Tìm x \(\left(x^2-9\right)=12x+1\)
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HB
1
14 tháng 3 2018
pt <=> x^4-18x^2+81-12x=1
<=> x^4-18x^2-12x+80 = 0
<=> (x^4-4x^2)-(14x^2-28x)-(40x-80) = 0
<=> (x-2).(x^3+2x^2-14x-40) = 0
<=> (x-2).[(x^3-4x^2)+(6x^2-24x)+(10x-40)] = 0
<=> (x-2).(x-4).(x^2+6x+10) = 0
<=> (x-2).(x-4) = 0 ( vì x^2+6x+10 > 0 )
<=> x-2=0 hoặc x-4=0
<=> x=2 hoặc x=4
Vậy S={2;4}
Tk mk nha
28 tháng 9 2019
a)\(\left(x-1\right)^3+3\left(x+1\right)^2=\left(x^2-2x+4\right)\left(x+2\right)\)
\(\Leftrightarrow x^3-3x^2+3x-1+3\left(x^2+2x+1\right)=x^3+8\)
\(\Leftrightarrow-3x^2+3x+3x^2+6x+3=9\)
\(\Leftrightarrow9x=6\Leftrightarrow x=\frac{2}{3}\)
b) \(x^2-4=8\left(x-2\right)\)
\(\Leftrightarrow x^2-4=8x-16\)
\(\Leftrightarrow x^2-8x+12=0\)
\(\Leftrightarrow x^2-2x-6x+12=0\)
\(\Leftrightarrow x\left(x-2\right)-6\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=2\end{cases}}\)
28 tháng 9 2019
c) \(x^2-4x+4=9\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)^2=9\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)^2-9\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-11\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-11=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=11\end{cases}}\)
d) \(4x^2-12x+9=\left(5-x\right)^2\)
\(\Leftrightarrow\left(2x-3\right)^2=\left(5-x\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=5-x\\2x-3=x-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{8}{3}\\x=-2\end{cases}}\)
19 tháng 4 2020
b, \(\frac{1}{x\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+9\right)}=\frac{1}{3}\left(27-\frac{1}{x+9}\right)\) (ĐKXĐ: x \(\ne\) 0; x \(\ne\) -3; x \(\ne\) -6; x \(\ne\) -9)
\(\Leftrightarrow\) \(\frac{1}{3}\)(\(\frac{1}{x}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+9}\)) = \(\frac{1}{3}\)(27 - \(\frac{1}{x+9}\))
\(\Leftrightarrow\) \(\frac{1}{3}\)(\(\frac{1}{x}-\frac{1}{x+9}\)) = \(\frac{1}{3}\)(27 - \(\frac{1}{x+9}\))
\(\Leftrightarrow\) \(\frac{1}{3}\)(\(\frac{1}{x}-\frac{1}{x+9}\)) - \(\frac{1}{3}\)(27 - \(\frac{1}{x+9}\))
\(\Leftrightarrow\) \(\frac{1}{3}\)(\(\frac{1}{x}-\frac{1}{x+9}-27+\frac{1}{x+9}\)) = 0
\(\Leftrightarrow\) \(\frac{1}{3}\)(\(\frac{1}{x}-27\)) = 0
\(\Leftrightarrow\) \(\frac{1}{x}-27\) = 0
\(\Leftrightarrow\) x = \(\frac{1}{27}\) (TM ĐKXĐ)
Vậy S = {\(\frac{1}{27}\)}
Chúc bn học tốt!!
19 tháng 4 2020
a, \(\frac{5x-3}{50x^2-2}+\frac{5x-9}{12x-60x^2}+\frac{1}{12x}=\frac{8x-5}{80x^2+16x}\) (ĐKXĐ: x \(\ne\) \(\pm\)\(\frac{1}{5}\); x \(\ne\) 0)
\(\Leftrightarrow\) \(\frac{5x-3}{2\left(5x-1\right)\left(5x+1\right)}+\frac{-5x+9}{12x\left(5x-1\right)}+\frac{1}{12x}=\frac{8x-5}{16x\left(5x+1\right)}\)
\(\Leftrightarrow\) \(\frac{24x\left(5x-3\right)\left(5x+1\right)}{48x\left(5x-1\right)\left(5x+1\right)}+\frac{-4\left(5x+1\right)\left(5x-9\right)}{48x\left(5-1x\right)\left(5x+1\right)}+\frac{4\left(5x-1\right)\left(5x+1\right)}{48x\left(5x-1\right)\left(5x+1\right)}=\frac{3\left(8x-5\right)\left(5x-1\right)}{48x\left(5x-1\right)\left(5x+1\right)}\)
\(\Leftrightarrow\) 24x(5x - 3) - 4(5x + 1)(5x - 9) + 4(5x - 1)(5x + 1) = 3(8x - 5)(5x - 1)
\(\Leftrightarrow\) 120x2 - 72x - 100x2 + 160x + 36 + 100x2 - 4 = 120x2 - 99x + 15
\(\Leftrightarrow\) 120x2 - 120x2 - 100x2 + 100x2 - 72x + 160x + 99x = 15 - 36 + 4
\(\Leftrightarrow\) 187x = -17
\(\Leftrightarrow\) x = \(\frac{-1}{11}\) (TM ĐKXĐ)
Vậy S = {\(\frac{-1}{11}\)}
Chúc bn học tốt!! (Đã được kiểm chứng không sai :)
24 tháng 7 2016
a) 4x2-12x=9
<=> 4x(x-3)=9
<=> 4x=9 hoặc x-3=9
=> x=4/9 => x=12
b) 3.(x2-4)-5x(x+2)=0
<=> 3(x-2)(x+2)-5x(x+2)=0
<=> (x+2)(3x-6-5x)=0
<=> (x+2)(-2x-6)=0
<=> x+2=0 hoặc -2x-6=0
=> x=-2 => x=-3
31 tháng 10 2021
a: \(=\dfrac{\left(x^4-y^4\right)^2}{x^2+y^2}=\left(x^2-y^2\right)^2\cdot\left(x^2+y^2\right)\)
b: \(=\dfrac{\left(4x+3\right)\left(16x^2-12x+9\right)}{16x^2-12x+9}=4x+3\)
30 tháng 3 2017
\(\left(x^2+1\right)^2+3x\left(x^2+1\right)+2x^2=0\)
\(\Leftrightarrow\left(x^2+1\right)^2+2\left(x^2+1\right)^2\frac{3x}{2}+\frac{9x^2}{4}-\frac{x^2}{4}=0\)
\(\Leftrightarrow\left(x^2+1+\frac{3x}{2}\right)^2-\left(\frac{x}{2}\right)^2=0\)
\(\Leftrightarrow\left(x^2+1+\frac{3x}{2}-\frac{x}{2}\right)\left(x^2+1+\frac{3x}{2}+\frac{x}{2}\right)=0\)
\(\Leftrightarrow\left(x^2+x+1\right)\left(x^2+2x+1\right)=0\)
\(\forall x,\)\(x^2+x+1=x^2+2x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\)
\(\Rightarrow x^2+2x+1=0\)
\(\Leftrightarrow\left(x+1\right)^2=0\)
\(\Leftrightarrow x=-1\)
Vậy tập nghiệm của pt là S={-1}
23 tháng 10 2016
a) \(4x^2-12x=-9\)
\(\Leftrightarrow4x^2-12x+9=0\)
\(\Leftrightarrow\left(2x-3\right)^2=0\)
\(\Leftrightarrow2x-3=0\Leftrightarrow x=\frac{3}{2}\)
b) \(\left(5-2x\right)\left(2x+7\right)=4x^2-25\)
\(\Leftrightarrow\left(5-2x\right)\left(2x+7\right)+\left(25-4x^2\right)=0\)
\(\Leftrightarrow\left(5-2x\right)\left(2x+7\right)+\left(5-2x\right)\left(5+2x\right)=0\)
\(\Leftrightarrow\left(5-2x\right)\left(2x+7+5+2x\right)=0\)
\(\Leftrightarrow\left(5-2x\right)\left(4x+12\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-3\end{array}\right.\)
c)\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow\left(x+3\right)x\left(x-2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-3\\x=0\\x=2\end{array}\right.\)
d) \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
\(\Leftrightarrow\left[2\left(2x+7\right)-3\left(x+3\right)\right]\left[2\left(2x+7\right)+3\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-5\\x=-\frac{23}{17}\end{array}\right.\)
\(\left(x^2-9\right)=12x+1\)
\(\Leftrightarrow\)\(x^2-9-12x-1=0\)
\(\Leftrightarrow\)\(x^2-12x-10=0\)
\(\Leftrightarrow\)\(x^2-12x+36-46=0\)
\(\Leftrightarrow\)\(\left(x-6\right)^2=46\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-6=\sqrt{46}\\x-6=-\sqrt{46}\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\sqrt{46}+6\\x=-\sqrt{46}+6\end{cases}}\)
Vậy..
P/S: mk cx ko bít đúng hay sai nữa, bn tham khảo nha