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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 \(x-21=0\Rightarrow x=21\)
LS
4
19 tháng 8 2016
Ta có : \(3x=4y;\frac{x}{4}=\frac{y}{3}\)
\(4y=5z;\frac{y}{5}=\frac{z}{4}\)
Qui đồng : \(\frac{x}{4}=\frac{y}{3};\frac{y}{5}=\frac{y}{4}\)
\(\frac{x}{20}=\frac{y}{15}=\frac{z}{12}\)
Áp dụng tích chất của dãy tỉ số bằng nhau , ta có :
\(\frac{x}{20}=\frac{y}{15}=\frac{z}{12};\frac{x+y+z}{20+15+12}=\frac{47}{47}=1\)
\(\Leftrightarrow\frac{x}{20}=1\Rightarrow x=20\)
\(\Leftrightarrow\frac{b}{15}=1\Rightarrow b=15\)
\(\Rightarrow\frac{z}{12}=1\Rightarrow z=12\)
19 tháng 8 2016
Theo đề bài, ta có:
3x=4y=5z và x+y+z=47
- \(3x=4y\Rightarrow\frac{x}{4}=\frac{y}{3}\)
- \(4y=5z\Rightarrow\frac{y}{5}=\frac{z}{4}\)
\(\Leftrightarrow\frac{x}{4}=\frac{y}{3};\frac{y}{5}=\frac{z}{4}\)
\(\Leftrightarrow\frac{x}{20}=\frac{y}{15};\frac{y}{15}=\frac{z}{12}\)
Áp dụng tính chất của dãy tỉ số bằng nhau:
\(\frac{x}{20}=\frac{y}{15}=\frac{z}{12}=\frac{x+y+z}{20+15+12}=\frac{47}{47}=1\)
- \(\frac{x}{20}=1.20=20\)
- \(\frac{y}{15}=1.15=15\)
- \(\frac{z}{12}=1.12=12\)
Vậy x=20,y=15,z=12
^...^ 
^_^
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 = \(\dfrac{89}{7}\)
NP
4
7 tháng 9 2017
\(\left(2-3x\right)^2+11=47\)
\(\Leftrightarrow\left(2-3x\right)^2=36\)
\(\Leftrightarrow\orbr{\begin{cases}\left(2-3x\right)^2=6^2\\\left(2-3x\right)^2=\left(-6\right)^2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2-3x=6\\2-3x=-6\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x=-4\\3x=8\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{4}{3}\\x=\frac{8}{3}\end{cases}}\)
7 tháng 9 2017
(2 - 3x)2 + 11 = 47
=> (2 - 3x)2 = 47 - 11 = 36
=> (2 - 3x)2 = 62
=> 2 - 3x = 6
=> 3x = 2 - 6 = -4
=> x = -4 : 3
=> x = \(-\frac{4}{3}\)
16 tháng 5 2017
a/ \(5\dfrac{5}{47}+\dfrac{37}{53}+2,7-\dfrac{5}{47}+\dfrac{6}{53}\)
= \(\dfrac{240}{47}-\dfrac{5}{47}+\dfrac{37}{53}+\dfrac{6}{53}+2,7\)
=\(\left(\dfrac{240}{47}-\dfrac{5}{47}\right)+\left(\dfrac{37}{53}+\dfrac{6}{53}\right)+2,7\)
= 5 + \(\dfrac{43}{53}\) + 2,7 = \(\dfrac{4511}{530}\)
b/ \(42\dfrac{1}{6}:\left(-1\dfrac{3}{5}\right)-52\dfrac{1}{6}:\left(-1\dfrac{3}{5}\right)\)
= \(\left(42\dfrac{1}{6}-52\dfrac{1}{6}\right):\left(-1\dfrac{3}{5}\right)\)
= \(-10:\left(-1\dfrac{3}{5}\right)\) =25/4
FT
10 tháng 11 2017
\(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}+\dfrac{51-x}{49}=-5\)
\(\Rightarrow\dfrac{59-x}{41}+1+\dfrac{57-x}{43}+1+\dfrac{55-x}{45}+1+\dfrac{53-x}{47}+1+\dfrac{51-x}{49}+1=0\)\(\Rightarrow\dfrac{100-x}{41}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}+\dfrac{100-x}{49}=0\)
\(\Rightarrow\left(100-x\right)\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\right)=0\)
\(\Rightarrow100-x=0\Rightarrow x=100\)
PT
11 tháng 8 2017
\(\dfrac{21}{47}+\dfrac{9}{45}+\dfrac{26}{47}+\dfrac{4}{5}\)
\(=\left(\dfrac{21}{47}+\dfrac{26}{47}\right)+\left(\dfrac{9}{45}+\dfrac{4}{5}\right)\)
\(=1+1\)
\(=2\)
8 tháng 10 2017
\(a.\)
\(A=5\dfrac{4}{23}.27\dfrac{3}{47}+5\dfrac{4}{23}.\left(-4\dfrac{3}{47}\right)\)
\(A=5\dfrac{4}{23}\left(27\dfrac{3}{47}-4\dfrac{3}{47}\right)\)
\(A=5\dfrac{4}{23}\left(27-4\right)\)
\(A=5\dfrac{4}{23}.23\)
\(A=119\)
\(b.\)
\(B=2^3+3.1-2^{-2}.4+\left(-2^2:\dfrac{1}{2}\right).8\)
\(B=2^3+3-\dfrac{1}{4}.4+\left(-8\right).8\)
\(B=2^3+3-1-64\)
\(B=-54\)
\(0,\left(47\right)=\dfrac{47}{x}\\ \Leftrightarrow\dfrac{47}{99}=\dfrac{47}{x}\\ \Leftrightarrow x=99\)
=>47/x=47/99
hay x=99