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\(y\times3+\dfrac{y}{2}+\dfrac{y}{4}=1\dfrac{1}{2}\\ \Rightarrow y\times3+y\times\dfrac{1}{2}+y\times\dfrac{1}{4}=\dfrac{3}{2}\\ \Rightarrow y\times\left(3+\dfrac{1}{2}+\dfrac{1}{4}\right)=\dfrac{3}{2}\\ \Rightarrow y\times\dfrac{15}{4}=\dfrac{3}{2}\\ \Rightarrow y=\dfrac{3}{2}:\dfrac{15}{4}\\ \Rightarrow y=\dfrac{2}{5}\)
\(\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right).y=\frac{2}{3}\)
\(\frac{1}{2}.\left(1-\frac{1}{3}\right)+\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+\frac{1}{2}.\left(\frac{1}{7}-\frac{1}{9}\right)+\frac{1}{2}.\left(\frac{1}{9}-\frac{1}{11}\right).y=\frac{2}{3}\)
\(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right).y=\frac{2}{3}\)
\(\frac{1}{2}.\left(1-\frac{1}{11}\right).y=\frac{2}{3}\)
\(\left(1-\frac{1}{11}\right).y=\frac{4}{3}\)
\(\frac{10}{11}.y=\frac{4}{3}\)
\(\Rightarrow y=\frac{22}{15}\)
\(2\frac{3}{4}.\frac{1}{2}-\frac{1}{2}.\frac{3}{7}+\frac{1}{3}\)
\(=\frac{11}{4}.\frac{1}{2}-\frac{1}{2}.\frac{3}{7}+\frac{1}{3}\)
\(=\frac{11}{8}-\frac{3}{14}+\frac{1}{3}\)
\(=\frac{251}{168}\)
Bài 1 : a, thực hiện phép tính :
\(2\frac{3}{4}×\frac{1}{2}-\frac{1}{2}×\frac{3}{7}+\frac{1}{3}\)
\(=\frac{11}{4}×\frac{1}{2}-\frac{1}{2}×\frac{3}{7}+\frac{1}{3}\)
\(=\frac{1}{2}×\left(\frac{11}{4}-\frac{3}{7}\right)+\frac{1}{3}\)
\(=\frac{1}{2}×\frac{65}{28}+\frac{1}{3}\)
\(=\frac{65}{56}+\frac{1}{3}\)
\(=\frac{251}{168}\)
b , Tìm x biết :
a, 435- ( x + 16 ) = 425 : 17
435 - ( x + 16 ) = 25
x + 16 = 435 - 25
x + 16 = 410
x = 410 - 16
x = 394
Vậy x = 394
b, ( x + 3/4 ) × 7/4 = 5 - 7/6
( x + 3/4 ) × 7/4 = 23/6
x + 3/4 = 23/6 : 7/4
x + 3/4 = 23/6 × 4/7
x + 3/4 = 46/21
x = 46/21 - 3/4
x = 121/84
Vậy x = 121/84
Ta có: \(\frac{y}{3}+y\cdot\frac{2}{3}=\frac{y}{12}+\frac{11}{6}\)
=> \(\frac{y}{3}+\frac{2y}{3}=\frac{y}{12}+\frac{22}{12}\)
=> \(\frac{y+2y}{3}=\frac{y+22}{12}\)
=> \(\frac{3y}{3}=\frac{y+22}{12}\)
=> \(y=\frac{y+22}{12}\)
=> y + 22 = 12y
=> y - 12y = 22
=> -11y = 22
=> y = 22 : (-11) = -2
Vậy y = -2
\(\frac{2}{3}\cdot y-\frac{12}{3}:\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\right)=\frac{1}{3}\)\(\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}\right)=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\left(\frac{3-1}{1\cdot3}+\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+\frac{9-7}{7\cdot9}+\frac{11-9}{9\cdot11}+\frac{13-11}{11\cdot13}\right)=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\left(1+\frac{1}{3}-\frac{1}{3}+\frac{1}{5}-\frac{1}{5}+\frac{1}{7}-\frac{1}{7}+\frac{1}{9}-\frac{1}{9}+\frac{1}{11}-\frac{1}{11}+\frac{1}{13}\right)\)\(=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\left(\frac{1}{1}+\frac{1}{3}\right)=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\frac{4}{3}\)\(=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4\cdot\frac{3}{4}=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-3=\frac{1}{3}\)
\(\frac{2}{3}\cdot y=\frac{1}{3}+3\)
\(\frac{2}{3}\cdot y=\frac{10}{3}\)
\(y=\frac{10}{3}:\frac{2}{3}\)
y=5
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
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
1
a,80+40-82+30=68
b,32x56+45-32x32=813
2
a12/7:x+2/3=7/5
12/7:x=7/5+2/3
12/7:x=31/15
x=31/15.12/7
x=124/35
`16 + 3 xx y = 7`
`3 xx y = 7 - 16`
`3 xx y = -9`
`y = -9 : 3`
`y = (-9)/3`
`y = -3
Vậy `y = -3`
_____________________________
`40 - 2 xx ( y- 1)=12`
`2 xx (y-1)=40-12`
`2 xx (y-1)=28`
`y-1=28:2`
`y-1=14`
`y=14+1`
`y=15`
Vậy `y =15`