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13 tháng 4 2020
a) Ta có: \(\left(x-3\right)\left(x-4\right)-2\left(3x-2\right)=\left(4-x\right)^2\)
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)-2\left(3x-2\right)-\left(x-4\right)^2=0\)
\(\Leftrightarrow\left(x-4\right)\left[\left(x-3\right)-\left(x-4\right)\right]-2\left(3x-2\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-3-x+4\right)-6x+4=0\)
\(\Leftrightarrow x-4-6x+4=0\)
\(\Leftrightarrow-5x=0\)
mà -5<0
nên x=0
Vậy: x=0
L
20 tháng 1 2019
a) <=> \(6x^2-5x+3-2x+3x\left(3-2x\right)=0\)
<=> \(6x^2-5x+3-2x+9x-6x^2=0\)
<=> \(2x+3=0\)
<=> \(x=\frac{-3}{2}\)
b) <=> \(10\left(x-4\right)-2\left(3+2x\right)=20x+4\left(1-x\right)\)
<=> \(10x-40-6-4x=20x+4-4x\)
<=> \(6x-46-16x-4=0\)
<=> \(-10x-50=0\)
<=> \(-10\left(x+5\right)=0\)
<=> \(x+5=0\)
<=> \(x=-5\)
c) <=> \(8x+3\left(3x-5\right)=18\left(2x-1\right)-14\)
<=> \(8x+9x-15=36x-18-14\)
<=> \(8x+9x-36x=+15-18-14\)
<=> \(-19x=-14\)
<=> \(x=\frac{14}{19}\)
d) <=>\(2\left(6x+5\right)-10x-3=8x+2\left(2x+1\right)\)
<=> \(12x+10-10x-3=8x+4x+2\)
<=> \(2x-7=12x+2\)
<=> \(2x-12x=7+2\)
<=> \(-10x=9\)
<=> \(x=\frac{-9}{10}\)
e) <=> \(x^2-16-6x+4=\left(x-4\right)^2\)
<=> \(x^2-6x-12-\left(x-4^2\right)=0\)
<=> \(x^2-6x-12-\left(x^2-8x+16\right)=0\)
<=> \(x^2-6x-12-x^2+8x-16=0\)
<=> \(2x-28=0\)
<=> \(2\left(x-14\right)=0\)
<=> x-14=0
<=> x=14
3 tháng 9 2016
1) \(\frac{8xy\left(3x-1\right)^3}{12x^3\left(1-3x\right)}=-\frac{8xy\left(3x-1\right)^3}{12x^3\left(3x-1\right)}=-\frac{2y\left(3x-1\right)^2}{3x^2}\)
2) \(\frac{5x^3+5x}{x^4-1}=\frac{5x\left(x^2+1\right)}{\left(x^2+1\right)\left(x^2-1\right)}=\frac{5x}{x^2-1}\)
3) \(\frac{9-\left(x+5\right)^2}{x^2+4x+4}=\frac{\left(3-x-5\right)\left(3+x+5\right)}{\left(x+2\right)^2}=\frac{-\left(x+2\right)\left(x+8\right)}{\left(x+2\right)^2}=-\frac{x+8}{x+2}\)
3) \(\frac{32x-8x^2+2x^3}{x^3+64}=\frac{2x\left(16-4x+x^2\right)}{\left(x+4\right)\left(x^2-4x+16\right)}=\frac{2x}{x+4}\)
XO
17 tháng 8 2020
a) \(5x^2-2x\left(3x+\frac{3}{2}\right)=-x^2-3x=-x\left(x+3\right)=-3\left(3+3\right)=-18\)
b) \(3x\left(x-4y\right)-\frac{12}{5}y\left(y-5x\right)=3x^2-\frac{12}{5}y^2=3\left(x^2-\frac{4}{5}y^2\right)\)
\(=3\left(4^2-\frac{4}{5}.5^2\right)=3.\left(-4\right)=-12\)
c) \(\left(x-2\right)^2-\left(x+7\right)\left(x-7\right)=x^2-4x+4-x^2+49=-4x+53=-4.3+53=41\)
d) \(x^2+12x+36=\left(x+6\right)^2=\left(64+6\right)^2=70^2=4900\)
e) \(\left(x-3\right)^2-\left(x-4\right)\left(x+4\right)=x^2-6x+9-x^2+16=-6x+25=-6\left(-1\right)+25\)
= 31
f) \(\left(3x+2y\right)^2-4y\left(3x+y\right)=9x^2+12xy+4y^2-12xy-4y^2=9x^2=9\left(-\frac{1}{3}\right)^2=1\)
\(\frac{3x-23}{\left(3x-3\right).\left(x+4\right)}-\frac{x-3}{x^2+5x+4}\left(ĐKXĐ:x\ne1;x\ne-4\right)\)
\(=\frac{3x-23}{\left(3x-3\right).\left(x+4\right)}-\frac{x-3}{x^2+4x+x+4}\)
\(=\frac{3x-23}{\left(3x-3\right).\left(x+4\right)}-\frac{x-3}{\left(x^2+4x\right)+\left(x+4\right)}\)
\(=\frac{3x-23}{\left(3x-3\right).\left(x+4\right)}-\frac{x-3}{x.\left(x+4\right)+\left(x+4\right)}\)
\(=\frac{3x-23}{\left(3x-3\right).\left(x+4\right)}-\frac{x-3}{\left(x+1\right).\left(x+4\right)}\)
\(=\frac{\left(3x-23\right).\left(x+1\right)}{\left(3x-3\right).\left(x+1\right).\left(x+4\right)}-\frac{\left(3x-3\right).\left(x-3\right)}{\left(3x-3\right).\left(x+1\right).\left(x+4\right)}\)
\(=\frac{3x^2+3x-23x-23}{\left(3x-3\right).\left(x+1\right).\left(x+4\right)}-\frac{3x^2-9x-3x+9}{\left(3x-3\right).\left(x+1\right).\left(x+4\right)}\)
\(=\frac{3x^2+3x-23x-23}{\left(3x-3\right).\left(x+1\right).\left(x+4\right)}+\frac{-\left(3x^2-9x-3x+9\right)}{\left(3x-3\right).\left(x+1\right).\left(x+4\right)}\)
\(=\frac{3x^2+3x-23x-23-3x^2+9x+3x-9}{\left(3x-3\right).\left(x+1\right).\left(x+4\right)}\)
\(=\frac{-8x-32}{\left(3x-3\right).\left(x+1\right).\left(x+4\right)}\)
\(=\frac{-8.\left(x+4\right)}{\left(3x-3\right).\left(x+1\right).\left(x+4\right)}\)
\(=\frac{-8}{\left(3x-3\right).\left(x+1\right)}.\)