\(x\left(x-7\right)\left(x-3\right)=0\)

\(\hept{\begin{cases}x=0\\x-7=0\\x-3=0\end{cases}}\)

\(\hept{\begin{cases}x=0\\x=0+7\\x=0+3\end{cases}}\)

\(\hept{\begin{cases}x=0\\x=7\\x=3\end{cases}}\)

\(\Rightarrow x=0;7;3\)

a , x.(2x+7)=0

(=) x = 0

     2x + 7 = 0 

(=) x = 0

     2x = -7

(=) x = 0

      x = -7/2

Mấy câu bạn hỏi có người hỏi rồi bạn tự tham khảo nhé

Công nhận là "NHẠT" .

Nhại thật !

\(1,\Leftrightarrow x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=9\\x=0\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\Leftrightarrow-4x=7\Leftrightarrow x=-\dfrac{7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\Leftrightarrow5x=15\Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(x-7\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{4}{3}\end{matrix}\right.\)

\(7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ 8,\Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=4\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow\left(4x-3\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\Leftrightarrow-10x=3\Leftrightarrow x=-\dfrac{3}{10}\)

\(1,\Leftrightarrow x\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\\ \Leftrightarrow-4x=7\\ \Leftrightarrow x=\dfrac{-7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\\ \Leftrightarrow5x=15\\ \Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)

\(5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(3x+4\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=7\end{matrix}\right.\\ 7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

\(8,\Leftrightarrow10x\left(x-4\right)+2\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\\ \Leftrightarrow-5x=0\\ \Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

\(11,\Leftrightarrow\left(2x-3\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\\ \Leftrightarrow-10x=3\\ \Leftrightarrow x=-\dfrac{3}{10}\)

a) \(x\left(2x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x+7=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x=-7\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{7}{2}\end{matrix}\right.\)

b) \(x\left(2x+7\right)>0\)

\(TH1:\left\{{}\begin{matrix}x>0\\2x+7>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x>0\\x>-\dfrac{7}{2}\end{matrix}\right.\) \(\Leftrightarrow x>0\)

\(TH2:\left\{{}\begin{matrix}x< 0\\2x+7< 0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x< 0\\x< -\dfrac{7}{2}\end{matrix}\right.\) \(\Leftrightarrow x< -\dfrac{7}{2}\)

Vậy \(x>0\) hay \(x< -\dfrac{7}{2}\)

c) \(x\left(2x+7\right)< 0\)

\(TH1:\left\{{}\begin{matrix}x>0\\2x+7< 0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x>0\\x< -\dfrac{7}{2}\end{matrix}\right.\) (Vô lý nên loại)

\(TH2:\left\{{}\begin{matrix}x< 0\\2x+7>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x< 0\\x>-\dfrac{7}{2}\end{matrix}\right.\) \(\Leftrightarrow-\dfrac{7}{2}< x< 0\)

Vậy \(-\dfrac{7}{2}< x< 0\)

a) x=0 hoặc x+7=0

suy ra x=0 hoặc x=-7

b) x+12=0 hoặc x-3=0

x=-12 hoặc x=3

c) x=0 hoặc x+2=0 hoặc 7-x=0

x=0 hoặc x=-2 hoặc x=7

d) x-1=0 hoặc x+2=0 hoặc -x-3=0

suy ra x=1 hoặc x=-2 hoặc x=-3

Bài làm

x( x + 7 ) = 0

<=> x = 0 hoẵ x + 7 = 0

=> x = 0 hoặc x = -7

Vậy x = 0 hoặc x = -7

( x + 12 )( x - 3 ) = 0

<=> x + 12 = 0 hoặc x - 3 = 0

=> x = -12 hoặc x = 3

Vậy x = -12 hoặc x = 3

( -x + 5 )( 3 - x ) = 0

<=> -x + 5 = 0 hoặc 3 - x = 0

=> x = 5 hoặc x = 3

Vậy x = 5 hoặc x = 3

x( 2 + x )( 7 - x ) = 0

<=> x = 0 hoặc 2 + x = 0 hoặc 7 - x = 0

=> x = 0 hoặc x = -2 hoặc x = 7

Vậy x = 0 hoặc x = -2 hoặc x j 7

( x - 1 )( x + 2 )( -x - 3 ) = 0

<=> ( x - 1 ) = 0 hoặc x + 2 = 0 hoặc ( -x - 3 ) = 0

<=> x = 1 hoăc x = -2 hoặc x = ( -3) 

Vậy x = 1 hoặc x = 2 hoặc x = -3

9:0=0,7*7=49,9*7=63,7:0=0,8*9=72,0*3=0,0:9=0

9 : 0  = 0

7 x 7 = 49

9 x 7 =63

8 x 9 = 72

0 x 3 = 0

0 : 9 = 0

7 : 0 = 0

học tốt nha

1) -12+3.(-x+7)=-18

   3.(-x+7)=-18+12

   3.(x+7)=-6

x+7=-6:3

x+7=-2

x=-2-7

x=-9

Trl :

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