Tìm Y: (y+1/3)+(y+1/6)+(y+1/12)+(y+1/24)=1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
3 tháng 7 2021
Giải:
a) \(\dfrac{-5}{8}=\dfrac{x}{16}\)
\(\Rightarrow x=\dfrac{16.-5}{8}=-10\)
\(\dfrac{3x}{9}=\dfrac{2}{6}\)
\(\Rightarrow3x=\dfrac{2.9}{6}=3\)
\(\Rightarrow x=1\)
b) \(\dfrac{x+3}{15}=\dfrac{1}{3}\)
\(\Rightarrow x+3=\dfrac{1.15}{3}=5\)
\(\Rightarrow x=2\)
\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)
\(\Rightarrow2x+1=\dfrac{6.7}{2}=21\)
\(\Rightarrow x=10\)
c) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow\dfrac{4}{x-6}=\dfrac{-12}{18}\)
\(\Rightarrow x-6=\dfrac{18.4}{-12}=-6\)
\(\Rightarrow x=0\)
\(\Rightarrow\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow y=\dfrac{-12.24}{18}=-16\)
\(\dfrac{3-x}{-12}=\dfrac{16}{y+1}=\dfrac{192}{-72}\)
\(\Rightarrow\dfrac{3-x}{-12}=\dfrac{192}{-72}\)
\(\Rightarrow3-x=\dfrac{192.-12}{-72}=32\)
\(\Rightarrow x=-29\)
\(\Rightarrow\dfrac{16}{y+1}=\dfrac{192}{-72}\)
\(\Rightarrow y+1=\dfrac{16.-72}{192}=-6\)
d) \(\dfrac{-2}{3}< \dfrac{x}{5}< \dfrac{-1}{6}\)
\(\Rightarrow\dfrac{-20}{30}< \dfrac{6x}{30}< \dfrac{-5}{30}\)
\(\Rightarrow6x\in\left\{-18;-12;-6\right\}\)
\(\Rightarrow x\in\left\{-3;-2;-1\right\}\)
\(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\)
\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\)
\(\Rightarrow5x\in\left\{-5;0;5;10\right\}\)
\(\Rightarrow x\in\left\{-1;0;1;2\right\}\)
e) \(\dfrac{x+46}{20}=x\dfrac{2}{5}\)
\(\Rightarrow\dfrac{x+46}{20}=x+\dfrac{2}{5}\)
\(\Rightarrow\dfrac{x+46}{20}=\dfrac{5x+2}{5}\)
\(\Rightarrow5.\left(x+46\right)=20.\left(5x+2\right)\)
\(\Rightarrow5x+230=100x+40\)
\(\Rightarrow5x-100x=40-230\)
\(\Rightarrow-95x=-190\)
\(\Rightarrow x=-190:-95\)
\(\Rightarrow x=2\)
\(y\dfrac{5}{y}=\dfrac{86}{y}\)
\(\Rightarrow y+\dfrac{5}{y}=\dfrac{86}{y}\)
\(\Rightarrow\dfrac{y^2+5}{y}=\dfrac{86}{y}\)
\(\Rightarrow y^2+5=86\)
\(\Rightarrow y^2=86-5\)
\(\Rightarrow y^2=81\)
\(\Rightarrow\left[{}\begin{matrix}y=9\\y=-9\end{matrix}\right.\)
Chúc bạn học tốt!
TL
3 tháng 2 2019
a,+) 4/x=12/15
=>15*4=12*x
=>60=12*x
=>x=5;có 4/x=y/45
=>4/5=y/45
=>36/45=y/45
=>y=36
Vậy x=5:y=36
b,3/x=1/y=6/24
=>3/x=1/y=1/4
có: 1/y=1/4
=>y=4
có: 1/4=3/x
=>1x=3*4
=>x=12
Vây x=12
y=4
hok tot
3 tháng 2 2019
a) \(\frac{4}{5}=\frac{12}{15}=\frac{36}{45}\)
Tự giải quyết nốt =))
7 tháng 8 2016
a)100-9:(372:3.y-1)-14=83
<=>100-9:(124y-1)-14=83
<=>86-9:(124y-1)=83
<=>9:(124y-1)=3
<=>124y-1=3
<=>124y=4
<=>y=\(\frac{1}{31}\approx0,3223\)
b)7260-120:24.y+924=528,3
<=>8184-5y=528,3
<=>5y=7655,7
<=>y=1531,14
c)2000-52:(615:3:y-15)-14=1984
<=>1986-52:(205:y-15)=1984
<=>52:(205:y-15)=2
<=>205:y-15=26
<=>205:y=41
<=>y=5
d)y+y.1/3+5/18=7/18
<=>4/3y=1/9
<=>y=1/12
e)13/15-(5/21+y)7/12=7/10
<=>7/12.(5/21+y)=1/6
<=>5/36+7/12y=1/6
<=>7/12y=1/36
<=>y=1/21\(\approx\)0,4762
h)y+2.y+3.y+...+10.y=49,5
<=>55y=49,5
<=>y=0,9
i)y.(1975/8.9+1885/9.10+1755/10.11+1579/11.12)=1/24
<=>7991,364899y=1/24
<=>y=33/6329161
tích nha mỏi tay quá
VL
0
YN
4 tháng 2 2023
\(\dfrac{4}{x}=\dfrac{12}{15}=\dfrac{y}{45}\)
Ta có:
\(12:4=3\)
\(\Rightarrow\dfrac{4}{x}=\dfrac{4\times3}{x\times3}=\dfrac{12}{15}\)
Mà \(15:3=5\)
\(\Rightarrow x=5\)
Lại có:
\(45:15=3\)
\(\Rightarrow\dfrac{12}{15}=\dfrac{12\times3}{15\times3}=\dfrac{y}{45}\)
Mà \(12\times y=36\)
\(\Rightarrow y=36\)
Vậy \(x=5;y=36\).
\(\dfrac{3}{x}=\dfrac{1}{y}=\dfrac{6}{24}\)
Ta có:
\(\dfrac{3}{x}=\dfrac{6}{24}\)
\(6\times x=3\times24\)
\(6\times x=72\)
\(x=72:6\)
\(x=12\)
\(\Rightarrow x=12\)
Lại có:
\(\dfrac{3}{12}=\dfrac{1}{y}\)
\(3\times y=1\times12\)
\(3\times y=12\)
\(y=12:3\)
\(y=4\)
\(\Rightarrow y=4\)
Vậy \(x=12;y=4\).
23 tháng 3 2020
Do x và y là hai đại lượng tỉ lệ thuận và \(x_1,x_2\)là hai giá trị khác nhau của x;\(y_1,y_2\)là hai giá trị tương ứng của y nên
\(\frac{y_1}{y_2}=\frac{x_1}{x_2}=\frac{2y_1}{2y_2}=\frac{3x_1}{3x_2}=\frac{2y_1+3x_1}{2y_2+3x_2}\)
Vì \(x_2=-6,y_2=-3\)và \(2y_1+3x_1=24\)nên ta có :
\(\frac{y_1}{-3}=\frac{x_1}{-6}=\frac{2y_1+3x_1}{2\cdot\left(-3\right)+3\cdot\left(-6\right)}=\frac{24}{-24}=-1\)
=> \(y_1=\left(-1\right)\cdot\left(-3\right)=3;x_1=\left(-1\right)\cdot\left(-6\right)=6\)
\(4y+\dfrac{5}{8}=1\Leftrightarrow4y=1-\dfrac{5}{8}=\dfrac{3}{8}\Leftrightarrow y=\dfrac{3}{8}:4=\dfrac{3}{32}\)
\(\left(y+\dfrac{1}{3}\right)+\left(y+\dfrac{1}{6}\right)+\left(y+\dfrac{1}{12}\right)+\left(y+\dfrac{1}{24}\right)=1\)
\(\Rightarrow y+\dfrac{1}{3}+y+\dfrac{1}{6}+y+\dfrac{1}{12}+y+\dfrac{1}{24}=1\)
\(\Rightarrow\left(y+y+y+y\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{24}\right)=1\)
\(\Rightarrow4y+\dfrac{15}{24}=1\)
\(\Rightarrow4y=\dfrac{9}{24}\)
\(\Rightarrow y=\dfrac{3}{32}\)