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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
cho hình vẽ:
Biết diện tích hình vuông là 80cm2. Tìm diện tích hình tròn
Ai đúng mình tich 3 cái luôn
4 x y + (1/3 + 1/9 + 1/27 + 1/81) = 56/81
4 x y + (27/81 + 9/81 + 3/81 + 1/81) = 56/81
4 x y + 40/81 = 56/81
4 x y = 56/81 - 40/81 = 16/81
y = 4/81
\(\left(y+\frac{1}{3}\right) +\left(y+\frac{1}{9}\right)+ \left(y+\frac{1}{27}\right)+\left(y+\frac{1}{81}\right)=\frac{56}{81}\)
\(y\cdot4+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\right)=\frac{56}{81}\)
\(y\cdot4+\frac{40}{81}=\frac{56}{81}\)
\(y\cdot4=\frac{56}{81}-\frac{40}{81}\)
\(y\cdot4=\frac{16}{81}\)
\(y=\frac{16}{81}:4\)
\(y=\frac{4}{81}\)
Tìm y :
( y + 1/3 ) + ( y + 1/9 ) + ( y + 1/27 ) + ( y + 1/81 ) = 56/81
y + 1/3 + y + 1/9 + y + 1/27 + y + 1/81 = 56/81
y x 4 + ( 1/3 + 1/9 + 1/27 +1/81 ) = 56/81
y x 4 + ( 27/81 + 9/81 + 3/81 + 1/81 ) = 56/81
y x 4 + 40/81 = 56/81
y x 4 = 56/81 - 40/81
y x 4 = 16/81
y = 16/81 : 4
y = 4/81
\(\frac{63+y}{81+y}=1-\frac{1}{8}\)
\(\Rightarrow\frac{63+y}{81+y}=\frac{7}{8}\)
\(\Rightarrow(63+y).8=\left(81+y\right).7\)
\(\Rightarrow504+8y=567+7y\)
\(\Rightarrow567-504=8y-7y\)
\(\Rightarrow y=63\)
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