Tìm y:
a, y + 847 x 2 = 1953 - 74 b, y : (7 x 18) = 5839 + 8591
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\(a,y\times2,8+5,2\times y=48\\ y\times\left(2,8+5,2\right)=48\\ y\times8=48\\ y=\dfrac{48}{8}=6\\ ---\\ b,y\times12,25-y+y\times2,75=1050\\ y\times\left(12,25-1+2,75\right)=1050\\ y\times14=1050\\ y=\dfrac{1050}{14}=75\)
a: \(y\cdot2,8+y\cdot5,2=48\)
=>\(y\left(2,8+5,2\right)=48\)
=>\(8y=48\)
=>\(y=\dfrac{48}{8}=6\)
b: \(y\cdot12,25-y+y\cdot2,75=1050\)
=>\(y\left(12,25-1+2,75\right)=1050\)
=>\(y\cdot14=1050\)
=>\(y=\dfrac{1050}{14}=75\)
\(a,y\times27+y\times30+y\times44-y=10500\\\Rightarrow y\times\left(27+30+44-1\right)=10500\\ \Rightarrow y\times100=10500\\ \Rightarrow y=10500:100\\ \Rightarrow y=105\\ b,y\times285+115\times y=400\\ \Rightarrow y\times\left(285+115\right)=40\\ \Rightarrow y\times400=400\\ y=400:400\\ \Rightarrow y=1\)
\(a,\Rightarrow y\times\left(27+30+44-1\right)=10500\\ \Rightarrow y\times100=10500\\ \Rightarrow y=105\\ b,\Rightarrow y\times\left(285+115\right)=400\\ \Rightarrow y\times400=400\\ \Rightarrow y=1\)
Bài 2:
a: \(\dfrac{4\cdot7}{9\cdot32}=\dfrac{4}{32}\cdot\dfrac{7}{9}=\dfrac{1}{8}\cdot\dfrac{7}{9}=\dfrac{7}{72}\)
b: \(\dfrac{3\cdot21}{14\cdot15}=\dfrac{63}{210}=\dfrac{3}{10}\)
c: \(\dfrac{9\cdot6-9\cdot3}{18}=\dfrac{9\cdot3}{18}=\dfrac{3}{2}\)
d: \(\dfrac{17\cdot15-17}{3-20}=\dfrac{17\cdot14}{-17}=-14\)
e: \(=\dfrac{26\cdot5}{26\cdot35}=\dfrac{5}{35}=\dfrac{1}{7}\)
f: \(=\dfrac{49\left(1+7\right)}{49}=8\)
a) (y + 5) x 2020 = 205 x 2020
(y + 5) x 2020 = 414100
y + 5 = 414100 : 2020 = 205
y = 205 - 5 = 200
a) y= 200
b) y-45600 = 1600 × 4 ×25
y-45600=160000
y= 114400
a: =>y=1,25x3,14=3,925
b: =>y:0,24=15-7,8=7,2
hay y=1,728
c: =>10y=36,7
hay y=3,67
d: =>6:y=1,5
hay y=4
e: =>4y+0,2y=44,1
=>4,2y=44,1
hay y=10,5
Tìm y:
\(a,y+847\times2=1953-74\\ y+847\times2=1879\\ y+1694=1897\\ y=1897-1694\\ y=185\\ b,y:\left(7\times18\right)=5839+8591\\ y:126=14430\\ y=14430\times126\\ y=1818180.\)
a/ y + 847 x 2 = 1953 - 74
y + 1694 = 1879
y = 1879 - 1694
y = 185
b/y : (7 x 18) = 5839 + 8591
y : 126 = 14430
y = 14430 x 126
y = 1818180
#AvoidMe