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31 tháng 5 2018
3) \(x^2-7x+6=0\)
\(\Leftrightarrow x^2-6x-x+6=0\)
\(\Leftrightarrow x\left(x-6\right)-\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=1\end{matrix}\right.\)
S=\(\left\{6;1\right\}\)
\(\)
8 tháng 4 2018
a) \(x^4+2x^3-12x^2-13x+42=0\)
\(\Leftrightarrow x^4+3x^3-x^3-3x^2-9x^2-27x+14x+42=0\)
\(\Leftrightarrow x^3\left(x+3\right)-x^2\left(x+3\right)-9x\left(x+3\right)+14\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^3-x^2-9x+14\right)=0\)
ND
8 tháng 4 2018
\(x^4+2x^3+5x^2+4x-12=0\)
\(\Leftrightarrow x^4-x^3+3x^3-3x^2+8x^2-8x^2+12x-12=0\)
\(\Leftrightarrow x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+3x^2+8x+12\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+2x^2+x^2+2x+6x+12\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)=0\)
Ta có:
\(x^2+x+6=x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{23}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}>0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy...........
H
3
NC
23 tháng 8 2020
a) \(\left(x^2+4x+3\right)\left(x^2-5x+6\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+3\right)\left(x-2\right)\left(x-3\right)=0\)
=> \(\orbr{\begin{cases}x+1=0\\x+3=0\end{cases}}\) hoặc \(\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-1\\x=-3\end{cases}}\) hoặc \(\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
Vậy tập nghiệm PT \(S=\left\{-3;-1;2;3\right\}\)
b) \(\left(x^2-7x+12\right)\left(x^2+8x+7\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)\left(x+1\right)\left(x+7\right)=0\)
=> \(\orbr{\begin{cases}x-3=0\\x-4=0\end{cases}}\) hoặc \(\orbr{\begin{cases}x+1=0\\x+7=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=3\\x=4\end{cases}}\) hoặc \(\orbr{\begin{cases}x=-1\\x=-7\end{cases}}\)
Vậy tập nghiệm PT \(S=\left\{-7;-1;3;4\right\}\)
A
23 tháng 8 2020
a, \(\left(x^2+4x+3\right)\left(x^2-5x+6\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+3\right)\left(x-3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1;-3\\x=3;2\end{cases}}\)
b, \(\left(x^2-7x+12\right)\left(x^2+8x+7\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-3\right)\left(x+1\right)\left(x+7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=4;3\\x=-1;-7\end{cases}}\)
13 tháng 4 2020
a) (x-1)2=2(x2-1)
<=> x2-2x+1=2x2-2
<=> x2-2x+1-2x2+2=0
<=> -x2-2x+3=0
<=> -x2+3x-x+3=0
<=> -x(x-3)-(x-3)=0
<=> (x-3)(-x-1)=0
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\-x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\-x=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3\\x=-1\end{cases}}}\)
14 tháng 3 2017
Bài 2
Ta có :
\(x^2+5x+6=\left(x+2\right)\left(x+3\right)\)
\(x^2+7x+12=\left(x+3\right)\left(x+4\right)\)
\(x^2+9x+20=\left(x+4\right)\left(x+5\right)\)
Khi đó:
\(\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}=\dfrac{3}{40}\)
=> \(\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}=\dfrac{3}{40}\)
=> \(\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}=\dfrac{3}{40}\)
=> \(\dfrac{1}{x+2}-\dfrac{1}{x+5}=\dfrac{3}{40}\)
Giải phương trình ta được x = 3
\(\frac{12}{x^2-4}-\frac{x+1}{x-2}+\frac{x+7}{x+2}=0\)
ĐKXĐ : \(x\ne\pm2\)
\(\Leftrightarrow\frac{12}{\left(x-2\right)\left(x+2\right)}-\frac{x+1}{x-2}+\frac{x+7}{x+2}=0\)
\(\Leftrightarrow\frac{12}{\left(x-2\right)\left(x+2\right)}-\frac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(x+7\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{12}{\left(x-2\right)\left(x+2\right)}-\frac{x^2+3x+2}{\left(x-2\right)\left(x+2\right)}+\frac{x^2+5x-14}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow12-\left(x^2+3x+2\right)+x^2+5x-14=0\)
\(\Leftrightarrow12-x^2-3x-2+x^2+5x-14=0\)
\(\Leftrightarrow2x-4=0\)
\(\Leftrightarrow2x=4\)
\(\Leftrightarrow x=2\)( không tmđk )
=> Phương trình vô nghiệm
\(\frac{12}{x^2-4}-\frac{x+1}{x-2}+\frac{x+7}{x+2}=0\left(đk:x\ne2;-2\right)\)
\(\Leftrightarrow\frac{12}{\left(x-2\right)\left(x+2\right)}-\frac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(x+7\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow12-\left(x^2+3x+3\right)+\left(x^2+5x-14\right)=0\)
\(\Leftrightarrow12-x^2+x^2-3x+5x-3-14=0\)
\(\Leftrightarrow2x-17+12=0\Leftrightarrow2x-5=0\Leftrightarrow x=\frac{5}{2}\left(tmđk\right)\)