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\(19,96+4,19-24,15:\left(x:\frac{1}{4}-\frac{1}{4}\right)=23,15\)
\(24,15-24,15:\left(x:\frac{1}{4}-\frac{1}{4}\right)=23,15\)
\(24,15:\left(x:\frac{1}{4}-\frac{1}{4}\right)=1\)
\(x:\frac{1}{4}-\frac{1}{4}=24,15\)
\(4x=24,4\)
\(x=6,1\)
\(\text{Bài này cx đơn giản thôi!}\)
\(4.19-24.15:\left(x:\frac{1}{4}-\frac{1}{4}\right)=23.15-19.96\)
\(4.19-24.15:\left(x:\frac{1}{4}-\frac{1}{4}\right)=3.19\)
\(24.15:\left(x:\frac{1}{4}-\frac{1}{4}\right)=4.19-3.19\)
\(24.15:\left(x:\frac{1}{4}-\frac{1}{4}\right)=1\)
\(x:\frac{1}{4}-\frac{1}{4}=24.15:1\)
\(x:\frac{1}{4}-\frac{1}{4}=24.15\)
\(x:\frac{1}{4}=24.15+\frac{1}{4}\)
\(x:\frac{1}{4}=24.4\)
\(x=24.4.\frac{1}{4}\)
\(x=6.1\)
a) 187 - {[497 - ( 8 x X + 11) : X] : 3 - 78} = 150
=> {[497 - ( 8 x X + 11) : X] : 3 - 78} = 187 - 150
=> {[497 - (8 x X + 11) : X] : 3 - 78} = 37
=> [497 - (8 x X +11): X ] : 3 - 78 = 37
=> [497 - (8 x X + 11) : X] : 3 = 115
=> 497 - ( 8 x X + 11) : X = 345
=> (8 x X + 11) : X = 497 - 345 = 152
=> 8X + 11 = 152X
=> 152X - 8X = 11
=> 144X = 11
=> X = 11/144
b) 19,96 + 4,19 - 24,15 : \(\left(x:\frac{1}{4}-\frac{1}{4}\right)=23,15\)
=> 19,96 + 4,19 - 24,15 : \(\left(x\cdot4-\frac{1}{4}\right)=23,15\)
=> 24,15 - 24,15 : \(\left(x\cdot4-\frac{1}{4}\right)\)= 23,15
=> 24,15 : \(\left(x\cdot4-\frac{1}{4}\right)\)= 1
=> \(x\cdot4-\frac{1}{4}=24,15\)
=> \(x\cdot4=24,15+\frac{1}{4}=24,4\)
=> x = 24,4 : 4 = 6,1
Còn câu c tương tự
*19,96+4,19-24,15:(1/4-1/4) *252/x=84/97
=19,96+4,19-24,15:0 x=252x97:84
=19,96+4,19-0 x=291
=24,15-0 * x-2/255=114/153
=24,15 x-2/255=38/51
x=38/51+2/255
x=64/85
a ) ko rõ đề
b )\(\frac{252}{x}=\frac{84}{97}\)
=> 252 . 97 = x . 84
=> x . 84 = 24 444
x = 24 444 : 84
x = 291
Vậy x = 291
c ) x - \(\frac{2}{255}=\frac{114}{153}\)
x = 114/153 + 2/255
x = 64/85
Vậy x = 64/85
\(2.THPT\)
\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{98.99}+\frac{9}{99.100}\)
\(A=9\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(A=9\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=9\left(1-\frac{1}{100}\right)\)
\(A=9.\frac{99}{100}\)
\(A=\frac{891}{100}\)
\(B=\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{93.95}\)
\(B=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{93}-\frac{1}{95}\)
\(B=\frac{1}{5}-\frac{1}{95}\)
\(B=\frac{18}{95}\)
\(D=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(D=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)
\(D=\frac{1}{2}-\frac{1}{28}\)
\(D=\frac{13}{28}\)
Ta có: B = 22010 - 22009 - 22008 -......- 2 -1
=> B = 22010 - (1 + 2 + 22 + ..... + 22009)
Đặt A = 1 + 2 + 22 + .... + 22009
=> 2A = 2 + 22 + .... + 22010
=> 2A - A = 22010 - 1
=> A = 22010 - 1
Vậy B = 22010 - (22010 - 1)
=> B = 22010 - 22010 + 1
=> B = 1
Ta có: B = 22010 - 22009 - 22008 -......- 2 -1
=> B = 22010 - (1 + 2 + 22 + ..... + 22009)
Đặt A = 1 + 2 + 22 + .... + 22009
=> 2A = 2 + 22 + .... + 22010
=> 2A - A = 22010 - 1
=> A = 22010 - 1
Vậy B = 22010 - (22010 - 1)
=> B = 22010 - 22010 + 1
=> B = 1