Đề: Giải
a)\(\frac{3x-2}{9x+1}<\frac{1}{3}\)
b)\(\frac{x-342}{15}+\frac{x-323}{17}+\frac{x-300}{19}+\frac{x-273}{21}=10\)
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DV
10 tháng 12 2016
= \(\left[\frac{x.\left(x+3\right)}{\left(x+3\right).\left(x^2+9\right)}+\frac{3}{x+9}\right]:\left[\frac{1}{x-3}-\frac{6x}{\left(x-3\right)\left(x^2+9\right)}\right]\) ]
\(=\frac{x+3}{x^2-9}.\frac{\left(x-3\right).\left(x^2+9\right)}{x^2+9-6x}\)
= \(\frac{\left(x-3\right).\left(x+3\right)}{\left(x-3\right)^2}\)
= \(\frac{x+3}{x-3}\)
k mik nhé. Plssss~
DT
26 tháng 6 2016
\(\left(\frac{x^2+3x}{x^3+3x^2+9x+27}\right)\): \(\left(\frac{1}{x-3}-\frac{6x}{x^3-3x^2+9x-27}\right)\)
=\(\left[\frac{x\left(x+3\right)}{x^2\left(x+3\right)+9\left(x+3\right)}\right]\):\(\left[\frac{1}{x-3}-\frac{6x}{x^2\left(x-3\right)+9\left(x-3\right)}\right]\)
=\(\left[\frac{x\left(x-3\right)}{\left(x^2+9\right)\left(x-3\right)}\right]\):\(\left[\frac{1}{x-3}-\frac{6x}{\left(x^2+9\right)\left(x-3\right)}\right]\)
=\(\frac{x}{x^2+9}\):\(\left[\frac{x^2+9}{\left(x-3\right)\left(x^2+9\right)}-\frac{6x}{\left(x-3\right)\left(x^2+9\right)}\right]\)
=\(\frac{x}{x^2+9}\):\(\frac{\left(x-3\right)^2}{\left(x-3\right)\left(x^2+9\right)}\)
=\(\frac{x}{x^2+9}\):\(\frac{x-3}{x^2+9}\)
=\(\frac{x}{x^2+9}\).\(\frac{x^2+9}{x-3}\)
=\(\frac{x}{x-3}\)
26 tháng 4 2021
a, \(\left|2x-1\right|-7=0\Leftrightarrow\left|2x-1\right|=7\)
Với \(x\ge\frac{1}{2}\)phương trình có dạng :
\(2x-1=7\Leftrightarrow x=4\)( tm )
Với \(x< \frac{1}{2}\)phương trình có dạng :
\(-2x+1=7\Leftrightarrow x=-3\)( tm )
Vậy tập nghiệm của phương trình là S = { -3 ; 4 }
26 tháng 4 2021
b, \(\frac{9x^2}{2\left(1-9x^2\right)}=\frac{3x}{6x-2}-\frac{1+9x}{3+9x}\)ĐK : \(x\ne\pm\frac{1}{3}\)
\(\Leftrightarrow-\frac{9x^2}{2\left(3x-1\right)\left(3x+1\right)}=\frac{3x}{2\left(3x-1\right)}-\frac{1+9x}{3\left(3x+1\right)}\)
\(\Leftrightarrow\frac{-27x^2}{6\left(3x-1\right)\left(3x+1\right)}=\frac{9x\left(3x+1\right)}{6\left(3x-1\right)\left(3x+1\right)}-\frac{2\left(1-9x\right)\left(3x+1\right)}{6\left(3x-1\right)\left(3x+1\right)}\)
\(\Leftrightarrow-27x^2=27x^2-9x-2\left(3x-27x^2\right)\)
\(\Leftrightarrow108x^2-15x=0\Leftrightarrow3x\left(36x-5\right)=0\Leftrightarrow x=0;x=\frac{5}{36}\)( tm )
Vậy tập nghiệm của phương trình là S = { 0 ; 5/36 }
TT
0
KR
2
5 tháng 5 2019
ĐKXĐ : \(x\ne\pm\frac{2}{3}\)
Suy ra : \(\left(3x+2\right)^2-6\left(3x-2\right)=9x^2\)
\(\Leftrightarrow9x^2+12x+4-18x+12=9x^2\)
\(\Leftrightarrow-6x=-16\)
\(\Leftrightarrow x=\frac{8}{3}\left(TM\right)\)
Vậy pt có tập nghiệm là \(S=\left\{\frac{8}{3}\right\}\)
NV
18 tháng 4 2017
\(1.\frac{7x-3}{x-1}=\frac{2}{3}\) ( \(x\ne1\))
\(\Leftrightarrow\frac{3\left(7x-1\right)}{3\left(x-1\right)}=\frac{2\left(x-1\right)}{3\left(x-1\right)}\)
\(\Rightarrow3\left(7x-3\right)=2\left(x-1\right)\)
\(\Leftrightarrow21x-9=2x-2\)
\(\Leftrightarrow19x=7\)
\(\Leftrightarrow x=\frac{7}{19}\)
\(2.\frac{5x-1}{3x+2}=\frac{5x-7}{3x-1}\)
\(\Leftrightarrow\frac{\left(5x-1\right)\left(3x-1\right)}{\left(3x+2\right)\left(3x-1\right)}=\frac{\left(5x-7\right)\left(3x+2\right)}{\left(3x-1\right)\left(3x+2\right)}\)
\(\Rightarrow\left(5x-1\right)\left(3x-1\right)=\left(5x-7\right)\left(3x+2\right)\)
\(\Leftrightarrow15x^2-5x-3x+1=15x^2+10x-21x-14\)
\(\Leftrightarrow15x^2-8x+1=15x^2-11x-14\)
\(\Leftrightarrow\left(15x^2-15x^2\right)+\left(-8x+11x\right)=-14-1\)
\(\Leftrightarrow3x=-15\)
\(\Leftrightarrow x=-5\)
\(3.\frac{1-x}{x+1}+3=\frac{2x+3}{3x-1}\)
\(\Leftrightarrow\frac{\left(1-x\right)\left(3x-1\right)}{\left(x+1\right)\left(3x-1\right)}+\frac{3\left(x+1\right)\left(3x-1\right)}{\left(x+1\right)\left(3x-1\right)}=\frac{\left(2x+3\right)\left(x+1\right)}{\left(3x-1\right)\left(0+1\right)}\)
\(\Rightarrow\left(1-x\right)\left(3x-1\right)+3\left(x+1\right)\left(3x-1\right)=\left(2x+3\right)\left(x+1\right)\)
\(\Leftrightarrow3x-1-3x^2+x+3\left(3x^2-x+3x-1\right)=2x^2+2x+3x+3\)
\(\Leftrightarrow3x-1-3x^2+x+9x^2-3x+9x-3=2x^2+2x+3x+3\)
\(\Leftrightarrow6x^2+10x-4=2x^2+5x+3\)
\(\Leftrightarrow\left(6x^2-2x^2\right)+\left(10x-5x\right)=7\)
\(\Leftrightarrow4x^2+5x-7=0\)
\(\Leftrightarrow\left(2x\right)^2+4x.\frac{5}{4}+\frac{16}{25}+\frac{191}{25}=0\)
\(\Leftrightarrow\left(2x+\frac{5}{4}\right)^2-\frac{191}{25}=0\)
\(\left(2x+\frac{5}{4}\right)^2>0\)
\(\Rightarrow\left(2x+\frac{5}{4}\right)^2+\frac{191}{25}>0\)
=> PT vô nghiệm
\(4.\frac{1-6x}{x-2}+\frac{9x+4}{x+2}=\frac{x\left(3x-2\right)+1}{x^2-4}\)
\(\Leftrightarrow\frac{\left(1-6x\right)\left(x+2\right)}{x^2-4}+\frac{\left(9x+4\right)\left(x-2\right)}{x^2-4}=\frac{2\left(3x-2\right)+1}{x^2-4}\)
\(\Rightarrow\left(1-6x\right)\left(x+2\right)+\left(9x+4\right)\left(x-2\right)=3\left(3x-2\right)+1\)
\(\Leftrightarrow x+2-6x^2-12x+9x^2-18x+4x-8=3x^2-2x+1\)
\(\Leftrightarrow3x^2-25x-6=3x^2-2x+1\)
\(\Leftrightarrow\left(3x^2-3x^2\right)+\left(-25x+2x\right)+\left(-6-1\right)=0\)
\(\Leftrightarrow-23x-7=0\)
\(\Leftrightarrow-23x=7\)
\(\Leftrightarrow x=\frac{-7}{23}\)
\(5.\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4}\)
\(\Leftrightarrow\frac{\left(3x+2\right)^2}{9x^2-4}-\frac{6\left(3x-2\right)}{9x^2-4}=\frac{9x^2}{9x^2-4}\)
\(\Rightarrow\left(3x+2\right)^2-6\left(3x-2\right)=9x^2\)
\(\Leftrightarrow9x^2+12x+4-18x+12=9x^2\)
\(\Leftrightarrow\left(9x^2-9x^2\right)+\left(12x-18x\right)+\left(4+12\right)=0\)
\(\Leftrightarrow-6x+16=0\)
\(\Leftrightarrow-6x=-16\)
\(\Leftrightarrow x=\frac{16}{6}\)
\(6.1+\frac{1}{x+2}=\frac{12}{8-x^3}\)
\(\Leftrightarrow\frac{\left(x+2\right)\left(8-x^3\right)}{\left(x+2\right)\left(8-x^3\right)}+\frac{1\left(8-x^3\right)}{\left(x+2\right)\left(8-x^3\right)}=\frac{12\left(x+2\right)}{\left(x+2\right)\left(8-x^3\right)}\)
\(\Rightarrow\left(x+2\right)\left(8-x^3\right)+1\left(8-x^3\right)=12\left(x+2\right)\)
\(\Leftrightarrow8x+x^4+16+2x^3+8-x^3=12x+24\)
\(\Leftrightarrow x^4+\left(2x^3-x^3\right)+\left(8x-12x\right)+\left(16-24\right)=0\)
\(\Leftrightarrow x^4+x^3-4x-8=0\)
\(\Leftrightarrow\left(x^4-4x\right)+\left(x^3-8\right)=0\)
Đến đấy mk tắc r xl bạn nhé