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ND
1
MT
30 tháng 6 2015
a). (x - 2) (x - 3) + (x - 2) - 1 =0
<=>(x-2)(x-3+1)-1=0
<=>(x-2)(x-2)-1=0
<=>(x-2)2-1=0
<=>(x-2-1)(x-2+1)=0
<=>(x-3)(x-1)=0
<=>x-3=0 hoặc x-1=0
<=>x=3 hoặc x=1
vậy S={3;1}
b). 6x^3 + x^2 = 2x
<=>6x3+x2-2x=0
<=>x(6x2+x-2)=0
<=>x(6x2-3x+4x-2)=0
<=>x[3x(2x-1)+2(2x-1)]=0
<=>x(2x-1)(3x+2)=0
<=>x=0 hoặc 2x-1=0 hoặc 3x-2=0
<=>x=0 hoặc x=1/2 hoặc x=2/3
vậy S={0;1/2;2/3}
N
Giải pt : a) 2/-x2+6x-8 - x-1/x-2 = x+3/x-4
b) 2/x3-x2-x+1 = 3/1-x2 - 1/x+1
c) x+2/x-2 - 2/x2-2x = 1/x
3
PN
10 tháng 5 2021
a,\(\frac{2}{-x^2+6x-8}-\frac{x-1}{x-2}=\frac{x+3}{x-4}\left(đkxđ:x\ne2;4\right)\)
\(< =>\frac{-2}{\left(x-2\right)\left(x-4\right)}-\frac{\left(x-1\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}=\frac{\left(x+3\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}\)
\(< =>-2-\left(x^2-5x+4\right)=x^2+x-5\)
\(< =>-x^2+5x-6-x^2-x+5=0\)
\(< =>-2x^2+4x-1=0\)
\(< =>2x^2-4x+1=0\)
đến đây thì pt bậc 2 dể rồi
PN
10 tháng 5 2021
\(\frac{2}{x^3-x^2-x+1}=\frac{3}{1-x^2}-\frac{1}{x+1}\left(đkxđ:x\ne\pm1\right)\)
\(< =>\frac{2}{x^2\left(x-1\right)-\left(x-1\right)}=\frac{3}{1-x^2}-\frac{1}{x+1}\)
\(< =>\frac{2}{\left(x^2-1\right)\left(x-1\right)}=-\frac{3}{x^2-1}-\frac{1}{x+1}\)
\(< =>\frac{2}{\left(x+1\right)\left(x-1\right)^2}=\frac{-3\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)}\)
\(< =>2+3x-3+x^2-2x+1=0\)
\(< =>x^2+x=0< =>x\left(x+1\right)=0< =>\orbr{\begin{cases}x=-1\left(loai\right)\\x=0\left(tm\right)\end{cases}}\)
8 tháng 4 2020
\(\left(x-1\right)^2-\left(x+1\right)^2=2\left(x+3\right)\)
\(\Leftrightarrow\left(x-1+x+1\right)\left(x-1-x-1\right)=2\left(x+3\right)\)
\(\Leftrightarrow2x\left(-2\right)=2\left(x+3\right)\)
\(\Leftrightarrow-4x=2x+6\)
\(\Leftrightarrow-6x=6\)
\(\Leftrightarrow x=-1\)
2) \(\left(2x-1\right)^2-\left(2x+1\right)^2=4\left(x-3\right)\)
\(\Leftrightarrow\left(2x-1+2x+1\right)\left(2x-1-2x-1\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow4x\left(-2\right)-4x+12=0\)
\(\Leftrightarrow-12x=-12\)
\(\Leftrightarrow x=1\)
3)\(\left(2x+3\right)^2-\left(2x+3\right)\left(2x-4\right)+\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(2x+3\right)\left(2x+3-2x+4\right)+\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow7\left(2x+3\right)+x^2-4x+4=0\)
\(\Leftrightarrow x^2+10x+25=0\)
\(\Leftrightarrow\left(x+5\right)^2=0\)
\(\Leftrightarrow x=-5\)
4) \(8x^3-\left(x+1\right)^3=3x-3\)
\(\Leftrightarrow8x^3-\left(x^3+3x+3x^2+1\right)-3x+3=0\)
\(\Leftrightarrow7x^3-3x^2-6x+2=0\)
\(\Leftrightarrow\left(x-1\right)\left(7x^2+4x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{-2+3\sqrt{2}}{7}\\x=\frac{-2-3\sqrt{2}}{7}\end{matrix}\right.\)
5)\(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)
\(\Leftrightarrow\left(3x\right)^3-2^3-\left(\left(3x\right)^3-1^3\right)=x-4\)
\(\Leftrightarrow27x^3-8-\left(27x^3-1\right)=x-4\)
\(\Leftrightarrow-7=x-4\)
\(\Leftrightarrow x=-3\)
N
1
30 tháng 6 2022
\(\Leftrightarrow\dfrac{x^2+2x+1-1}{x+1}+\dfrac{x^2+8x+16+4}{x+4}=\dfrac{x^2+4x+4+2}{x+2}+\dfrac{x^2+6x+9+3}{x+3}\)
\(\Leftrightarrow x+1-\dfrac{1}{x+1}+x+4+\dfrac{4}{x+4}=x+2+\dfrac{2}{x+2}+x+3+\dfrac{3}{x+3}\)
\(\Leftrightarrow2x+5-\dfrac{1}{x+1}+\dfrac{4}{x+4}=2x+5+\dfrac{2}{x+2}+\dfrac{3}{x+3}\)
=>-x-4+4x+4=2x+6+3x+6
=>3x=5x+12
=>-2x=12
hay x=-6(nhận)
KN
24 tháng 2 2018
\(\dfrac{x^2+2x+2}{x+1}+\dfrac{x^2+8x+20}{x+4}=\dfrac{x^2+4x+6}{x+2}+\dfrac{x^2+6x+12}{x+3}\)\(\Leftrightarrow\)\(\dfrac{x^2+2x+1+1}{x+1}+\dfrac{x^2+8x+16+4}{x+4}=\dfrac{x^2+4x+4+2}{x+2}+\dfrac{x^2+6x+9+3}{x+3}\)
\(\Leftrightarrow\) \(\dfrac{\left(x+1\right)^2+1}{x+1}+\dfrac{\left(x+4\right)^2+4}{x+4}=\dfrac{\left(x+2\right)^2+2}{x+2}+\dfrac{\left(x+3\right)^2+3}{x+3}\)
\(\Leftrightarrow\) \(x+1+\dfrac{1}{x+1}+x+4+\dfrac{4}{x+4}=x+2+\dfrac{2}{x+2}+x+3+\dfrac{3}{x+3}\)
\(\Leftrightarrow\) \(\dfrac{1}{x+1}\) + \(\dfrac{4}{x+4}\) - \(\dfrac{2}{x+2}\) - \(\dfrac{3}{x+3}\) = x + 2 + x + 3 - x - 1 - x - 4
\(\Leftrightarrow\) \(\dfrac{1}{x+1}\) + \(\dfrac{4}{x+4}\) - \(\dfrac{2}{x+2}\) - \(\dfrac{3}{x+3}\) = 0
\(\Leftrightarrow\) \(\dfrac{1}{x+1}\) + \(\dfrac{4}{x+4}\) = \(\dfrac{2}{x+2}\) + \(\dfrac{3}{x+3}\)
\(\Leftrightarrow\) \(\dfrac{x+4}{\left(x+1\right)\left(x+4\right)}\) + \(\dfrac{4\left(x+1\right)}{\left(x+1\right)\left(x+4\right)}\) = \(\dfrac{2\left(x+3\right)}{\left(x+3\right)\left(x+2\right)}\) + \(\dfrac{3\left(x+2\right)}{\left(x+2\right)\left(x+3\right)}\)
\(\Leftrightarrow\) \(\dfrac{x+4+4x+4}{x^2+5x+4}\) = \(\dfrac{2x+6+3x+6}{x^2+5x+6}\)
\(\Leftrightarrow\) \(\dfrac{5x+8}{x^2+5x+4}\) = \(\dfrac{5x+12}{x^2+5x+6}\)
Đặt 5x + 8 = y; x2 + 5x + 4 = t, ta có:
\(\dfrac{y}{t}\) = \(\dfrac{y+4}{t+2}\)
\(\Leftrightarrow\) \(\dfrac{y\left(t+2\right)}{t\left(t+2\right)}\) = \(\dfrac{t\left(y+4\right)}{t\left(t+2\right)}\)
\(\Leftrightarrow\) yt + 2y = yt + 4t
\(\Leftrightarrow\) 2y = 4t
\(\Leftrightarrow\) 2(5x + 8) = 4(x2 + 5x + 4)
\(\Leftrightarrow\) 10x + 16 = 4x2 + 20x + 16
\(\Leftrightarrow\) 16 - 16 = 4x2 + 20x - 10x
\(\Leftrightarrow\) 0 = 4x2 + 10x
\(\Leftrightarrow\) 2x(2x + 5) = 0
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\end{matrix}\right.\)
CHÚC BN HOK TỐT...
NT
1
15 tháng 1 2018
\(\Leftrightarrow9x^2-6x+1-10x-5+12x^2+6x-6x-3=x-1\)
\(\Leftrightarrow21x^2-17x-6=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1,075\\x=-0,266\end{cases}}\)