K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
19 tháng 7 2021
\(\dfrac{x+35}{65}+\dfrac{x+39}{61}=\dfrac{x+43}{57}+\dfrac{x+47}{53}\)
\(\Leftrightarrow\dfrac{x+35}{65}+1+\dfrac{x+39}{61}+1=\dfrac{x+43}{57}+1+\dfrac{x+47}{53}+1\)
\(\Leftrightarrow\dfrac{x+100}{65}+\dfrac{x+100}{61}-\dfrac{x+100}{57}-\dfrac{x+100}{53}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{65}+\dfrac{1}{61}-\dfrac{1}{57}-\dfrac{1}{53}\ne0\right)=0\Leftrightarrow x=-100\)
19 tháng 7 2021
Ta có:
\(\dfrac{x+35}{65}+\dfrac{x+39}{61}=\dfrac{x+43}{57}+\dfrac{x+47}{53}\\ \Rightarrow\left(\dfrac{x+35}{65}+1\right)+\left(\dfrac{x+39}{61}+1\right)=\left(\dfrac{x+43}{57}+1\right)+\left(\dfrac{x+47}{53}+1\right)\\ \Rightarrow\dfrac{x+100}{53}+\dfrac{x+100}{61}=\dfrac{x+100}{57}+\dfrac{x+100}{53}\\ \Rightarrow\left(x+100\right)\left(\dfrac{1}{65}+\dfrac{1}{61}-\dfrac{1}{57}-\dfrac{1}{53}\right)=0\)
Ta thấy:
\(\dfrac{1}{65}< \dfrac{1}{57}\\ \dfrac{1}{61}< \dfrac{1}{53}\\ \Rightarrow\left(\dfrac{1}{65}+\dfrac{1}{62}\right)-\left(\dfrac{1}{57}+\dfrac{1}{53}\right)< 0\)
Hay \(\dfrac{1}{65}+\dfrac{1}{62}-\dfrac{1}{57}-\dfrac{1}{53}\ne0\)
\(\Rightarrow x+100=0\\ \Rightarrow x=0-100\\ \Rightarrow x=-100\)
Vậy \(x=-100\)
M
1
9 tháng 8 2017
\(\dfrac{x+1}{3}=\dfrac{3x+3}{9}\)
\(\dfrac{y+2}{-4}=\dfrac{2y+4}{-8}\)
\(\dfrac{t+3}{5}=\dfrac{4t+12}{20}\)
Do đó : \(\dfrac{3x+3}{9}=\dfrac{2y+4}{-8}=\dfrac{4t+12}{20}=\dfrac{3x+2y+4t+19}{21}=\dfrac{47+19}{21}=\dfrac{22}{7}\)
Suy ra : x = \(\dfrac{59}{7}\); y = \(-\dfrac{102}{7}\); z = \(\dfrac{89}{7}\)
30 tháng 8 2017
\(\dfrac{x-1}{50}+\dfrac{x-2}{49}=\dfrac{x-3}{48}+\dfrac{x-4}{47}\)
\(\Rightarrow\dfrac{x-1}{50}-1+\dfrac{x-2}{49}-1=\dfrac{x-3}{48}-1+\dfrac{x-4}{47}-1\)
\(\Rightarrow\dfrac{x-51}{50}+\dfrac{x-51}{49}=\dfrac{x-51}{48}+\dfrac{x-51}{47}\)
\(\Rightarrow\dfrac{x-51}{50}+\dfrac{x-51}{49}-\dfrac{x-51}{48}-\dfrac{x-51}{47}=0\)
\(\Rightarrow\left(x-51\right)\left(\dfrac{1}{50}+\dfrac{1}{49}-\dfrac{1}{48}-\dfrac{1}{47}\right)=0\)
Vì \(\dfrac{1}{50}+\dfrac{1}{49}-\dfrac{1}{48}-\dfrac{1}{47}\ne0\) nên \(x-51=0\Rightarrow x=51\)
\(\dfrac{x+25}{6}+\dfrac{x+20}{11}+\dfrac{x+16}{15}+3=0\)
\(\Rightarrow\dfrac{x+25}{6}+1+\dfrac{x+20}{11}+1+\dfrac{x+16}{15}+1=0\)
\(\Rightarrow\dfrac{x+31}{6}+\dfrac{x+31}{11}+\dfrac{x+31}{15}=0\)
\(\Rightarrow\left(x+31\right)\left(\dfrac{1}{6}+\dfrac{1}{11}+\dfrac{1}{15}\right)=0\)
Vì \(\dfrac{1}{6}+\dfrac{1}{11}+\dfrac{1}{15}\ne0\) nên \(x+31=0\Rightarrow x=-31\)
\(\dfrac{x-15}{6}+\dfrac{x-10}{11}=\dfrac{x-3}{18}+\dfrac{x-7}{14}\)
\(\Rightarrow\dfrac{x-15}{6}-1+\dfrac{x-10}{11}-1=\dfrac{x-3}{18}-1+\dfrac{x-7}{14}-1\)
\(\Rightarrow\dfrac{x-21}{6}+\dfrac{x-21}{11}=\dfrac{x-21}{18}+\dfrac{x-21}{14}\)
\(\Rightarrow\dfrac{x-21}{6}+\dfrac{x-21}{11}-\dfrac{x-21}{18}-\dfrac{x-21}{14}=0\)
\(\Rightarrow\left(x-21\right)\left(\dfrac{1}{6}+\dfrac{1}{11}-\dfrac{1}{18}-\dfrac{1}{14}\right)=0\)
Vì \(\dfrac{1}{6}+\dfrac{1}{11}-\dfrac{1}{18}-\dfrac{1}{14}\ne0\) nên \(x-21=0\Rightarrow x=21\)
NM
0
HV
1
HG
2
24 tháng 7 2015
\(\frac{x+1}{3}=\frac{y+2}{-4}=\frac{z-3}{5}=\frac{3x+3}{9}=\frac{2y+4}{-8}=\frac{4z-12}{20}=\frac{3x+3+2y+4+4z-12}{-8+9+20}=\frac{42}{21}=2\)
=>x+1=6=>x=5
y+2=2.(-4)=-8=>y=-10
z-3=10=>x=13
vậy x=5;y=-10;z=13
MT
24 tháng 7 2015
áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{x+1}{3}=\frac{y+2}{-4}=\frac{z-3}{5}=\frac{3.\left(x+1\right)+2.\left(y+2\right)+4.\left(z-3\right)}{3.3+2.\left(-4\right)+4.5}\)
\(=\frac{3x+3+2y+4+4z-12}{9-8+20}=\frac{\left(3x+2y+4z\right)+\left(3+4-12\right)}{21}\)
\(=\frac{47-5}{21}=2\)
suy ra: \(\frac{x+1}{3}=2\Rightarrow x+1=6\Rightarrow x=5\)
\(\frac{x+2}{-4}=2\Rightarrow x+2=-8\Rightarrow x=-6\)
\(\frac{z-3}{5}=2\Rightarrow z-3=10\Rightarrow z=13\)
NP
2
1 tháng 10 2016
a) \(4x^3+15=47\)
\(\Rightarrow4x^3=32\)
\(\Rightarrow x^3=8\)
\(\Rightarrow x^3=2^3\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
b) \(4.2^x-3=125\)
\(\Rightarrow4.2^x=128\)
\(\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
Vậy \(x=5\)
1 tháng 10 2016
a ) \(4x^3+15=47\)
\(\Leftrightarrow4x^3=32\)
\(\Leftrightarrow x^3=8\)
\(\Leftrightarrow x^3=2^3\)
\(\Leftrightarrow x=3\)
\(4.2^x-3=125\)
\(\Leftrightarrow4.2^x=128\)
\(\Leftrightarrow2^x=32\)
\(\Leftrightarrow2^x=2^5\)
\(\Leftrightarrow x=5\)
\(0,\left(47\right)=\dfrac{47}{x}\\ \Leftrightarrow\dfrac{47}{99}=\dfrac{47}{x}\\ \Leftrightarrow x=99\)
=>47/x=47/99
hay x=99