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\(\frac{1}{3}\) + \(\frac{5}{6}\): \(\left(x-2\frac{1}{5}\right)\)= \(\frac{3}{4}\)
<=> \(\frac{5}{6}\):\(\left(x-2\frac{1}{5}\right)\)= \(\frac{3}{4}\)- \(\frac{1}{3}\)
<=> \(\frac{5}{6}\) : \(\left(x-2\frac{1}{5}\right)\) = \(\frac{5}{12}\)
<=> \(\left(x-2\frac{1}{5}\right)\) = \(\frac{5}{6}\) : \(\frac{5}{12}\)
,<=> \(\left(x-2\frac{1}{5}\right)\)= 2
<=. x = 2 + \(\frac{11}{5}\)
<=> x = \(\frac{21}{5}\)
\(\frac{3}{4}+\frac{3}{28}+\frac{3}{70}+\frac{3}{130}+....+\frac{3}{418}+\frac{3}{550}\)
\(\Leftrightarrow\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{19.22}+\frac{3}{22.25}\)
\(\Leftrightarrow\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{19}-\frac{1}{22}+\frac{1}{22}-\frac{1}{25}\)
\(\Leftrightarrow\frac{1}{1}-\frac{1}{25}=\frac{24}{25}\)
Nhớ k cho m nhé!
a, 7/5 : x + 3/2 = 16/3
7/5 : x = 16/3 - 3/2
7/5 : x = 23/6
x = 7/5 : 23/6
x = 42/115
b, x : 1/5 + 1/7 = 3/5 . 18/21
x : 1/5 + 1/7 = 18/35
x : 1/5 = 18/35 - 1/7
x : 1/5 = 13/35
x = 13/35 . 1/5
x = 13/175
c, x - 1 và 1/3 : 2 = 5/7
x - 4/3 : 2 = 5/7
x - 4/3 = 5/7 . 2
x - 4/3 = 10/7
x = 10/7 + 4/3
x = 58/21
d, x + 2 và 3/5 . 1/6 = 35/36
x + 13/5 . 1/6 = 35/36
x + 13/5 = 35/36 : 1/6
x + 13/5 = 35/6
x = 35/6 - 13/5
x = 97/30
e, ( x + 3/2 ) : 2 = 7/10 + 1/5
( x + 3/2 ) : 2 = 9/10
x + 3/2 = 9/10 . 2
x + 3/2 = 9/5
x = 9/5 - 3/2
x = 3/10
a) \(\frac{18}{35};\frac{28}{35};\frac{31}{25};\frac{32}{25}\)
CÁCH LÀM NHƯ SAU :
(7/28 + 1/28) + 1/70 + 1/130 + 1/x.(x+3)
8/28 + 1/70 +1/130 +1/x.(x+3)
2/7+1/70+1/130+1/x.(x+3)
(20/70 +1/70)+1/130+1/x.(x+3)
3/10+1/130+1/x.(x+3)
39/130+1/130+1/x.(x+3)
4/13+1/x.(x+3)
Đến đây bn tự làm hộ mình vớ. chúc hok tốt k cho mình nhé
\(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+\frac{1}{x\left(x+3\right)}\)
\(=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{x\left(x+3\right)}\)
\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{x}-\frac{1}{x+3}\right)\)
\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{13}+\frac{1}{x}-\frac{1}{x+3}\right)\)
\(=\frac{1}{3}\left(\frac{12}{13}+\frac{1}{x}-\frac{1}{x+3}\right)\)
\(=\frac{1}{3}.\frac{12}{13}+\frac{1}{3}.\frac{1}{x}-\frac{1}{3}.\frac{1}{x+3}\)
\(=\frac{4}{13}+\frac{1}{3x}-\frac{1}{3x+3}\)
\(=\frac{4}{13}+\frac{1}{3x}-\frac{1}{3x+3}\)
\(=\frac{4}{13}+\frac{1}{3x}=\frac{1}{3x+3}\)
\(=\frac{4}{13}+\frac{1}{3x}=\frac{1}{3x+3}\)
\(=\frac{4}{13}+\frac{1}{3x}=\frac{1}{3}.\frac{1}{x+3}\)
\(=\frac{4}{13}=\frac{1}{3}.\frac{1}{x+3}-\frac{1}{3x}\)
\(=\frac{4}{13}=\frac{1}{3}.\frac{1}{x+3}-\frac{1}{3}.\frac{1}{x}\)
\(=\frac{4}{13}=\frac{1}{3}\left(\frac{1}{x+3}-\frac{1}{x}\right)\)
\(=\frac{4}{13}:\frac{1}{3}=\frac{1}{x+1}-\frac{1}{x}\)
\(=\frac{12}{13}=\frac{1}{x+1}-\frac{1}{x}\)
\(=\frac{12}{13}=\frac{x-\left(x+1\right)}{\left(x+1\right)x}\)
\(=\frac{12}{13}=-\frac{1}{x^2+x}\)
\(\Leftrightarrow=12\left(x^2+x\right)=13.\left(-1\right)\)
\(=12\left(x^2+x\right)=-13\)
\(=x^2+x=-\frac{13}{12}\)
\(=x\left(x+1\right)=-\frac{13}{12}\)
.... Chiụ
b) D = \(\frac{3}{4}+\frac{3}{8}+\frac{3}{70}+\frac{3}{130}+\frac{3}{208}+\frac{3}{304}\)
D = \(3\left(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+\frac{1}{208}+\frac{1}{304}\right)\)
D = \(3\left(\frac{1}{1x4}+\frac{1}{4x7}+\frac{1}{7x10}+\frac{1}{10x13}+\frac{1}{13x16}+\frac{1}{16x19}\right)\)
D = \(\frac{1}{1}-\frac{1}{19}=\frac{18}{19}\)
Chắc vậy