60% ✖ y + \(\dfrac{3}{5}\) ✖ y + y = 19,8
60% ✖ y + \(\dfrac{3}{5}\) ✖ y + y = 19,8
Tìm y
y ✖ 2 + y ✖ \(\dfrac{1}{5}\) = \(1\dfrac{3}{5}\) giúp mik với mik cần rất chi là gấp
\(\Rightarrow y\times\left(2+\dfrac{1}{5}\right)=\dfrac{8}{5}\\ \Rightarrow y\times\dfrac{11}{5}=\dfrac{8}{5}\\ \Rightarrow y=\dfrac{8}{5}:\dfrac{11}{5}=\dfrac{8}{5}\times\dfrac{5}{11}=\dfrac{8}{11}\)
( 1 - \(\dfrac{1}{2}\) ) ✖ ( 1 - \(\dfrac{1}{3}\) ) ✖ ( 1 - \(\dfrac{1}{4}\) ) ✖ ( 1 - \(\dfrac{1}{5}\) ) ✖ ( 1 - \(\dfrac{1}{6}\) ) ✖ ( 1 - \(\dfrac{1}{7}\) ) ✖ (1 - \(\dfrac{1}{8}\))
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{7}{8}=\dfrac{1}{8}\)
`= 1/2 xx 2/3 xx...xx 7/8`
`= 1/8`
\(=\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{6}{7}\times\dfrac{7}{8}\\ =\dfrac{1}{8}\)
Tính cách thuận tiện nhất. a, 1152- 96 ✖ y= (96 ✖ y -96 ✖ y) ✖ 96 ✖ y
\(1152-96\cdot y=\left(96\cdot y-96\cdot y\right)\cdot96\cdot y\)(dấu nhân viết là ".")
\(1152-96\cdot y=1\cdot96\cdot y\)
\(1152=96\cdot y+96\cdot y\)
\(1152=192\cdot y\)
\(y=\frac{1152}{192}\)
\(y=6\)
tìm y:
1,5 ✖︎ y + y =2,5 + 3 ✖︎ 2,5
(1,5+1).y=2,5+7,5
2,5.y=10
y=10:2,5
y=4
1,5 x y + y = 2,5 + 3 ✖︎ 2,5
1,5 x y + y = 2,5 + 7,5
1,5 x y + y = 10
y x ( 1,5 + 1) = 10
y x 2,5 = 10
y = 10 : 2,5
y = 4
1,5 x y + y = 2,5 + 3 x 2,5
1,5 x y + y = 2,5 + 7,5.
y x ( 1,5 + 1) = 10
\(=>y=\dfrac{10}{1,5+1}=\dfrac{10}{2,5}=4\)
Tìm y :
102,5 : ( y - 69,3 ) = 5
Tính giá trị biểu thức
a) 365,4 : ( 25,2 + 18,3 ) - ( 40 - 32,3 ) ✖ 0,2
b) \(\dfrac{2}{5}\): ( \(\dfrac{4}{5}\) - \(\dfrac{1}{2}\) ) + \(\dfrac{3}{4}\)
Bài 1:
=>y-69,3=20,5
hay y=89,8
Bài 2:
a: \(=365.4:43.5-7.7\cdot0.2\)
=8,4-1,54=6,86
b: \(=\dfrac{2}{5}:\dfrac{8-5}{10}+\dfrac{3}{5}=\dfrac{2}{5}\cdot\dfrac{10}{3}+\dfrac{3}{5}=\dfrac{20+9}{15}=\dfrac{29}{15}\)
B=(1-\(\dfrac{1}{3}\)) ✖(1-\(\dfrac{1}{6}\))✖(1-\(\dfrac{1}{10}\))✖(1-\(\dfrac{1}{15}\))✖....✖(1-\(\dfrac{1}{780}\))
\(B=\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{6}\right)\cdot\left(1-\dfrac{1}{10}\right)\cdot\left(1-\dfrac{1}{15}\right)\cdot...\cdot\left(1-\dfrac{1}{780}\right)\)
\(B=\left(1-\dfrac{1}{3+6+10+15+...+780}\right)\)
\(B=\left(1-\dfrac{1}{\left(780-3\right)\div3+1}\right)\)
\(B=\left(1-\dfrac{1}{260}\right)\)
\(B=\dfrac{259}{260}\)
cho x,y,z ≠0 và x-y-z=0. Tính giá trị của biểu thức
B=(1- \(\dfrac{2}{2}\))✖ (1- \(\dfrac{x}{y}\))✖ (1 + \(\dfrac{y}{z}\))
\(\dfrac{-5}{13}\)✖\(\dfrac{3}{7}\)-\(\dfrac{2}{17}\)✖\(\dfrac{8}{13}\)+\(\dfrac{5}{13}\)✖\(\dfrac{1}{7}\)
Giải hộ mình nhé có cả lời giải đấy
ok.mình cho 1 tim
Ta có: \(\dfrac{-5}{13}\cdot\dfrac{3}{7}-\dfrac{2}{17}\cdot\dfrac{8}{13}+\dfrac{5}{13}\cdot\dfrac{1}{7}\)
\(=\dfrac{5}{13}\left(-\dfrac{3}{7}+\dfrac{1}{7}\right)-\dfrac{2}{17}\cdot\dfrac{8}{13}\)
\(=\dfrac{5}{13}\cdot\dfrac{-2}{7}-\dfrac{2}{17}\cdot\dfrac{8}{13}\)
\(=\dfrac{-10}{91}-\dfrac{16}{221}\)
\(=\dfrac{-282}{1547}\)
\(\dfrac{-5}{13}.\dfrac{3}{7}-\dfrac{2}{17}.\dfrac{8}{13}+\dfrac{5}{13}.\dfrac{1}{7}\)
\(=\dfrac{5}{13}.\dfrac{-3}{7}-\dfrac{2}{17}.\dfrac{8}{13}+\dfrac{5}{13}.\dfrac{1}{7}\)
\(=\dfrac{5}{13}.\left(\dfrac{-3}{7}+\dfrac{1}{7}\right)+\dfrac{-2}{17}.\dfrac{8}{13}\)
\(=\dfrac{5}{13}.\dfrac{-2}{7}+\dfrac{-2}{13}.\dfrac{8}{17}\)
\(=\dfrac{-2}{13}.\dfrac{5}{7}+\dfrac{-2}{13}.\dfrac{8}{17}\)
\(=\dfrac{-2}{13}.\left(\dfrac{5}{7}+\dfrac{8}{17}\right)\)
\(=\dfrac{-2}{13}.\dfrac{141}{119}\)
\(=\dfrac{-282}{1547}\)