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a)\(\frac{25.4-0,5\cdot40\cdot0,2\cdot20\cdot0,25}{1+2+8+...+129+156}\)
\(=\frac{100-100}{1+2+8+...+156}\)
\(=\frac{0}{1+2+8+...+156}\)
\(=0\)
b)\(\frac{0,5\cdot40-0,5\cdot20\cdot8\cdot0,1\cdot0,25\cdot10}{128:8.16.4\left(4+52:4\right)}=\frac{20-20}{128:8.16.4.\left(4+52:4\right)}=\frac{0}{128:8.16.4.\left(4+52:4\right)}=0\)
= \(\frac{120-0.25x2x20x2x5x0.2x0.25x1-20}{A}\)
A là :
Bước 1: Số số hạng có trong dãy số : (37 - 1 ) : 4 + 1 = 9)
Bước 2: Tỏng của dãy số đó: (37 + 1) x 9 (9 là số số hạng có trong dãy số) : 2 = 342
Vậy A là 342.
Ta có: \(\frac{120-0.25x2x20x2x5x0.2x0.25x1-20}{342}\)
Bước sau bạn tự tính.
Chúc bạn suy luận đúng
\(\frac{1}{2}:0,5-\frac{1}{4}:0,25+\frac{1}{8}:0,125+25\%+\frac{3}{4}\)
\(=1-1+1+1\)
\(=0+2=2\)
lưu ý là\(\frac{1}{2}=0,5\)
mấy số kia cũng bằng nhau nên chia ra cũng = 1
tính nhẩm được mà
\(\frac{2006x125+1000}{126x2006-1006}=\frac{2006x125+1000}{125x2006-2006-1006}=\frac{2006x125+1000}{125x2006+1000}=1\)
=\(\frac{2006x125+1000}{125x2006+2006-1006}\)
=\(\frac{2006x125+1000}{125x2006+1000}\)
=1
\(\frac{2006\cdot125+1000}{126\cdot2006-1006}=\frac{2006\cdot\left(126-1\right)+1000}{126\cdot2006-1006}\)
\(=\frac{2006\cdot126-2006+1000}{126\cdot2006-1006}\)
\(=\frac{2006\cdot126-1006}{126\cdot2006-1006}\)
\(=1\)
\(\frac{2006x125+1000}{126x2006-1006}\)
\(=\frac{2006x125+1000}{\left(125+1\right)x2006-1006}\)
\(=\frac{2006x125+1000}{125x2006+1x2006-1006}\)
\(=\frac{2006x125+1000}{125x2006+2006-1006}\)
\(=\frac{2006x125+1000}{125x2006+1000}\)
\(=1\)
\(\frac{2x-4,36}{0,125}=0,25.42,9-11,7.0,25+0,25.0,8\)
\(\Leftrightarrow\frac{2x-4,36}{0,125}=0,25.\left(42,9-11.7+0,8\right)\)
\(\Leftrightarrow\frac{2x-4,36}{0,125}=0,25.32\)
\(\Leftrightarrow\frac{2x-4,36}{0,125}=8\)
\(\Leftrightarrow2x-4,36=1\)
\(\Leftrightarrow2x=5,36\)
\(\Leftrightarrow x=2,68\)
b) \(N=\frac{1}{1.5}+\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+...+\frac{1}{2005.2010}\)
\(\Leftrightarrow N=\frac{1}{5}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{2005}-\frac{1}{2010}\right)\)
\(\Leftrightarrow N=\frac{1}{5}\left(1-\frac{1}{2010}\right)\)
\(\Leftrightarrow N=\frac{1}{5}.\frac{2009}{2010}=\frac{2009}{10050}\)
Bài 1:
a)\(\frac{2\cdot x-4,36}{0,125}=0,25\cdot42,9-11,7\cdot0,25+0,25\cdot0,8\)
\(\frac{2\cdot x-4,36}{0,125}=0,25\cdot\left(42,9-11,7+0,8\right)\)
\(\frac{2\cdot x-4,36}{0,125}=0,25\cdot32\)
\(\frac{2\cdot x-4,36}{0,125}=8\)
\(2\cdot x-4,36=8\cdot0,125\)
\(2\cdot x-4,36=1\)
\(2\cdot x=1+4,36\)
\(2\cdot x=5,36\)
\(x=\frac{5,36}{2}=2,68\)
b) \(N=\frac{1}{1\cdot5}+\frac{1}{5\cdot10}+\frac{1}{10\cdot15}+\frac{1}{15\cdot20}+...+\frac{1}{2005\cdot2010}\)
\(4N=\frac{4}{1\cdot5}+\frac{4}{5\cdot10}+\frac{4}{10\cdot15}+\frac{4}{15\cdot20}+...+\frac{4}{2005\cdot2010}\)
\(4N=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{2005}-\frac{1}{2010}\)
\(4N=1-\frac{1}{2010}=\frac{2009}{2010}\)
\(N=\frac{2009}{2010}\div4=\frac{2009}{8040}\)
Bài 2:
a) ( x + 5,2 ) : 3,2 = 4,7 ( dư 0,5 )
\(x+5,2=4,7\cdot3,2+0,5\)
\(x+5,2=15,54\)
\(x=15,54-5,2=10,34\)
b)\(A=\frac{4047991-2010\cdot2009}{4050000-2011\cdot2009}\)
\(A=\frac{4047991-2010\cdot2009}{4050000-2009-2010\cdot2009}\)
\(A=\frac{4047991-2010\cdot2009}{4047991-2010\cdot2009}=1\)
Bài 3:
a) \(104,5\cdot x-14,1\cdot x+9,6\cdot x=25\)
\(x\cdot\left(104,5-14,1+9,6\right)=25\)
\(x\cdot100=25\)
\(x=\frac{25}{100}=\frac{1}{4}=0,25\)
b) \(T=\frac{2009\cdot2010+2000}{2011\cdot2010-2020}\)
\(T=\frac{2009\cdot2010+2000}{2009\cdot2010+4020-2020}\)
\(T=\frac{2009\cdot2010+2000}{2009\cdot2010+2000}=1\)
DẤU . ĐÓ LÀ DẤU NHÂN NHA