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PN
14 tháng 8 2020
.a, \(\frac{x+1}{999}+\frac{x+2}{998}=\frac{x+3}{997}+\frac{x+4}{996}\)
.\(< =>\frac{x+1}{999}+1+\frac{x+2}{998}+1=\frac{x+3}{997}+1+\frac{x+4}{996}+1\)
.\(< =>\frac{x+1}{999}+\frac{999}{999}+\frac{x+2}{998}+\frac{998}{998}=\frac{x+3}{997}+\frac{997}{997}+\frac{x+4}{996}+\frac{996}{996}\)
.\(< =>\frac{x+1+999}{999}+\frac{x+2+998}{998}=\frac{x+3+997}{997}+\frac{x+4+996}{996}\)
.\(< =>\frac{x+1000}{999}+\frac{x+1000}{998}-\frac{x+1000}{997}-\frac{x+1000}{996}=0\)
.\(< =>\left(x+1000\right)\left(\frac{1}{999}+\frac{1}{998}-\frac{1}{997}-\frac{1}{996}\right)=0\)
.Do \(\frac{1}{999}+\frac{1}{998}-\frac{1}{997}-\frac{1}{996}\ne0\)
.Suy ra \(x+1000=0\Leftrightarrow x=-1000\)
.b, \(\frac{x+1}{1001}+\frac{x+2}{1002}=\frac{x+3}{1003}+\frac{x+4}{1004}\)
.\(< =>\frac{x+1}{1001}-1+\frac{x+2}{1002}-1=\frac{x+3}{1003}-1+\frac{x+4}{1004}-1\)
.\(< =>\frac{x+1}{1001}-\frac{1001}{1001}+\frac{x+2}{1002}-\frac{1002}{1002}=\frac{x+3}{1003}-\frac{1003}{1003}+\frac{x+4}{1004}-\frac{1004}{1004}\)
.\(< =>\frac{x+1-1001}{1001}+\frac{x+2-1002}{1002}=\frac{x+3-1003}{1003}+\frac{x+4-1004}{1004}\)
.\(< =>\frac{x-1000}{1001}+\frac{x+1000}{1002}-\frac{x+1000}{1003}-\frac{x+1000}{1004}=0\)
.\(< =>\left(x-1000\right)\left(\frac{1}{1001}+\frac{1}{1002}-\frac{1}{1003}-\frac{1}{1004}\right)=0\)
.Do \(\frac{1}{1001}+\frac{1}{1002}-\frac{1}{1003}-\frac{1}{1004}\ne0\)
.Suy ra \(x-1000=0\Leftrightarrow x=1000\)
6 tháng 10 2017
a) \(\frac{2}{3}=\frac{-10}{x}\)
\(\Rightarrow2x=-30\)
\(\Rightarrow x=-15\)
6 tháng 10 2017
b) -2|x - 1| = \(\frac{-3}{4}\)
\(\Rightarrow\)|x - 1| = \(\frac{3}{8}\)
\(\Rightarrow\)x - 1 = \(\frac{3}{8}\)hoặc\(\frac{-3}{8}\)
\(\Rightarrow\)x = \(1\frac{3}{8}\)hoặc\(1\frac{-3}{8}\)
28 tháng 9 2016
d) \(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)
<=> \(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}-\frac{x+10}{2000}-\frac{x+11}{1999}-\frac{x+12}{1998}=0\)
<=> \(\left(\frac{x+1}{2009}+1\right)+\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)-\left(\frac{x+10}{2000}+1\right)-\left(\frac{x+11}{1999}+1\right)-\left(\frac{x+12}{1998}+1=0\right)\)
<=> \(\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)
<=>\(\left(x+2010\right).\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)
<=> x+2010 = 0 vì \(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\ne0\)
<=> x = -2010
21 tháng 7 2017
\(\frac{x}{2}=\frac{y}{3};\frac{y}{4}=\frac{z}{5}\)và x + y -z = 10
\(\frac{x}{2}=\frac{y}{3}=\frac{1}{4}.\frac{x}{2}=\frac{1}{4}.\frac{y}{3}\)\(=\frac{x}{8}=\frac{y}{12}\)
\(\frac{y}{4}=\frac{z}{5}=\frac{1}{3}.\frac{y}{4}=\frac{1}{3}.\frac{z}{5}=\frac{y}{12}=\frac{z}{15}\)
\(\Leftrightarrow\frac{x}{8}=\frac{y}{12}=\frac{z}{15}\)và x + y - z = 10
Theo tính chất dãy tỉ số bằng nhau:
\(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x+y-z}{8+12-15}=\frac{10}{5}=2\)
* \(\frac{x}{8}=2\Rightarrow x=2.8=16\)
* \(\frac{y}{12}=2\Rightarrow y=2.12=24\)
* \(\frac{z}{5}=2\Rightarrow z=2.5=10\)
Vậy...
21 tháng 7 2017
Ý mk nhầm chút xíu nhé! Cko sorry!
* \(\frac{z}{15}=2\Rightarrow z=2.15=30\)
... :( Xl
TY
20 tháng 3 2019
\(\frac{x+1}{5}+\frac{x+2}{4}=\frac{x+3}{3}+\frac{x+4}{2}\)\(\Rightarrow\left(\frac{x+1}{5}+1\right)+\left(\frac{x+2}{4}+1\right)=\left(\frac{x+3}{3}+1\right)+\left(\frac{x+4}{2}+1\right)\)
\(\Rightarrow\frac{x+1+5}{5}+\frac{x+2+4}{4}=\frac{x+3+3}{3}+\frac{x+4+2}{2}\)
\(\Rightarrow\frac{x+6}{5}+\frac{x+6}{4}-\frac{x+6}{3}-\frac{x+6}{2}=0\)
\(\Rightarrow\left(x+6\right)\cdot\frac{1}{5}+\left(x+6\right)\cdot\frac{1}{4}-\left(x+6\right)\cdot\frac{1}{3}-\left(x+6\right)\cdot\frac{1}{2}=0\)
\(\Rightarrow\left(x+6\right)\cdot\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Rightarrow x+6=0\Rightarrow x=-6\)
Các bạn nhớ cho mk nha
26 tháng 7 2016
B1:
a) \(\frac{x+4}{x+3}=\frac{x+9}{x+4}\)
-->(x+4)(x+4)=(x+3)(x+9)
\(x^2\)+4x+4x+16=\(x^2\)+9x+3x+27
\(x^2-x^2\)+4x+4x-9x-3x= - 16+27
- 4x=11
x=\(\frac{-4}{11}\)
b) \(\frac{x-5}{x+3}=\frac{x-4}{x+6}\)
-->(x-5)(x+6)=(x+3)(x-4)
\(x^2\)+6x-5x-30=\(x^2\)-4x+3x-12
\(x^2-x^2\)+6x-5x+4x-3x=30-12
2x=18
x=9
c)\(\frac{3x-1}{3x}=\frac{2x-1}{2x+1}\)
--> (3x-1)(2x+1)=3x.(2x-1)
\(6x^2\)+3x-2x-1=\(6x^2\)-3x
\(6x^2-6x^2\)+3x-2x+3x=1
4x=1
x=\(\frac{1}{4}\)
11 tháng 10 2016
a) \(\frac{x-3}{x+5}=\frac{5}{7}\)
\(\Rightarrow\left(x-3\right).7=\left(x+5\right).5\)
\(\Rightarrow7x-21=5x+25\)
\(\Rightarrow7x-5x=21+25\)
\(\Rightarrow2x=46\)
\(\Rightarrow x=23\)
Vậy \(x=23\)
11 tháng 10 2016
b) \(\frac{7}{x-1}=\frac{x+1}{9}\)
\(\Rightarrow\left(x-1\right).\left(x+1\right)=7.9\)
\(\Rightarrow\left(x-1\right)x-\left(x+1\right)=7.9\)
\(\Rightarrow x^2-x-x-1=63\)
\(\Rightarrow x^2-1=63\)
\(\Rightarrow x^2=64\)
\(\Rightarrow x=8\) hoặc \(x=-8\)
Vậy \(x=8\) hoặc \(x=-8\)
c) \(\frac{x+4}{20}=\frac{5}{x+4}\)
\(\Rightarrow\left(x+4\right)^2=100\)
\(\Rightarrow x+4=\pm10\)
+) \(x+4=10\Rightarrow x=6\)
+) \(x+4=-10\Rightarrow x=-16\)
Vậy \(x\in\left\{6;-16\right\}\)
30 tháng 1 2020
b) \(\left|5x-3\right|-x=7\)
\(\Rightarrow\left|5x-3\right|=7+x\)
\(\Rightarrow\orbr{\begin{cases}5x-3=7+x\\5x-3=-\left(7+x\right)\end{cases}\Rightarrow\orbr{\begin{cases}5x-3=7+x\\5x-3=-7-x\end{cases}\Rightarrow}\orbr{\begin{cases}5x-x=7+3\\5x+x=-7+3\end{cases}}}\)
\(\Rightarrow\orbr{\begin{cases}4x=10\\6x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{2}{3}\end{cases}}}\)
Vậy ....................
30 tháng 1 2020
Bạn ơi !!! ý A tham khảo tại link này nè :
https://h.vn/hoi-dap/question/394208.html
~ Học tốt ~
17 tháng 9 2016
\(\frac{x+1}{x-2}=\frac{3}{4}\) ( \(ĐKXĐ\) : \(x\ne2\) )
\(\Leftrightarrow\left(x+1\right).4=\left(x-2\right).3\)
\(\Leftrightarrow4x+4=3x-6\)
\(\Leftrightarrow4x-3x=-6-4\)
\(\Leftrightarrow x=-10\)
b ) \(\frac{2x-3}{x+1}=\frac{4}{7}\left(ĐKXĐ:x\ne1\right)\)
\(\Leftrightarrow7\left(2x-3\right)=4\left(x+1\right)\)
\(\Leftrightarrow14x-21=4x+4\)
\(\Leftrightarrow10x=25\)
\(\Leftrightarrow x=\frac{5}{2}\)
<=>(x-3)*(x+3)=(x-4)*(x+4)
<=>x(-3+3)=x(-4+4)
<=>x*0=x*0
=>x=0
thank bn nhìu