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\(\left(y+1\right)+\left(Y+3\right)+...+\left(y+99\right)=2550\)
\(\left(y+y+..+y\right)+\left(1+3+5+...+99\right)\)
\(50y+\frac{\left(99+1\right).50}{2}=2550\)
\(50y+2500=2550\)
\(\Rightarrow50y=50\)
\(\Rightarrow y=50:50=1\)
a) 186 - 3 x y = 99
3 x y = 186 - 99
3 x y = 87
y = 87/3
b) y :2 + y:3 + y:4 = 15
y x 1/2 + y x 1/3 + y x 1/4 = 15
y x ( 1/2 + 1/3 + 1/4) =15
y x 13/12 = 15
y = 15 : 13/12
y = 180/13
Tìm y biết
a 186 -3 x y = 99
3 x y = 186 - 99
3 x y = 87
y = 87 : 3
y = 29
b y :2 + y : 3 + y : 4 = 15
\(\frac{y}{2}+\frac{y}{3}+\frac{y}{4}=15\)
\(\frac{6\times y}{12}+\frac{4\times y}{12}+\frac{3\times y}{12}=\frac{180}{12}\)
\(6\times y+4\times y+3\times y=180\)
\(\left(6+4+3\right)\times y=180\)
\(13\times y=180\)
\(y=180:13\)
\(y=\frac{180}{13}\)
MÌNH GHI RA TỪNG CHI TIẾT RỒI NHA , CHÚC BẠN HỌC TỐT !!!
\(\frac{2}{3}\cdot y-\frac{12}{3}:\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\right)=\frac{1}{3}\)\(\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}\right)=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\left(\frac{3-1}{1\cdot3}+\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+\frac{9-7}{7\cdot9}+\frac{11-9}{9\cdot11}+\frac{13-11}{11\cdot13}\right)=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\left(1+\frac{1}{3}-\frac{1}{3}+\frac{1}{5}-\frac{1}{5}+\frac{1}{7}-\frac{1}{7}+\frac{1}{9}-\frac{1}{9}+\frac{1}{11}-\frac{1}{11}+\frac{1}{13}\right)\)\(=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\left(\frac{1}{1}+\frac{1}{3}\right)=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\frac{4}{3}\)\(=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4\cdot\frac{3}{4}=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-3=\frac{1}{3}\)
\(\frac{2}{3}\cdot y=\frac{1}{3}+3\)
\(\frac{2}{3}\cdot y=\frac{10}{3}\)
\(y=\frac{10}{3}:\frac{2}{3}\)
y=5
(y + 1) + (y + 2) + (y + 3) + .... + (y + 10) = 2012
<=> y + 1 + y + 2 + y + 3 + .... + y + 10 = 2012
<=> ( y + y + .. + y) + ( 1 + 2 + 3 + ... + 10 ) = 2012
<=> 10y + 55 = 2012
<=> 10y = 2012 - 55 = 1957
=> y = 1957 : 10 = 195.7
[ y + 1 ] + [ y + 2 ] + [ y + 3 ] + .......... + [ y + 10 ] = 2012
y x 10 + ( 1 + 2 + 3 + ............ + 10 ) = 2012
Từ 1 đến 10 có số số hạng là:
( 10 - 1 ) : 1 + 1 = 10 ( số )
Tổng các số từ 1 đến 10 là:
( 10 + 1 ) x 10 : 2 = 55
Thay vào ta có:
y x 10 + 55 = 2012
y x 10 = 2012 - 55
y x 10 = 1957
y = 1957 : 10
y = 195,7
(y+1) +(y+2) +(y+3) +... +(y+50)= 685
có (50-1) + 1= 50 số y
=> 50y + 1+2+3+...+50
=> 50y + (50+1) x 50 : 2
=> 50y + 1275 = 685
=> 50y = -590
=> y = -11,8
\(\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right).y=\frac{2}{3}\)
\(\frac{1}{2}.\left(1-\frac{1}{3}\right)+\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+\frac{1}{2}.\left(\frac{1}{7}-\frac{1}{9}\right)+\frac{1}{2}.\left(\frac{1}{9}-\frac{1}{11}\right).y=\frac{2}{3}\)
\(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right).y=\frac{2}{3}\)
\(\frac{1}{2}.\left(1-\frac{1}{11}\right).y=\frac{2}{3}\)
\(\left(1-\frac{1}{11}\right).y=\frac{4}{3}\)
\(\frac{10}{11}.y=\frac{4}{3}\)
\(\Rightarrow y=\frac{22}{15}\)
\(\left(y+1\right)+\left(y+2\right)+\left(y+3\right)+...+\left(y+99\right)=2550\)
Có \(\left(99-1\right):1+1=99\)cặp
\(\left(y+y+y+...+y\right)+\left(1+2+3+...+99\right)=2550\)
\(99y+\left[\left(99+1\right)\times99:2\right]=2550\)
\(99y+4590=2550\)
\(99y=2550-4590\)
\(99y=-2040\)
\(y=\frac{-680}{33}\)
Đề hình như sai bạn ơi
#)Giải :
\(\left(y+1\right)+\left(y+2\right)+\left(y+3\right)+...+\left(y+99\right)=2550\)
\(\Rightarrow\left(y+y+y+...+y\right)+\left(1+2+3+...+99\right)=2550\)
\(\Rightarrow99y+\frac{\left(99+1\right).99}{2}=2550\)
\(\Rightarrow99y+4950=2550\)
\(\Leftrightarrow99y=-2400\)
\(\Leftrightarrow y=-24,2424242424.................\approx-24,24\)
P/s : Có j đó sai sai :v đề lỗi ak ?