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1/2-2y=9/20
=>2y=1/2-9/20=1/20
=>y=1/20:2=1/40
b,3/5:4/3:y=2+7/10=9/20:y=27/10
=>y=9/20:27/10=1/6
c,y+y*3/2-y*1/2=1/10
=>y(1+3/2-1/2)=1/10
=>2y=1/10
=>y=1/10:2=1/20
\(y\div\frac{2}{10}+y\div12,5\%+4\div\frac{1}{y}+y+y\div0,5=\frac{20}{3}\)
\(\Rightarrow y\times5+y\times8+4\times y+y+y=\frac{20}{3}\)
\(\Rightarrow y\times\left(5+8+4+1+1\right)=\frac{20}{3}\)
\(\Rightarrow y\times19=\frac{20}{3}\)
\(\Rightarrow y=\frac{20}{3}\div19\)
\(\Rightarrow y=\frac{20}{57}\)
( y + 1 ) + ( y + 2 ) + ( y + 3 ) + ......+ ( y + 9 ) + ( y + 10 ) = 2012
Suy ra có 10 số y
Tổng các số từ 1 đến 10 là : ( 10 + 1 ) x 10 : 2 = 55
y x 10 + 55 = 2012
y x 10 = 2012 - 55
y x 10 = 1957
y = 1957 : 10
y = 195,7
(y+y+y+...+y)+(1+2+3+...+10)=2012
yx10+55=2012
yx10=1957
y=1957:10
y=195,7
yx10+1+2+3+4+5+6+7+8+9+10=2012
yx10+55 =2012
yx10 =2012-55
yx10 =1957
y =1957:10
y=195,7
yx10+1+2+3+4+5+6+7+8+9+10=2012
yx10+55 =2012
yx10 =2012-55
yx10 =1957
y =1957:10=195.7
hiệu 2 số là:
2-1=1
nên y sẽ là 1
nên ta có tổng là:
37x37x2=2738
đáp số:2738
37 x y + 1 + 2 + 3+ ...+37 = 37 x y + (1 + 37) : 2 x 37
= 37 x (y + 19)
2: y \(\times\) \(\dfrac{3}{5}\) = \(\dfrac{9}{10}\)
2:y = \(\dfrac{9}{10}\) : \(\dfrac{3}{5}\)
2: y = \(\dfrac{3}{2}\)
y = 2 : \(\dfrac{3}{2}\)
y = \(\dfrac{4}{3}\)
\(\dfrac{5}{4}\) - \(\dfrac{2}{5}\) : y = 1
\(\dfrac{2}{5}\) : y = \(\dfrac{5}{4}\) - 1
\(\dfrac{2}{5}\): y = \(\dfrac{1}{4}\)
y = \(\dfrac{2}{5}\) : \(\dfrac{1}{4}\)
y = \(\dfrac{8}{5}\)
\(\dfrac{3}{4}\) \(\times\) ( \(\dfrac{7}{2}\) - y) = \(\dfrac{3}{2}\)
\(\dfrac{7}{2}\) - y = \(\dfrac{3}{2}\) : \(\dfrac{3}{4}\)
\(\dfrac{7}{2}\) - y = 2
y = \(\dfrac{7}{2}\) - 2
y = \(\dfrac{3}{2}\)
=> Yx10+45=2012
Yx10=1967
Y=196,7
Ta có ( y +1 ) + (y + 2) + ......+ ( y +10 ) = 2012
=> (y + y + ..... + y) + ( 1 + 2 + ..... + 10 ) = 2012
=> y x 10 + 55 = 2012
=> y x 10 = 1957
=> y = 195,7