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a/ \(x^2-6x+5=0\)
\(x^2-5x-x+5=0\)
\(x\left(x-5\right)-\left(x-5\right)=0\)
\(\left(x-5\right)\left(x-1\right)=0\)
\(\orbr{\begin{cases}x-5=0\rightarrow x=5\\x-1=0\rightarrow x=1\end{cases}}\)
b/\(2x^2+7x+9=0\)
?!
c/ \(4x^2-7x+3=0\)
\(4x^2-4x-3x+3=0\)
\(4x\left(x-1\right)-3\left(x-1\right)=0\)
\(\left(x-1\right)\left(4x-3\right)=0\)
\(\orbr{\begin{cases}x-1=0\Rightarrow x=1\\4x-3=0\Rightarrow x=\frac{3}{4}\end{cases}}\)
d/ \(2\left(x+5\right)=2x+10\)
-,- mik ko rõ đề ạ, sai thì ibox ạ.Cảm ơn
\(x^2-6x+5=0\)
<=> \(x^2-x-5x+5=0\)
<=> \(x\left(x-1\right)-5\left(x-1\right)=0\)
<=> \(\left(x-1\right)\left(x-5\right)=0\)
<=> \(\left\{{}\begin{matrix}x-1=0\\x-5=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)
Vậy phương trình có nghiệm là x=1 và x=5
\(2x^2+7x-9=0\) ( nếu là 9 thì ko ra kq đc nên mình đổi thành -9 nha )
<=> \(2x^2-2x+9x-9=0\)
<=> \(2x\left(x-1\right)+9\left(x-1\right)=0\)
<=> \(\left(x-1\right)\left(2x+9\right)=0\)
<=> \(\left\{{}\begin{matrix}x-1=0\\2x+9=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=1\\x=\frac{-9}{2}\end{matrix}\right.\)
\(4x^2-7x+3=0\)
<=> \(4x^2-4x-3x+3=0\)
<=>\(4x\left(x-1\right)-3\left(x-1\right)=0\)
<=> \(\left(x-1\right)\left(4x-3\right)=0\)
<=> \(\left\{{}\begin{matrix}x-1=0\\4x-3=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=1\\x=\frac{3}{4}\end{matrix}\right.\)
\(2\left(x+5\right)=x^2+5x\)
<=> \(2\left(x+5\right)-x^2-5x=0\)
<=>\(2\left(x+5\right)-x\left(x+5\right)=0\)
<=>\(\left(x+5\right)\left(2-x\right)=0\)
<=>\(\left\{{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
A, => (x-2)(x+2)(x-3)(x-3)=0
\(\Leftrightarrow\left[\begin{matrix}x-2=0\\x+2=0\\x-3=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=2\\x=-2\\x=3\\x=-3\end{matrix}\right.\)
Vậy tập no của pt ..
B, \(\Leftrightarrow x^2+2x+5x+10=0\)
\(\Leftrightarrow x\left(x+2\right)+5\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x+2=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-2\\x=-5\end{matrix}\right.\)
Vậy ...............
C, \(\Leftrightarrow3x^2-6x-6x+12=0\)
\(\Leftrightarrow3x\left(x-2\right)-6\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x-6\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x-2=0\\3x-6=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=2\\x=2\end{matrix}\right.\)
Vậy tập no của pt là S{2}
1/
Ta có: 6x4 -x3-7x2+x+1=0
<=> 6x4-6x3+5x3-5x2-2x2+2x-x+1=0
<=> 6x3(x-1)+5x2(x-1)-2x(x-1)-(x-1)=0
<=> (x-1) ( 6x3+5x2-2x-1)=0
<=> ( x-1) ( 6x3-3x2+8x2-4x+2x-1)=0
<=> (x-1)\(\left[3x^2\left(2x-1\right)+4x\left(2x-1\right)+\left(2x-1\right)\right]\)=0
<=> (x-1) ( 2x-1) ( 3x2+4x+1)=0
<=> (x-1) ( 2x-1) (3x2+3x+x+1)=0
<=> (x-1) (2x-1) \(\left[3x\left(x+1\right)+\left(x+1\right)\right]\)=0
<=> (x-1)(2x-1)(x+1)(3x+1)=0
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\2x-1=0\\x+1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\2x=1\\x=-1\\3x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2}\\x=-1\\x=\dfrac{-1}{3}\end{matrix}\right.\)
vậy \(S=\left\{\pm1;\dfrac{1}{2};\dfrac{-1}{3}\right\}\)
\(6x^4-x^3-7x^2+x+1=0\)
\(\Leftrightarrow6x^4-6x^3+5x^3-5x^2-2x^2+2x-x+1=0\)
\(\Leftrightarrow6x^3\left(x-1\right)+5x^2\left(x-1\right)-2x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(6x^3+5x^2-2x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(6x^3+6x^2-x^2-x-x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[6x^2\left(x+1\right)-x\left(x+1\right)-\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(6x^2-x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(6x^2-3x+2x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(2x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\2x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=\dfrac{1}{2}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
\(x^2-6x+5=0\)
<=>\(x^2-x-5x+5=0\)
<=> \(x\left(x-1\right)-5\left(x-1\right)=0\)
<=> \(\left(x-1\right)\left(x-5\right)=0\)
<=> \(\left\{{}\begin{matrix}x-1=0\\x-5=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)
nãy làm rồi mà bạn ?