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( 285 - x - 0,15 ) : 0,25 = 60
284,85 - x = 15
x = 269,85
2010 \(x\)x =4062220
x = 2021,004975
5 x y -1952 =553
5 x y = 553 + 1952
5 x y = 2505
y = 2505 : 5
y = 501
y x 2011 - y = 2011 x 2009 + 2011 x 1
y x 2011 - y x 1 =2011 x 2009 + 2011x1
y x ( 2011 - 1) = 2011 x ( 2009 +1)
y x 2010 = 2011 x 2010
nhìn vào phép tính trên ta thấy cả 2 vế đều có 1 thừa số chung đó là 2010
mà 2 vế bằng nhau
nên y = 2011
bn nào đi qua thấy đúng cho mk 1 k nha. yêu mọi người moa
1
a)
5 x X - 1952 = 2500 - 1947
5 x X - 1952 = 553
5 x X = 553 + 1952
5 x X = 2505
X = 2505 : 5
X = 501
b)
X x 2011 - X = 2011 x 2009 + 2011
X x 2011 - X x 1 = 2011 x 2009 + 2011 x 1
X x ( 2011 + 1 ) = 2011 x ( 2009 + 1 )
X x 2012 = 2011 x 2010
X x 2012 = 4042110
X = 4042110 : 2012
X = \(_{\frac{2021055_{ }}{1006}}\)
a)(3/2-0,5)/x=7/2+1/4
(3/2-1/2)/x=14/4+1/4
1/x=15/4
x=1:15/4
x=4/15
b)(x*0,25+2010)*2011=(53+2010)*(2012-1)
(x*0,25+2010)*2011=2063*2011
=>0,25x+2010=2063
0,25x=2063-2010
0,25x=53
x=53/0,25
x=212
\(\frac{2x-4,36}{0,125}=0,25.42,9-11,7.0,25+0,25.0,8\)
\(\Leftrightarrow\frac{2x-4,36}{0,125}=0,25.\left(42,9-11.7+0,8\right)\)
\(\Leftrightarrow\frac{2x-4,36}{0,125}=0,25.32\)
\(\Leftrightarrow\frac{2x-4,36}{0,125}=8\)
\(\Leftrightarrow2x-4,36=1\)
\(\Leftrightarrow2x=5,36\)
\(\Leftrightarrow x=2,68\)
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) \(T=\frac{2009\cdot2010+2000}{2011\cdot2010-2020}\)
\(T=\frac{2009\cdot2010+2000}{2009\cdot2010+4020-2020}\)
\(T=\frac{2009\cdot2010+2000}{2009\cdot2010+2000}=1\)
Tìm x biết:
a) (15 × 19 – x – 0,15):0,25 = 15 : 0,25
b)\(2\frac{2}{3}-x=\frac{1}{3}\cdot\frac{6}{2}\)
a) (15 x 19 - x - 0,15) : 0,25 = 15 : 0,15
(285 - x - 0,15) : 0,25 = 60
(285 - x - 0,25) = 60 : 0,25
( 285 - x - 0,25) = 15
285 - x = 15 + 0,25
285 - x = 15,25
x = 285 - 15,25
x = 269,75
( x x 0,25 + 2000 ) x 2011 = ( 40 + 2000 ) x 2011
Vì cả hai vế đều có x 2011 nên ta khử cả hai x 2011
x x 0,25 + 2000 = 40 + 2000
Vì cả hai đều có + 2000 nên ta khử cả hai + 2000
x x 0,25 = 40
x = 40 : 0,25
x = 160