Bài 1 : Tìm X
a) \(\frac{2xX-4,36}{0,125}=\)0,25 x 42,9 - 11,7 x 0,25 + 0,25 x 0,8
b) 104,5 x X -14,1 x X + 9,6 x X = 25
c) \(\frac{5xX}{x-13}=\)49 x 0,5 - 7 x 0,5 + 0,25
d) 13,44 x X - ( 1,6 + x ) x 0,5 -3,7 x ( x - 0,9 ) = 13, 618
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\(\frac{2x-4,36}{0,125}=0,25.42,9-11,7.0,25+0,25.0,8\)
\(\Leftrightarrow\frac{2x-4,36}{0,125}=0,25.\left(42,9-11.7+0,8\right)\)
\(\Leftrightarrow\frac{2x-4,36}{0,125}=0,25.32\)
\(\Leftrightarrow\frac{2x-4,36}{0,125}=8\)
\(\Leftrightarrow2x-4,36=1\)
\(\Leftrightarrow2x=5,36\)
\(\Leftrightarrow x=2,68\)
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) \(T=\frac{2009\cdot2010+2000}{2011\cdot2010-2020}\)
\(T=\frac{2009\cdot2010+2000}{2009\cdot2010+4020-2020}\)
\(T=\frac{2009\cdot2010+2000}{2009\cdot2010+2000}=1\)
\(2x-\frac{4.36}{0.125}=0.15\times42.9-11.7\times0,25\times0.8\)
\(2x-\frac{872}{25}=\frac{819}{200}\)
\(2x=\frac{819}{200}+\frac{872}{25}=\frac{1559}{40}\)
\(x=\frac{1559}{40}\times\frac{1}{2}=\frac{1559}{80}\)
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
Tìm x,biết:
a)x*96+x:0,25=48,32
b)x:0,25+x+x*95=73,74
c)x:0,5+x:0,25+x+x*93=35,36
d)x:0,125+x+x*91=78,56
a: =>100x=48,32
hay x=0,4832
b: =>4x+x+95x=73,74
=>100x=73,74
hay x=0,7374
c: =>2x+4x+x+93x=35,36
=>100x=35,36
hay x=0,3536
a)Xx96+x:0,25=48,32
Xx96+Xx4 =48,32
X x(96+4) =48,32
Xx 100 =48,32
X =48,32:100
X = 0,4832
b)x:0,25+x+Xx95 =73,74
Xx4 +Xx1+Xx95= 73,74
X x(4+1+95) =73,74
X x 100 =73,74
X =73,74:100
X = 0,7374
c)x:0,5+x:0,25+x+Xx93=35,36
Xx2 + Xx4+Xx1+Xx93=35,36
Xx(2+4+1+93) = 35,36
Xx 100 = 35,36
X =35,36:100
X =0,3536
d)X:0,125+X+Xx91=78,56
Xx 8+Xx1+Xx91=78,56
Xx(8+1+91) =78,56
Xx 100 =78,56
X =78,56:100
X =0,7856
#)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~#
#)Next :
b) 104,5 x X - 14,1 x X + 9,6 x X = 25
( 104,5 - 14,1 + 9,6 ) x X = 25
100 x X = 25
X = 25 : 100
X = 0,25
#~Will~be~Pens~#