\(\frac{x}{2x-4}\)=\(\frac{x}{x+2}\)
Hay giai giup minh nhe!
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HP
2
LM
26 tháng 8 2016
7(2x - 5) - 5(7x - 2) + 2(5x - 7) = (x + 2) - (x + 4)
=> 14x - 35 - 35x + 10 + 10x -14 = x + 2 - x - 4
=> (14x - 35x + 10x) + (-35 + 10 - 14) = -2
=> -11x + (-39) = -2
=> -11x = -2 - (-39)
=> -11x = 37
=> x = \(\frac{-37}{11}\)
NP
Ng. Phạm Duy Anh
VIP
2 tháng 6 2023
7(2x - 5) - 5(7x - 2) + 2(5x - 7) = (x + 2) - (x + 4)
=> 14x - 35 - 35x + 10 + 10x -14 = x + 2 - x - 4
=> (14x - 35x + 10x) + (-35 + 10 - 14) = -6
=> -11x - 39 = -6
=> -11x = -6+39
=> -11x = 33
=> x = 33:(-11)
=> x = -3
KT
13 tháng 1 2019
\(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)
<=> \(\frac{x-5}{100}-1+\frac{x-4}{101}-1+\frac{x-3}{102}-1=\frac{x-100}{5}-1+\frac{x-101}{4}-1+\frac{x-102}{3}-1\)
<=> \(\frac{x-105}{100}+\frac{x-105}{101}+\frac{x-105}{102}=\frac{x-105}{5}+\frac{x-105}{4}+\frac{x-105}{3}\)
<=> \(\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)
Nhận thấy: \(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}+\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\ne0\)
=> \(x-105=0\)
<=> \(x=105\)
14 tháng 1 2019
\(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)
\(\Leftrightarrow\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}-\frac{x-100}{5}-\frac{x-101}{4}-\frac{x-102}{3}=0\)
\(\Leftrightarrow\left(\frac{x-5}{100}-1\right)+\left(\frac{x-4}{101}-1\right)+\left(\frac{x-3}{102}-1\right)-\left(\frac{x-100}{5}-1\right)-\left(\frac{x-101}{4}-1\right)-\left(\frac{x-102}{3}-1\right)=0\)
\(\Leftrightarrow\frac{x-105}{100}+\frac{x-105}{101}+\frac{x-105}{102}-\frac{x-105}{5}-\frac{x-105}{4}-\frac{x-105}{3}=0\)
\(\Leftrightarrow\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)
\(\Leftrightarrow x-105=0\left(Vì\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\ne0\right)\)
\(\Leftrightarrow x=105\)
14 tháng 2 2019
\(\frac{x-4}{2000}+\frac{x-3}{2001}+\frac{x-2}{2002}=\frac{x-2002}{2}+\frac{x-2001}{3}+\frac{x-2000}{4}\)
\(\Rightarrow\left(\frac{x-4}{2000}-1\right)+\left(\frac{x-3}{2001}-1\right)+\left(\frac{x-2}{2002}-1\right)=\left(\frac{x-2002}{2}-1\right)+\left(\frac{x-2001}{3}-1\right)+\left(\frac{x-2000}{4}-1\right)\)\(\Rightarrow\frac{x-2004}{2000}+\frac{x-2004}{2001}+\frac{x-2004}{2002}=\frac{x-2004}{2}+\frac{x-2004}{3}+\frac{x-2004}{4}\)
\(\Rightarrow\left(x-2004\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}\right)=\left(x-2004\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)\)
Với \(x-2004\ne0\)
\(\Rightarrow\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\left(KTM\right)\)
Với \(x-2004=0\)
\(\Rightarrow x=2004\)
N
20 tháng 6 2017
a) \(x-\frac{4}{5}=\frac{5}{7}\)
\(x=\frac{5}{7}+\frac{4}{5}=\frac{53}{35}\)
b) \(5x=-\frac{1}{5}\)
\(x=-\frac{1}{5}:5=-\frac{1}{25}\)
c) \(\frac{5}{3}-x=7+\frac{4}{5}\)
\(\frac{5}{3}-x=\frac{39}{5}\)
\(x=\frac{5}{3}-\frac{39}{5}=-\frac{92}{15}\)
d) \(-\frac{5}{11}+2x=\frac{7}{22}\)
\(2x=\frac{7}{22}+\frac{5}{11}\)
\(2x=\frac{17}{22}\)
\(x=\frac{17}{22}:2\)
\(x=\frac{17}{44}\)
\(x=-\frac{1}{5}:5\)
NN
20 tháng 6 2017
NÈ BẠN!!!
a) \(x-\frac{4}{5}=\frac{5}{7}\)
\(x=\frac{5}{7}+\frac{4}{5}=\frac{25}{35}+\frac{28}{35}=\frac{53}{35}\)
b) \(5x=-\frac{1}{5}+\frac{11}{5}\)
\(5x=2\)
\(x=\frac{2}{5}\)
c)\(\frac{5}{3}-x=7\)
\(x=\frac{5}{3}-7=\frac{5}{3}-\frac{21}{3}=-\frac{16}{3}\)
d) \(-\frac{5}{11}+2x=\frac{7}{22}\)
\(2x=\frac{7}{22}-\frac{-5}{11}=\frac{7}{22}-\frac{-10}{22}=\frac{17}{22}\)
\(x=\frac{17}{22}:2=\frac{17}{22}\cdot\frac{1}{2}=\frac{17}{44}\)
K CHO MÌNH NHA!!!
HT
23 tháng 11 2017
\(\left[\frac{-2}{5}x^3.\left(2x-1\right)^m+\frac{2}{5}x^{m+3}\right]:\left(\frac{-2}{5}x^3\right)\)
\(=\left[\frac{2}{5}x^3\left(2x+1\right)^m+\frac{2}{5}x^3.\left(\frac{2}{5}\right)^m\right]:\left(\frac{-2}{5}x^3\right)\)
\(=\left\{\frac{2}{5}x^3.\left[\left(2x+1\right)^m+\left(\frac{2}{5}\right)^m\right]\right\}:\left(\frac{-2}{5}x^3\right)\)
\(=\left\{\frac{2}{5}x^3.\left[2x+\frac{7}{5}\right]^m\right\}:\frac{-2}{5}x^3\)
\(=-\left(2x+\frac{7}{5}\right)^m\)
đến đây thì mình chịu
DL
2
S
6 tháng 3 2019
\(ĐK:x\ne2;x\ne0\)
\(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\Leftrightarrow\frac{\left(x+2\right)x}{x\left(x-2\right)}-\frac{x-2}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\Leftrightarrow\frac{x^2+2x-x+2}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\Leftrightarrow x^2+x+2=2\Leftrightarrow x\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x+1=0\Leftrightarrow x=-1\left(thoảmanx\right)\end{matrix}\right..Vậy:x=-1\)
6 tháng 3 2019
\(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\)
\(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)},x\ne2,x\ne0\)
\(\frac{x+2}{x-2}-\frac{1}{x}-\frac{2}{x\left(x-2\right)}=0\)
\(\frac{x\left(x+2\right)-x+2-2}{x\left(x-2\right)}=0\)
\(\frac{x^2+x}{x\left(x-2\right)}=0\)
\(\frac{x\left(x-1\right)}{x\left(x-2\right)}=0\)
\(\frac{x+1}{x-2}=0\)
x+1=0
\(\)x=-1(TM)
Xét 2 trường hợp x=0 và x khác 0
nếu x = 0 => phương trình đã đc giải
nếu x khác 0=> |x| khác 2
2x - 4 = x + 2 và khác 0
x - 4 = 2
x = 6