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\(\frac{2}{3}\cdot y-\frac{12}{3}:\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\right)=\frac{1}{3}\)\(\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}\right)=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\left(\frac{3-1}{1\cdot3}+\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+\frac{9-7}{7\cdot9}+\frac{11-9}{9\cdot11}+\frac{13-11}{11\cdot13}\right)=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\left(1+\frac{1}{3}-\frac{1}{3}+\frac{1}{5}-\frac{1}{5}+\frac{1}{7}-\frac{1}{7}+\frac{1}{9}-\frac{1}{9}+\frac{1}{11}-\frac{1}{11}+\frac{1}{13}\right)\)\(=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\left(\frac{1}{1}+\frac{1}{3}\right)=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\frac{4}{3}\)\(=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4\cdot\frac{3}{4}=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-3=\frac{1}{3}\)
\(\frac{2}{3}\cdot y=\frac{1}{3}+3\)
\(\frac{2}{3}\cdot y=\frac{10}{3}\)
\(y=\frac{10}{3}:\frac{2}{3}\)
y=5
x + y = 36 (1)
2x - y = 12
Áp dụng phương pháp cộng đại số
<=> x + 2x = 36 + 12
<=> 3x = 48
<=> x = 16
Thế x = 16 vào pt (1) => y = 20
y x 10 - y x \(\frac{1}{4}\)- y x\(\frac{3}{4}\)= 1,25
y x (10 - \(\frac{1}{4}\)- \(\frac{3}{4}\)) = 1,25
y x 9 = 1,25
y = 1,25 : 9
y = \(\frac{5}{36}\)
y x 25 + y x 1,5 - y x 6,5=2014
y x (25+1,5-6,5)=2014
y x 20 =2014
y=100,7
\(y+y\div0.25+y\div0,125+y\times2=15\)
\(y\times1+y\times4+y\times8+y\times2=15\)
\(y\times\left(1+4+8+2\right)=15\)
\(y\times15=15\)
\(y=15\div15\)
\(y=1\)
y + y : 0,25 + y : 0,125 + y x 2 = 15
y x 1 + y x 4 + y x 8 + y x 2 = 15
y x (1+4+8+2) = 15
y x 15 = 15
y = 15 : 15
y = 1
y x 35 =70
y = 70 : 35
y = 2
y.35=70
y =70:35
y =2
Vậy y=2
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