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NT
1 tháng 1 2018
\(\dfrac{x-1}{9}+\dfrac{x-2}{8}+\dfrac{x-3}{7}=\dfrac{x-9}{1}+\dfrac{x-8}{2}+\dfrac{x-7}{3}\\ \Leftrightarrow\dfrac{x-1}{9}-1+\dfrac{x-2}{8}-1+\dfrac{x-3}{7}-1=\dfrac{x-9}{1}-1+\dfrac{x-8}{2}-1+\dfrac{x-7}{3}-1\\ \Leftrightarrow\dfrac{x-10}{9}+\dfrac{x-10}{8}+\dfrac{x-10}{7}=\dfrac{x-10}{1}+\dfrac{x-10}{2}+\dfrac{x-10}{3}\\ \Leftrightarrow\left(x-10\right)\left(\dfrac{1}{9}+\dfrac{1}{8}+\dfrac{1}{7}-1-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\Leftrightarrow x-10=0\\ \Leftrightarrow x=10\)
HD
1 tháng 1 2018
Trừ 2 vế với 1:
\(\Rightarrow\dfrac{x-1}{9}+\dfrac{x-2}{8}+\dfrac{x-3}{7}+3=\dfrac{x-9}{1}+\dfrac{x-8}{2}+\dfrac{x-7}{3}+3\)
\(\Rightarrow\left(\dfrac{x-1}{9}-1\right)+\left(\dfrac{x-2}{8}-1\right)+\left(\dfrac{x-3}{7}-1\right)=\left(\dfrac{x-9}{1}-1\right)+\left(\dfrac{x-8}{2}-1\right)+\left(\dfrac{x-7}{3}-1\right)\)
\(\Rightarrow\left(\dfrac{x-1}{9}-\dfrac{9}{9}\right)+\left(\dfrac{x-2}{8}-\dfrac{8}{8}\right)+\left(\dfrac{x-3}{7}-\dfrac{7}{7}\right)=\left(\dfrac{x-9}{1}-\dfrac{1}{1}\right)+\left(\dfrac{x-8}{2}-\dfrac{2}{2}\right)+\left(\dfrac{x-7}{3}-\dfrac{3}{3}\right)\)
\(\Rightarrow\dfrac{x-10}{9}+\dfrac{x-10}{8}+\dfrac{x-3}{7}=\dfrac{x-10}{1}+\dfrac{x-10}{2}+\dfrac{x-10}{3}\)
\(\Rightarrow\dfrac{x-10}{9}+\dfrac{x-10}{8}+\dfrac{x-10}{7}-\dfrac{x-10}{1}-\dfrac{x-10}{2}-\dfrac{x-10}{3}\)
\(\Rightarrow\left(x-10\right)\left(\dfrac{1}{9}+\dfrac{1}{8}+\dfrac{1}{7}-1-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)
\(\Rightarrow\left(x-10\right)=0\)
\(\Rightarrow x=10\)
14 tháng 12 2018
a) \(\dfrac{x}{x-3}+\dfrac{9-6x}{x^2-3x}=\dfrac{x^2}{x\left(x-3\right)}+\dfrac{9-6x}{x\left(x-3\right)}=\dfrac{x^2-6x+9}{x\left(x-3\right)}=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}=\dfrac{x-3}{x}\)
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 3 2019
a/ ĐKXĐ: \(x\ne2;3\)
\(\dfrac{x+3}{x-2}+\dfrac{5}{\left(x-2\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\dfrac{\left(x+3\right)\left(x-3\right)+5}{\left(x-2\right)\left(x-3\right)}=0\)
\(\Leftrightarrow x^2-9+5=0\Leftrightarrow x^2=4\Rightarrow\left[{}\begin{matrix}x=-2\\x=2\left(l\right)\end{matrix}\right.\)
b/ ĐKXĐ: \(x\ne\pm\dfrac{3}{4}\)
\(\dfrac{12x^2+30x-21}{\left(4x-3\right)\left(4x+3\right)}+\dfrac{3x-7}{4x-3}-\dfrac{6x+5}{4x+3}=0\)
\(\Leftrightarrow12x^2+30x-21+\left(3x-7\right)\left(4x+3\right)-\left(6x+5\right)\left(4x-3\right)=0\)
\(\Leftrightarrow9x-27=0\Rightarrow x=3\)
c/ ĐKXĐ: \(x\ne-1;2\)
\(\dfrac{x+3}{\left(x+1\right)\left(x-2\right)}-\dfrac{4}{x+1}+\dfrac{2}{x-2}=0\)
\(\Leftrightarrow x+3-4\left(x-2\right)+2\left(x+1\right)=0\)
\(\Leftrightarrow-x+13=0\)
\(\Rightarrow x=13\)
31 tháng 12 2022
a: \(\Leftrightarrow4\left(6-x\right)-3x=6\left(2x+3\right)-12\)
=>24-4x-3x=12x+18-12
=>12x+6=-7x+24
=>19x=18
=>x=18/19
b: \(\Leftrightarrow-210x-6\left(x-3\right)-15x=30x+10\left(2x+1\right)\)
=>-225x-6x+18=30x+20x+10
=>-231x+18-50x-10=0
=>-281x=-8
=>x=8/281
c: \(\Leftrightarrow36-2\left(x+3\right)=-4x+1-x\)
=>36-2x-6=-5x+1
=>3x=1+6-36=5-36=-31
=>x=-31/3
d: \(\Leftrightarrow-30\left(x-3\right)+10\left(2x-7\right)=6\left(6-x\right)\)
=>-30x+90+20x-70=36-6x
=>-10x+20=36-6x
=>-4x=16
=>x=-4
23 tháng 12 2018
\(a.\dfrac{3x-2}{5}+\dfrac{x-1}{9}=\dfrac{14x-3}{15}-\dfrac{2x+1}{9}\\ \Leftrightarrow\dfrac{27x-18}{45}+\dfrac{5x-5}{45}=\dfrac{42x-9}{45}-\dfrac{10x+5}{45}\\ \Rightarrow27x-18+5x-5=42x-9-10x-5\\ \Leftrightarrow32x-23=32x-14\\ \Leftrightarrow0x=9\\ \Rightarrow Phươngtrìnhvônghiệm\\ \Rightarrow S=\phi\)
\(b.\dfrac{x+3}{2}-\dfrac{2-x}{3}-1=\dfrac{x+5}{6}\\ \Leftrightarrow\dfrac{3x-9}{6}-\dfrac{4-2x}{6}-\dfrac{6}{6}=\dfrac{x+5}{6}\\ \Rightarrow3x-9-4+2x-6=x+5\\ \Leftrightarrow5x-19=x+5\\ \Leftrightarrow4x=24\\ \Rightarrow x=6\\ \Rightarrow S=\left\{6\right\}\)
4 tháng 1 2019
\(c.\dfrac{x+5}{2010}+\dfrac{x+4}{2011}+\dfrac{x+3}{2012}+\dfrac{x+2}{2013}=-4\\ \Leftrightarrow\dfrac{x+5}{2010}+1+\dfrac{x+4}{2011}+1+\dfrac{x+3}{2012}+1+\dfrac{x+2}{2013}+1=-4+4\\ \Rightarrow\dfrac{2015+x}{2010}+\dfrac{2015+x}{2011}+\dfrac{2015+x}{2012}+\dfrac{2015+x}{2013}=0\\ \Leftrightarrow\left(2015+x\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}\right)=0\)
Do \(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}>0\)
nên \(2015+x=0\Rightarrow x=-2015\)
Câu d tương tự...thêm rồi chuyển vế sang :v
29 tháng 5 2022
a: \(=\dfrac{x}{y\left(x-y\right)}+\dfrac{2x-y}{y\left(x-y\right)}=\dfrac{x+2x-y}{y\left(x-y\right)}=\dfrac{3x-y}{y\left(x-y\right)}\)
b: \(=\dfrac{x\left(x+3\right)}{\left(x+3\right)^2}+\dfrac{3}{x-3}-\dfrac{6x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x}{x+3}+\dfrac{3}{x-3}-\dfrac{6x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x^2-3x+3x+9-6x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{x-3}{x+3}\)
c: \(=\dfrac{x+9}{\left(x-3\right)\left(x+3\right)}-\dfrac{3}{x\left(x+3\right)}\)
\(=\dfrac{x^2+9x-3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x^2+9x-3x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{x+3}{x-3}\)
d: \(=\dfrac{x^2-1-x^2+4}{x+1}=\dfrac{3}{x+1}\)
14 tháng 1 2019
a) \(\frac{6-x}{3}-\frac{x}{4}=\frac{3+2x}{2}-1\)
\(\frac{4\left(6-x\right)}{12}-\frac{3x}{12}=\frac{3+2x}{2}-\frac{2}{2}\)
\(\frac{24-4x-3x}{12}=\frac{3+2x-2}{2}\)
\(\frac{24-7x}{12}=\frac{2x+1}{2}\)
\(\Rightarrow2\left(24-7x\right)=12\left(2x+1\right)\)
\(\Rightarrow48-14x=24x+12\)
\(\Rightarrow24x+14x=48-12\)
\(\Rightarrow38x=36\)
\(\Rightarrow x=\frac{18}{19}\)
14 tháng 1 2019
b) \(-7x-\frac{x-3}{5}-\frac{x}{2}=x+\frac{2x+1}{3}\)
\(\frac{-70x}{10}-\frac{2\left(x-3\right)}{10}-\frac{5x}{10}=\frac{3x}{3}+\frac{2x+1}{3}\)
\(\frac{-70x-2x+6-5x}{10}=\frac{3x+2x+1}{3}\)
\(\frac{-77x+6}{10}=\frac{5x+1}{3}\)
\(\Rightarrow3\left(-77x+6\right)=10\left(5x+1\right)\)
\(\Leftrightarrow-231x+18=50x+10\)
\(\Leftrightarrow50x+231x=18-10\)
\(\Leftrightarrow281x=8\)
\(\Leftrightarrow x=\frac{8}{281}\)
Mấy câu kia tương tự
NN
15 tháng 11 2017
a) \(\dfrac{3}{x-4}-\dfrac{2}{4-x}\)
\(=\dfrac{3}{x-4}+\dfrac{2}{x-4}\)
\(=\dfrac{3+2}{x-4}\)
\(=\dfrac{5}{x-4}\)
b) \(\dfrac{7}{x-3}-\dfrac{4}{3-x}\)
\(=\dfrac{7}{x-3}+\dfrac{4}{x-3}\)
\(=\dfrac{7+4}{x-3}\)
\(=\dfrac{11}{x-3}\)
NN
15 tháng 11 2017
c) \(\dfrac{3}{x-5}-\dfrac{2}{x+2}\) MTC: \(\left(x-5\right)\left(x+2\right)\)
\(=\dfrac{3\left(x+2\right)}{\left(x-5\right)\left(x+2\right)}-\dfrac{2\left(x-5\right)}{\left(x-5\right)\left(x+2\right)}\)
\(=\dfrac{3\left(x+2\right)-2\left(x-5\right)}{\left(x-5\right)\left(x+2\right)}\)
\(=\dfrac{3x+6-2x+10}{\left(x-5\right)\left(x+2\right)}\)
\(=\dfrac{x+16}{\left(x-5\right)\left(x+2\right)}\)
d) \(\dfrac{9}{x-5}-\dfrac{6}{x^2-25}\)
\(=\dfrac{9}{x-5}-\dfrac{6}{\left(x-5\right)\left(x+5\right)}\) MTC: \(\left(x-5\right)\left(x+5\right)\)
\(=\dfrac{9\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\dfrac{6}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{9\left(x+5\right)-6}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{9x+45-6}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{9x+39}{\left(x-5\right)\left(x+5\right)}\)
\(\dfrac{x-7}{2010}+\dfrac{x-604}{471}+\dfrac{x-2}{403}=9\)
\(\Leftrightarrow\dfrac{x-7}{2010}-1+\dfrac{x-604}{417}-3+\dfrac{x-2}{403}-5=0\)
\(\Leftrightarrow\dfrac{x-2017}{2010}+\dfrac{x-2017}{471}+\dfrac{x-2017}{403}=0\)
\(\Leftrightarrow\left(x-2017\right)\left(\dfrac{1}{2010}+\dfrac{1}{471}+\dfrac{1}{403}\right)=0\)
Mà \(\dfrac{1}{2010}+\dfrac{1}{471}+\dfrac{1}{403}\ne0\)
\(\Rightarrow x-2017=0\Leftrightarrow x=2017\)