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#)Giải :
a) \(\frac{2xX-4,36}{0,125}=\)0,25 x 42,9 - 11,7 x 0,25 + 0,25 x 0,8
\(\frac{2xX-4,36}{0,125}=\)0,25 x ( 42,9 - 11,7 + 0,8 )
\(\frac{2xX-4,36}{0,125}=\)0,25 x 32
\(\frac{2xX-4,36}{0,125}=8\)
\(2xX-4,36=8x0,125\)
\(2xX-4,36=1\)
\(2xX=1+4,36\)
\(2xX=5,36\)
\(X=5,36:2\)
\(X=2,68\)
#~Will~be~Pens~#
\(\frac{2x-4,36}{0,125}=0,25.42,9-11,7.0,25+0,25.0,8\)
\(\Leftrightarrow\frac{2x-4,36}{0,125}=0,25.\left(42,9-11.7+0,8\right)\)
\(\Leftrightarrow\frac{2x-4,36}{0,125}=0,25.32\)
\(\Leftrightarrow\frac{2x-4,36}{0,125}=8\)
\(\Leftrightarrow2x-4,36=1\)
\(\Leftrightarrow2x=5,36\)
\(\Leftrightarrow x=2,68\)
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) \(T=\frac{2009\cdot2010+2000}{2011\cdot2010-2020}\)
\(T=\frac{2009\cdot2010+2000}{2009\cdot2010+4020-2020}\)
\(T=\frac{2009\cdot2010+2000}{2009\cdot2010+2000}=1\)
a: \(=27\left(77+15\right)+27\cdot8\)
\(=27\cdot92+27\cdot8=27\cdot100=2700\)
b: \(=75\cdot89+75\cdot9+2\cdot75=75\left(89+9+2\right)=75\cdot100=7500\)
c: \(=0.25\left(42.9-11.7+0.8\right)=0.25\cdot32=8\)
d: 738 phút=12,3 giờ
\(N=\frac{1}{1.5}+\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+...+\frac{1}{2005.2010}\)
\(N=\frac{1}{5}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{2005}-\frac{1}{2010}\right)\)
\(N=\frac{1}{5}.\left(1-\frac{1}{2010}\right)\)
\(N=\frac{1}{5}.\frac{2009}{2010}\)
\(N=\frac{2009}{10050}\)
a, x nhân 5 + x nhân 4 + x nhân 8 + x = 21,6
x nhân (5+4+8+1) = 21,6
x nhân 18 = 21,6
x = 21,6 : 18 = 1,2
B, = 1350 + 150 = 1500
k mk nha
a)x:0,2+x:0,25+x:0,125+x=21,6 x.5+x.4+x.8+x=21,6 x.(5+4+8+1)=21,6 x.18=21,6 x =21,6:18
x=1,2
b)0,15.9000+10.15=15.90+10.15
=15.(90+10)
=15.100
=1500
Bài 1:
Khoảng cách giữa các số hạng liền kề nhau trong tổng là : 0,5 - 0,2 = 0,3
Số số hạng là : ( 14,9 - 0,2 ) / 0,3 + 1 = 50 ( số hạng )
Tổng là : ( 14,9 + 0,2 ) * 50 / 2 = 377,5
Đ/s : 377,5
Bài 2:
387 * 0,25 - ( x / 4 + x / 0,125 ) / 0,5 = 14,25
96,75 - ( 0,25 * x + 8 * x ) * 2 = 14,25
8,25 * x * 2 = 96,75 - 14,25
16,5 * x = 82,5
x = 82,5/16,5 = 5
\(2x-\frac{4.36}{0.125}=0.15\times42.9-11.7\times0,25\times0.8\)
\(2x-\frac{872}{25}=\frac{819}{200}\)
\(2x=\frac{819}{200}+\frac{872}{25}=\frac{1559}{40}\)
\(x=\frac{1559}{40}\times\frac{1}{2}=\frac{1559}{80}\)