x nha
a) \(\dfrac{-2}{3}.\dfrac{x}{4}=\dfrac{1}{2}\)
\(\dfrac{x}{4}=\dfrac{1}{2}:\dfrac{-2}{3}\)
\(\dfrac{x}{4}=\dfrac{-3}{4}\)
⇒x=-3
b)\(\dfrac{x}{3}.\dfrac{5}{8}=\dfrac{-5}{12}\)
\(\dfrac{x}{3}=\dfrac{-5}{12}:\dfrac{5}{8}\)
\(\dfrac{x}{3}=\dfrac{-2}{3}\)
⇒x=-2
c)\(\dfrac{5}{6}.\dfrac{3}{x}=\dfrac{1}{4}\)
\(\dfrac{3}{x}=\dfrac{1}{4}:\dfrac{5}{6}\)
\(\dfrac{3}{x}=\dfrac{3}{10}\)
⇒x=10
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a) \(\dfrac{\left(x+\dfrac{3}{4}\right)\cdot\dfrac{7}{2}-\dfrac{1}{6}}{-\left(\dfrac{4}{5}+\dfrac{1}{3}\right)\cdot\dfrac{1}{2}+1}=2\dfrac{33}{52}\)
b)\(\dfrac{\left(5-\dfrac{2}{7}\right)\cdot\dfrac{7}{9}\cdot\dfrac{3}{5}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=5\dfrac{5}{21}\)
Giải:
a) \(\dfrac{\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}}{-\left(\dfrac{4}{5}+\dfrac{1}{3}\right).\dfrac{1}{2}+1}=2\dfrac{33}{52}\)
\(\Leftrightarrow\dfrac{\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}}{-\dfrac{17}{15}.\dfrac{1}{2}+1}=\dfrac{137}{52}\)
\(\Leftrightarrow\dfrac{\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}}{\dfrac{13}{30}}=\dfrac{137}{52}\)
\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}=\dfrac{137}{52}.\dfrac{13}{30}\)
\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}=\dfrac{137}{120}\)
\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}=\dfrac{137}{120}+\dfrac{1}{6}\)
\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}=\dfrac{157}{120}\)
\(\Leftrightarrow x+\dfrac{3}{4}=\dfrac{157}{120}:\dfrac{7}{2}\)
\(\Leftrightarrow x+\dfrac{3}{4}=\dfrac{157}{420}\)
\(\Leftrightarrow x=\dfrac{157}{420}-\dfrac{3}{4}\)
\(\Leftrightarrow x=-\dfrac{79}{210}\)
Vậy \(x=-\dfrac{79}{210}\).
b) \(\dfrac{\left(5-\dfrac{2}{7}\right).\dfrac{7}{9}.\dfrac{3}{5}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=5\dfrac{5}{21}\)
\(\Leftrightarrow\dfrac{\left(5-\dfrac{2}{7}\right).\dfrac{7}{15}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=\dfrac{110}{21}\)
\(\Leftrightarrow\dfrac{\dfrac{33}{7}.\dfrac{7}{15}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=\dfrac{110}{21}\)
\(\Leftrightarrow\dfrac{\dfrac{11}{5}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=\dfrac{110}{21}\)
\(\Leftrightarrow\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}=\dfrac{11}{5}:\dfrac{110}{21}\)
\(\Leftrightarrow\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}=\dfrac{21}{50}\)
\(\Leftrightarrow3x-\dfrac{5}{6}=\dfrac{21}{50}.\dfrac{1}{7}\)
\(\Leftrightarrow3x-\dfrac{5}{6}=\dfrac{3}{50}\)
\(\Leftrightarrow3x=\dfrac{3}{50}+\dfrac{5}{6}\)
\(\Leftrightarrow3x=\dfrac{67}{75}\)
\(\Leftrightarrow x=\dfrac{67}{75}:3\)
\(\Leftrightarrow x=\dfrac{67}{225}\)
Vậy \(x=\dfrac{67}{225}\).
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CÁC BẠN GIÚP MK NHA!!!
NGÀY MAI MK NỘP BÀI RỒI
AI TRẢ LỜI NHANH NHẤT
CHÍNH XÁC NHẤT VÀ RÕ RÀNG
THÌ MK TICK CHO NHA!!!
NHỚ TRẢ LỜI NHANH GIÙM MK NHA
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m.n giúp mk ik nếu đúng mk sẻ giúp m.n trả ơn mờ nếu bn nghĩ bn trong hoàn cảnh này bn hiểu đc cảm giác của mk nếu bn là bn của mk thì xinh hãy giúp mk ik mờ
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\(\left[\dfrac{3}{7}\cdot\dfrac{4}{15}+\dfrac{1}{3}\cdot\left(9^{15}\right)\right]^0\cdot\dfrac{1}{3}\cdot\dfrac{6^8}{12^4}\). Tính
[3/7.4/15+1/3.(9^15)]^0.1/3.6^8/12^4
= 1.1/3.(2.3)^8/(3.4)^4
= 1/3.2^8.3^8/3^4.4^4
= 1/3.2^8.3^8/3^4.2^8
= 1/3.3^8/3^4
= 1/3.3^4=27
(dấu . là nhân nha)
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Tính giá trị của các biểu thức sau1) A1+2+2^2+...+2^{2015}2) Bleft(dfrac{1}{4}-1right)cdotleft(dfrac{1}{9}-1right)cdotleft(dfrac{1}{16}-1right)cdotcdotcdotcdotcdotleft(dfrac{1}{400}-1right)3) Cleft(dfrac{1}{4cdot9}+dfrac{1}{9cdot14}+dfrac{1}{14cdot19}+...+dfrac{1}{44cdot49}right)cdotdfrac{1-3-5-7-...-49}{89}4) Ddfrac{2^{12}cdot3^5-4^6cdot9^2}{left(2^2cdot3right)^6+8^4cdot3^5}-dfrac{5^{10}cdot7^3-25^5cdot49^2}{left(125cdot7right)^3+5^9cdot14^3}5) Edfrac{dfrac{1}{2003}+dfrac{1}{2004}-dfrac{1}{2005...
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Tính giá trị của các biểu thức sau
1) \(A=1+2+2^2+...+2^{2015}\)
2) \(B=\left(\dfrac{1}{4}-1\right)\cdot\left(\dfrac{1}{9}-1\right)\cdot\left(\dfrac{1}{16}-1\right)\cdot\cdot\cdot\cdot\cdot\left(\dfrac{1}{400}-1\right)\)
3) \(C=\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)
4) \(D=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)
5) \(E=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{3}{2004}}\)
6) Cho 13+23+...+103=3025
Tính S= 23+43+63+...+203
Cho M = \(\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot...\cdot\dfrac{99}{100}\) ; N = \(\dfrac{2}{3}\cdot\dfrac{4}{5}\cdot\dfrac{6}{7}\cdot...\cdot\dfrac{100}{101}\).
Tính M \(\cdot\) N.
Ta có : M . N = \(\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot...\cdot\dfrac{99}{100}\cdot\dfrac{2}{3}\cdot\dfrac{4}{5}\cdot\dfrac{6}{7}\cdot...\cdot\dfrac{100}{101}\)
= \(\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot\dfrac{6}{7}\cdot...\cdot\dfrac{99}{100}\cdot\dfrac{100}{101}\)
= \(\dfrac{1}{101}\)
Vậy M . N = \(\dfrac{1}{101}\)
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