19,96+4,19-24,15:(x:\(\frac{1}{4}-\frac{1}{4}\))=23,15
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\(19,96+4,19-24,15:\left(x:\frac{1}{4}-\frac{1}{4}\right)=23,15\)
\(24,15-24,15:\left(x:\frac{1}{4}-\frac{1}{4}\right)=23,15\)
\(24,15:\left(x:\frac{1}{4}-\frac{1}{4}\right)=1\)
\(x:\frac{1}{4}-\frac{1}{4}=24,15\)
\(4x=24,4\)
\(x=6,1\)
\(\text{Bài này cx đơn giản thôi!}\)
\(4.19-24.15:\left(x:\frac{1}{4}-\frac{1}{4}\right)=23.15-19.96\)
\(4.19-24.15:\left(x:\frac{1}{4}-\frac{1}{4}\right)=3.19\)
\(24.15:\left(x:\frac{1}{4}-\frac{1}{4}\right)=4.19-3.19\)
\(24.15:\left(x:\frac{1}{4}-\frac{1}{4}\right)=1\)
\(x:\frac{1}{4}-\frac{1}{4}=24.15:1\)
\(x:\frac{1}{4}-\frac{1}{4}=24.15\)
\(x:\frac{1}{4}=24.15+\frac{1}{4}\)
\(x:\frac{1}{4}=24.4\)
\(x=24.4.\frac{1}{4}\)
\(x=6.1\)
1. \(\left(\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{15.17}+\frac{2}{17.19}+\frac{2}{19.21}\right)\times462-x=19\)
\(\left(\frac{13-11}{11.13}+\frac{15-13}{13.15}+\frac{17-15}{15.17}+\frac{19-17}{17.19}+\frac{21-19}{19.21}\right)\times462-x=19\)
\(\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+\frac{1}{17}-\frac{1}{19}+\frac{1}{19}-\frac{1}{21}\right)\times462-x=19\)
\(\left(\frac{1}{11}-\frac{1}{21}\right)\times462-x=19\)
\(\frac{10}{231}\times462-x=19\)
\(20-x=19\)
\(x=20-19\)
\(x=1\)
2.b \(187-[[497-(8\times x+11)\div x]\div3-78]=150\)
\(187-[[497-(\frac{8\times x}{x}+\frac{11}{x})]:3-78]=150\)
\(187-[(497-8-\frac{11}{x}):3-78]=150\)
\(187-[(489-\frac{11}{x}):3-78]=150\)
\(187-[\frac{489}{3}-\frac{33}{x}-78]=150\)
\(187-[163-\frac{33}{x}-78]=150\)
\(187-85+\frac{33}{x}=150\)
\(102+\frac{33}{x}=150\)
\(\frac{33}{x}=150-102\)
\(\frac{33}{x}=48\)
\(x=\frac{48}{33}=\frac{16}{11}\)
X=21
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*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
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
a/\(4,19\times50-132:\dfrac{2}{3}\)
\(=209,5-198\)
\(=11,5\)
b/\(637,38:18\times2,5\)
\(=35,41\times2,5\)
\(=88,525\)
c/\(56,32-13,4\times2,4\)
\(=56,32-32,16\)
\(=24,16\)
d/\(\dfrac{2}{5}:\left(\dfrac{4}{5}-\dfrac{1}{2}\right)\)
\(=\dfrac{2}{5}:\dfrac{3}{10}\)
\(=\dfrac{4}{3}\)
\(b)\) \(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{97.101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(1-\frac{1}{101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(\frac{100}{101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(100=2x+4\)
\(\Leftrightarrow\)\(2x=96\)
\(\Leftrightarrow\)\(48\)
Vậy \(x=48\)
Chúc bạn học tốt ~
\(a)\) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{47.49}=\frac{24}{x+1}\)
\(\Leftrightarrow\)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{47.49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{47}-\frac{1}{49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(1-\frac{1}{49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(\frac{48}{49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(49=x+1\)
\(\Leftrightarrow\)\(x=48\)
Vậy \(x=48\)
Chúc bạn học tốt ~