\(\frac{3x-2}{4}=\frac{36}{3x-2}\)
\(\frac{3x-2}{2}=\frac{x-3}{3}\)
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11 tháng 12 2019
\(a)=\frac{-2\left(x+3\right)}{x\left(1-3x\right)}.\frac{1-3x}{x\left(x+3\right)}\)
\(=\frac{-2}{x^2}\)
\(b)=\frac{\left(x+3\right)\left(x-3\right)}{x\left(x-3\right)}-\frac{x^2}{x\left(x-3\right)}+\frac{9}{x\left(x-3\right)}\)
\(=\frac{x^2-3x+3x-9-x^2+9}{x\left(x-3\right)}\)
\(=x\left(x-3\right)\)
\(c)=\frac{x+3}{\left(x-1\right)\left(x+1\right)}-\frac{1}{x\left(x+1\right)}\)
\(=\frac{\left(x+3\right).x}{x\left(x-1\right)\left(x+1\right)}-\frac{1.\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2+3x-x+1}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x\left(x+3\right)-\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x+3}{x+1}\)
# Sắp ik ngủ nên làm vậy hoi, ko chắc phần kq câu b và c đâu nha
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
KN
5 tháng 3 2020
\(\frac{3x-7}{5}=\frac{2x-1}{3}\)
\(\Leftrightarrow9x-21=10x-5\)
\(\Leftrightarrow-x=16\Leftrightarrow x=-16\)
\(\frac{4x-7}{12}-x=\frac{3x}{8}\)
\(\Leftrightarrow\frac{4x-7-12x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow\frac{-7-8x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow-56-64x=36x\)
\(\Leftrightarrow-56=100x\Leftrightarrow x=\frac{-14}{25}\)
KN
5 tháng 3 2020
\(\frac{x-2009}{1234}+\frac{x-2009}{5678}-\frac{x-2009}{197}=0\)
\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)=0\)
Vì \(\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)\ne0\)nên x - 2019 = 0
Vậy x = 2019
\(\frac{5x-8}{3}=\frac{1-3x}{2}\)
\(\Leftrightarrow10x-16=3-9x\)
\(\Leftrightarrow19x=19\Leftrightarrow x=1\)
20 tháng 7 2020
\(\frac{x^2-36}{2x+10}\cdot\frac{3}{6-x}=\frac{\left(x-6\right)\left(x+6\right)}{2x+10}\cdot\frac{3}{6-x}=-\frac{3\left(x+6\right)}{2x+10}=-\frac{3x+18}{2x+10}\)
\(\frac{x^2-4}{x^2-9}\cdot\frac{3x+9}{x+2}=\frac{\left(x-2\right)\left(x+2\right)}{\left(x+3\right)\left(x-3\right)}\cdot\frac{3\left(x+3\right)}{x+2}=\frac{3\left(x-2\right)}{x-3}\)
\(\frac{x^3-8}{5x+20}\cdot\frac{x^2+4x}{x^2+2x+4}=\frac{\left(x-2\right)\left(x^2+2x+4\right)}{5\left(x+4\right)}\cdot\frac{x\left(x+4\right)}{x^2+2x+4}=\frac{x\left(x-2\right)}{5}\)
\(\frac{4x+12}{\left(x+4\right)^2}:\frac{3x+9}{x+4}=\frac{4\left(x+3\right)}{\left(x+4\right)^2}\cdot\frac{x+4}{3\left(x+3\right)}=\frac{4}{3\left(x+4\right)}\)
18 tháng 8 2016
\(\left(8x^3-60x^2+150x-125\right)-\left(27x^3-108x^2+144x-64\right)+\left(x^3+3x^2+3x+1\right)=0\)
\(-18x^3+51x^2+9x-60=0\)
\(\left(2x-5\right)\left(x+1\right)\left(3x-4\right)=0\)
\(\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-1\\x=\frac{4}{3}\end{array}\right.\)
NN
26 tháng 11 2017
) \(\dfrac{x^3+8y^3}{2y+x}\)
\(=\dfrac{x^3+\left(2y\right)^3}{x+2y}\)
\(=\dfrac{\left(x+2y\right)\left[x^2+x.2y+\left(2y\right)^2\right]}{x+2y}\)
\(=x^2+2xy+4y^2\)
b) \(\dfrac{a-1}{2\left(a-4\right)}+\dfrac{a}{a-4}\) MTC: \(2\left(a-4\right)\)
\(=\dfrac{a-1}{2\left(a-4\right)}+\dfrac{2a}{2\left(a-4\right)}\)
\(=\dfrac{a-1+2a}{2\left(a-4\right)}\)
\(=\dfrac{3a-1}{2\left(a-4\right)}\)
c) \(\dfrac{x^3+3x^2y+3xy^2+y^3}{2x+2y}\)
\(=\dfrac{\left(x+y\right)^3}{2\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2}\)
d) \(\left(x-5\right)^2+\left(7-x\right)\left(x+2\right)\)
\(=\left(x^2-2.x.5+5^2\right)+\left(7x+14-x^2-2x\right)\)
\(=x^2-10x+25+7x+14-x^2-2x\)
\(=39-5x\)
e) \(\dfrac{3x}{x-2}-\dfrac{2x+1}{2-x}\)
\(=\dfrac{3x}{x-2}+\dfrac{2x+1}{x-2}\)
\(=\dfrac{3x+2x+1}{x-2}\)
\(=\dfrac{5x+1}{x-2}\)
h) \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x+6}{4-9x^2}\)
\(=\dfrac{1}{3x-2}-\dfrac{1}{3x+2}+\dfrac{3x+6}{9x^2-4}\)
\(=\dfrac{1}{3x-2}-\dfrac{1}{3x+2}+\dfrac{3x+6}{\left(3x-2\right)\left(3x+2\right)}\) MTC: \(\left(3x-2\right)\left(3x+2\right)\)
\(=\dfrac{3x+2}{\left(3x-2\right)\left(3x+2\right)}-\dfrac{3x-2}{\left(3x-2\right)\left(3x+2\right)}+\dfrac{3x+6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{\left(3x+2\right)-\left(3x-2\right)+\left(3x+6\right)}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{3x+2-3x+2+3x+6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{3x+10}{\left(3x-2\right)\left(3x+2\right)}\)
3x -2/4 = 36/3x-2
<=> (3x - 2 )2 = 36 × 4
<=> ( 3x -2)2 = 122
=> 3x - 2 = 12
<=> 3x = 14
<=> x = 14/3
b/ 3x-2 /2 = x-3/3
<=> (3x -2 )×3 = ( x-3) ×2
<=> 9x - 6 = 2x - 6
<=> 9x -2x = -6 +6
<=> 7x = 0
<=> x = 0
Vậy x = 0
1) 3x - 2/4 = 36/3x - 2
(3x - 2)(3x - 2) = 4.36
(3x - 2)2 = 144
(3x - 2)2 = 122
3x - 2 = -+12
3x - 2 = 12 hoặc 3x - 2 = -12
3x = 12 + 2 3x = -12 + 2
3x = 14 3x = -10
x = 14/3 x = -10/3