Tìm Y
15*Y-Y*4-Y*1=120
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c: Ta có: 4x=3y=3z
nên \(\dfrac{x}{\dfrac{1}{4}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{3}}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{\dfrac{1}{4}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{3}}=\dfrac{x+y+z}{\dfrac{1}{4}+\dfrac{1}{3}+\dfrac{1}{3}}=\dfrac{1975}{\dfrac{11}{12}}=\dfrac{23700}{11}\)
Do đó: \(\left\{{}\begin{matrix}x=\dfrac{5925}{11}\\y=\dfrac{7900}{11}\\z=\dfrac{7900}{11}\end{matrix}\right.\)
\((\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}).120+y:\frac{1}{3}=-4\)
Ta có: \(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\)
\(=\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\)
\(=\frac{1}{24}-\frac{1}{30}\)
\(=\frac{1}{120}\)
Thay vào ta có: \(\frac{1}{120}.120+y:\frac{1}{3}=-4\)
\(\implies 1+y:\frac{1}{3}=-4\)
\(\implies y:\frac{1}{3}=-5\)
\(\implies y=-5.\frac{1}{3}\)
\(\implies y=\frac{-5}{3}\). Vậy \(y=\frac{-5}{3}\)
~ Học tốt a~
\(\left(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\right).120+y:\frac{1}{3}=-4\)
\(\Rightarrow\left(\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\right).120+y:\frac{1}{3}=-4\)
\(\Rightarrow\left(\frac{1}{24}-\frac{1}{30}\right).120+y:\frac{1}{3}=-4\)
\(\Rightarrow\frac{1}{120}.120+y:\frac{1}{3}=-4\)
\(\Rightarrow1+y:\frac{1}{3}=-4\)
\(\Rightarrow y:\frac{1}{3}=-4-1\)
\(\Rightarrow y:\frac{1}{3}=-5\)
\(\Rightarrow y=-5.\frac{1}{3}\)
\(\Rightarrow y=\frac{-5}{3}\)
Vậy \(y=\frac{-5}{3}\)
_Chúc bạn học tốt_
\(\left(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\right).120+y:\frac{1}{3}=-4\)
\(\Leftrightarrow\)\(\left(\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\right).120+y:\frac{1}{3}=-4\)
\(\Leftrightarrow\)\(\left(\frac{1}{24}-\frac{1}{30}\right).120+y:\frac{1}{3}=-4\)
\(\Leftrightarrow\)\(1+3y=-4\)
\(\Leftrightarrow\)\(3y=-5\)
\(\Leftrightarrow\)\(y=-\frac{5}{3}\)
Vậy...
Ta có :
\(\left(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\right).120+y:\frac{1}{3}=-4\)
\(\Rightarrow\left(\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\right).120+y.3=-4\)
\(\Rightarrow\left(\frac{1}{24}-\frac{1}{30}\right).120+y.3=-4\)
\(\Rightarrow\left(\frac{5}{120}-\frac{4}{120}\right).120+y.3=-4\)
\(\Rightarrow\frac{1}{120}.120+y.3=-4\)
\(\Rightarrow1+y.3=-4\)
\(\Rightarrow3y=-4-1\)
\(\Rightarrow3y=-5\)
\(\Rightarrow y=-\frac{5}{3}\)
Vậy \(y=-\frac{5}{3}\)
y x 500 = 780 000
y = 780 000 : 500
y = 1560
y x 120 = 12000
y = 12000 : 120
y = 100
y : 145 = 318
y = 318 x 145
y = 46110
y : 213 = 456
y = 456 x 213
y= 97128
y x 70 = 238
y = 238 : 70
y = 3.4
y x 78 = 936
y = 936 : 78
y = 12
462 : y = 66
y = 462 : 66
y = 7
709 : y = 82
y = 709 : 82
y = 8.64634146341
y x 500 = 780 000
y = 780 000 : 500
y = 1560
y x 120 = 12000
y = 12000 : 120
y = 100
y : 145 = 318
y = 318 x 145
y = 46110
y : 213 = 456
y = 456 x 213
y= 97128
y x 70 = 238
y = 238 : 70
y = 3.4
y x 78 = 936
y = 936 : 78
y = 12
462 : y = 66
y = 462 : 66
y = 7
709 : y = 82
y = 709 : 82
y = 8.64634146341
y.15-4.y-y.1=120
(=) y.(15-4-1)=120
(=) y.10=120
(=) y=12
1) ( y-25):4 - 120 = 0 2) 120 - ( x+25 )x4 = 0
( y-25 ):4 = 0+120 (x+25) x4 =120-0
( y-25):4 =120 (x+25) x4 =120
( y-25)=120x4 (x+25)=120:4
y-25=480 x+25=30
y=480+25 x=30-25
y=505 x=5
k mk nha