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b)4y+(1/3+1/9+1/27)=56/81
4y+40/81=56/81
4y=56/81-40/81
4y=16/81
y=16/81:4
y=4/81
b) Tìm y:
( y + \(\frac{1}{3}\)) + ( y + \(\frac{1}{9}\)) + ( y + \(\frac{1}{27}\)) + ( y + \(\frac{1}{81}\)) = \(\frac{56}{81}\)
y + ( \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)) = \(\frac{56}{81}\)
y + \(\frac{27+9+3+1}{81}\) = \(\frac{56}{81}\)
y + \(\frac{40}{81}\) = \(\frac{56}{81}\)
y = \(\frac{56}{81}-\frac{40}{81}\)
y = \(\frac{16}{81}\)
Tl roy, ủng hộ nha! :)
Bài làm
Ta có: \(\frac{x}{5}=\frac{y}{4}\Rightarrow\frac{x}{5}=\frac{2y}{8}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{x}{5}=\frac{2y}{8}=\frac{x-2y}{5-8}=\frac{81}{-3}=-27\)
Do đó: \(\hept{\begin{cases}\frac{x}{5}=-27\\\frac{y}{4}=-27\end{cases}\Rightarrow\hept{\begin{cases}x=-135\\y=-108\end{cases}}}\)
Vậy x = -135, y = -108
# Học tốt #
Ta có: \(\frac{x}{5}=\frac{y}{4}\)và x-2y=81
\(\Rightarrow\frac{x}{5}=\frac{2y}{8}\)
Áp dụng t/chất của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{5}=\frac{2y}{8}=\frac{x-2y}{5-8}=\frac{81}{-3}=-27\)
+) x=-27.5=-135
+)\(\frac{y}{4}=-27\rightarrow y=-108\)
Vậy x=-135; y=-108
Hok tốt nha^^
1, \(\frac{1}{2}-\left(6\frac{5}{9}+x-\frac{117}{8}\right):\left(12\frac{1}{9}\right)=0\)
\(\left(\frac{6.9+5}{9}+x-\frac{117}{8}\right):\frac{12.9+1}{9}=\frac{1}{2}\)
( . là nhân nha)
\(\left(\frac{59}{9}-\frac{117}{8}+x\right):\frac{109}{9}=\frac{1}{2}\)
\(\frac{59}{9}-\frac{117}{8}+x=\frac{1}{2}\cdot\frac{109}{9}\)
\(\frac{59}{9}-\frac{117}{8}+x=\frac{109}{18}\)
\(x=\frac{109}{18}-\frac{59}{9}+\frac{117}{8}\)
\(x=\frac{113}{8}\)
( \(\left(y+\frac{1}{3}\right)+\left(y+\frac{2}{9}\right)+\left(y+\frac{1}{27}\right)+\left(y+\frac{1}{81}\right)=\frac{56}{81}\)
\(y+\frac{1}{3}+y+\frac{2}{9}+y+\frac{1}{27}+y+\frac{1}{81}=\frac{56}{81}\)
\(4y+\frac{1}{3}+\frac{2}{9}+\frac{1}{27}+\frac{1}{81}=\frac{56}{81}\)
\(4y+\frac{49}{81}=\frac{56}{81}\)
\(4y=\frac{7}{81}\)
y = 7/81:4
y = 7/324
\(6xy+\left(\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{7x8}\right)=\frac{29}{8}\)
Đăt \(A=\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{7x8}\)
\(\Rightarrow A=\frac{3-2}{2x3}+\frac{4-3}{3x4}+\frac{5-4}{4x5}+...+\frac{8-7}{7x8}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{8}=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)
\(\Rightarrow6xy+A=6xy+\frac{3}{8}=\frac{29}{8}\Rightarrow6xy=\frac{26}{8}\Rightarrow y=\frac{26}{8x6}\)
- 2^y + 2^x - 224 = 0
- ( 2^y - 2^x + 224 ) = 0
2^y - 2^x + 224 = 0
Tìm cá số nguyên dương x,y biết:
\(2^x\)\(-\) \(2^y\)= \(224\)
=> \(2^x-2^y=2^{10}\)
=> \(2^y=2^{10}\)
=> y = 10
=> \(2^x=2^{10}+2^{10}\)
=> \(2^x=2^{11}\)
=> x = 11
Vậy x = 11; y = 10
\(\frac{63+y}{81+y}=1-\frac{1}{8}\)
\(\Rightarrow\frac{63+y}{81+y}=\frac{7}{8}\)
\(\Rightarrow(63+y).8=\left(81+y\right).7\)
\(\Rightarrow504+8y=567+7y\)
\(\Rightarrow567-504=8y-7y\)
\(\Rightarrow y=63\)
1 / 7/4 - y . 5/6 = 1/2 + 1/3
7/4 - 5/6y = 5/6
5/6y = 7/4 - 5/6
5/6y = 11/12
y = 11/12 : 5/6
y = 11/10
2 . 136 là số chia 9 dư 1
=> y = 136 - 1 = 135
3 . Dựa theo quy luật của dãy thì số tiếp theo là :
1/16 : 2 = 1/32
1: y = \(\frac{11}{10}\)
2: Y = 135
3: Số hạng tiếp theo của dãy: \(\frac{1}{32}\)
9/4-y*5/6=1/2+3/2
9/4-y*5/6=2
y*5/6=9/4-2/1
y*5/6=9/4-8/4
y*5/6=1/4
y=1/4:5/6
y=3/10
81 + 81 x 5 = \(\dfrac{y+160}{y}\) + 405
81 + 405 = 1 + \(\dfrac{160}{y}\) + 405
81 + 405 - 1 - 405 = \(\dfrac{160}{y}\)
80 = \(\dfrac{160}{y}\)
160 : 80 = y
2 = y
81 + 81 x 5 = \(\dfrac{y+160}{y}\) + 405
486 = \(\dfrac{y+160}{y}\) + 405
486 y = y +160 + 405 y
80 y = 160
y = 160 : 80
y = 2