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

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

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

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

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

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

1) \(x^2-9x=0\Rightarrow x\left(x-9\right)=0\Rightarrow x=0;9\)

2) \(x\left(x-4\right)-x^2=7\Rightarrow-4x=7\Rightarrow x=-\dfrac{7}{4}\)

3) \(3x+2\left(x-5\right)=5\Rightarrow5x-10=5\Rightarrow5x=15\Rightarrow x=3\)

4) \(25x^2-1=0\Rightarrow x^2=\dfrac{1}{25}\Rightarrow x=\pm\dfrac{1}{5}\)

5) \(3x\left(x-2\right)-5\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(3x-5\right)=0\Rightarrow x=2;\dfrac{5}{3}\)

6) \(3x\left(x-7\right)+4\left(x-7\right)\Rightarrow\left(3x+4\right)\left(x-7\right)=0\Rightarrow x=-\dfrac{4}{3};7\)

7) \(4x^2-9=0\Rightarrow x^2=\dfrac{9}{4}\Rightarrow x=\pm\dfrac{3}{2}\)

8) \(10x\left(x-4\right)+2x-8=0\Rightarrow2\left(x-4\right)\left(5x+1\right)=0\Rightarrow x=4;-\dfrac{1}{5}\)

9) \(x\left(2x-5\right)-2x^2=0\Rightarrow x\left(2x-5-2x=0\right)\Rightarrow x=0\)

10) \(2x^2-4x=0\Rightarrow2x\left(x-2\right)=0\Rightarrow x=0;2\)

11) \(2x\left(3-4x\right)+3\left(4x-3\right)=0\Rightarrow2x\left(4x-3\right)-3\left(4x-3\right)=0\Rightarrow\left(4x-3\right)\left(2x-3\right)=0\Rightarrow x=\dfrac{3}{4};\dfrac{3}{2}\)

12) \(2x\left(x-5\right)-2x^2=3\Rightarrow-10x=3\Rightarrow x=-\dfrac{3}{10}\)

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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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

x.(x+7)=0

=>x =0 hoặc x+7=0

                     x=0-7

                     x=(-7)

Vậy x thuộc {0;-7}

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

=>x+12=0 hoặc x-3=0

 x=0-12              x=0+3

x=(-12)              x=3

Vậy x thuộc {-12;3}

Các câu khác bn lm tương tự nha

7 x 3 = 21

7 x 8 = 56

7 x 2 = 14

7 x 1 = 7

7 x 4 = 35

7 x 6 = 42

7 x 10 = 70

0 x 7 = 0

7 x 7 = 49

7 x 4 = 28

7 x 9 = 63

7 x 0 = 0

7x3=21

7x8=56

7x2=14

7x1=7

7x5=35

7x6=42

7x10=70

0x7=0

7x7=49

7x4=28

7x9=63

7x0=0

7 x 1 = 7

7 x 8 = 56

7 x 6 = 42

7 x 5 = 35

7x 2 = 14

7 x 9 = 63

7 x 4 = 28

0 x 7 = 0

7x 3 = 21

7 x 7 = 49

7 x 0 = 0

7 x 10 = 70

7 x 2 = 14       7 x 5 = 35       7 x 6 = 42       0 x 7 = 0

7 x 4 = 28       7 x 3 = 21       7 x 9 = 63       7 x 0 = 0

7 x 8 = 56       7 x 1 = 7       7 x 10 =70       1 x 7 = 7

đăng cho vui à

Làm theo công thức: tích bằng 0 thì một trong x thừa số bằng 0 rồi xét các trường hợp

1. x ( x + 7 ) = 0
( 1 ) x = 0
( 2 ) x + 7 = 0 => x = -7
S = { -7 ; 0 }

2. ( x + 12 ) ( x - 3 ) = 0
( 1 ) x + 12 = 0 => x = -12
( 2 ) x - 3 = 0 => x = 3
S = { -12 ; 3 }

3. ( -x + 5 ) ( 3 - x ) = 0
( 1 ) -x + 5 = 0 => -x = -5 => x = 5
( 2 ) 3 - x = 0 => x = 3
S = { 3 ; 5 }

4. x ( 2 + x ) ( 7 - x ) = 0
( 1 ) x = 0
( 2 ) 2 + x = 0 => x = -2
( 3 ) 7 - x = 0 => x = 7
S = { -2 ; 0 ; 7 }

5. ( x - 1 ) ( x + 2 ) ( -x - 3 ) = 0
( 1 ) x - 1 = 0 => x = 1
( 2 ) x + 2 = 0 => x = -2
( 3 ) -x - 3 = 0 => -x = 3 => x = -3
S = { -3 ; -2 ; 1 }

\(1,x.\left(x+7\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-7\end{cases}}}\)

\(2,\left(x+12\right).\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+12=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-12\\x=3\end{cases}}}\)

\(3,\left(-x+5\right).\left(3-x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}-x+5=0\\3-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=3\end{cases}}}\)

4/   \(x.\left(2+x\right).\left(7-x\right)=0\)

   \(\hept{\begin{cases}x=0\\2+x=0\\7-x=0\end{cases}}\) =>   \(\hept{\begin{cases}x=0\\x=-2\\x=7\end{cases}}\)

Vậy \(x=\left\{0,-2,7\right\}\)

5/   \(\left(x-1\right).\left(x+2\right).\left(-x-3\right)=0\)

\(\hept{\begin{cases}x-1=0\\x+2=0\\-x-3=0\end{cases}}\)=>   \(\hept{\begin{cases}x=1\\x=-2\\x=-3\end{cases}}\)

Bài 1

1.(x-3)(x+2)-x(x-7)=15

\(\Leftrightarrow x^2+2x-3x-6-x^2+7x=15\)

\(\Leftrightarrow-6+6x=15\)

\(\Leftrightarrow6x=15+6\) =21

\(\Rightarrow x=\dfrac{21}{6}=3,5\)

2.(x-5)(x+5)+x(3-x)=20

\(\Leftrightarrow x^2-25+3x-x^2=20\)

\(\Leftrightarrow-25+3x=20\)

\(\Leftrightarrow3x=20+25=45\)

\(\Rightarrow x=\dfrac{45}{3}=15\)

3.(x-7)²-x(2+x)=-7

\(\Leftrightarrow x^2-14x+49-2x-x^2=-7\)

\(\Leftrightarrow-16x+49=-7\)

\(\Leftrightarrow-16x=-7-49=-56\)

\(\Rightarrow x=\dfrac{-56}{-16}=\dfrac{7}{2}=3,5\)

Tiếp bài 1

4.(x-4)²-(x+4)(x-4)=-16

\(\Leftrightarrow x^2-8x+16-x^2-16=-16\)

\(\Leftrightarrow-8x=-16\)

\(\Rightarrow x=\dfrac{-16}{-8}=2\)

5.(x-5)(x+5)-x(2-3x)=4x²-7

\(\Leftrightarrow x^2-25-2x+3x^2=4x^2-7\)

\(\Leftrightarrow4x^2-25-2x+3x^2=4x^2-7\)

\(\Leftrightarrow4x^2-4x^2-2x=-7+25\)

\(\Leftrightarrow-2x=18\)

\(\Rightarrow x=\dfrac{18}{-2}=-9\)

B2

1. 2x(x-2)+3x-6=0

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

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

\(\Rightarrow\left[{}\begin{matrix}x-2=0\Leftrightarrow x=2\\2x+3=0\Leftrightarrow x=\dfrac{-3}{2}\end{matrix}\right.\)

2.3x(x-5)-5x+25=0

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

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

\(\Rightarrow\left[{}\begin{matrix}x-5=0\Leftrightarrow x=5\\3x-5=0\Leftrightarrow x=\dfrac{5}{3}\end{matrix}\right.\)

3. x(x+7)-x-7=0

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

=>(x+7)(x-1)=0

\(\Rightarrow\left[{}\begin{matrix}x+7=0\Leftrightarrow x=-7\\x-1=0\Leftrightarrow x=1\end{matrix}\right.\)

1. x(x + 7) = 0

=> x = 0

x + 7 = 0 => x =  -7

Vậy x = 0 ; -7

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

x + 12 = 0 => x = -12

x - 3 = 0 => x = 3

Vậy x = -12 ; 3

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

-x + 5 = 0 => -x = -5 => x = 5

3 - x = 0 => x = 3

Vậy z = 5 ; 3

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

=> x = 0

2 + x = 0 => x = -2

7 - x = 0 => x = 7

Vậy x = 0 ; -2 ; 7

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

x - 1 = 0 => x = 1

x + 2 = 0 => x = -2

-x - 3 = 0 => -x = 3 => x = -3

Vậy x = 1 ; -2 ; -3

\(x+\left(x+7\right)=0\)

+) \(x=0\)

+) \(x+7=0=>x=-7\)

Vậy x=0 hoặc x=-7

\(\left(x+12\right).\left(x-3\right)=0\)

+) \(x+12=0=>x=-12\)

+) \(x-3=0=>x=3\)

Vậy x=-12 hoặc x=3

\(x.\left(2+x\right).\left(7-x\right)=0\)

+) \(x=0\)

+) \(2+x=0=>x=-2\)

+) \(7-x=0=>x=7\)

Vậy x=0 hoặc x=-2 hoặc x=7

\(\left(x-1\right).\left(x+2\right).\left(x+3\right)=0\)

+) \(x-1=0=>x=1\)

+) \(x+2=0=>x=-2\)

+) \(x+3=0=>x=-3\)

Vậy x=1 hoặc x=-2 hoặc x=-3

1) x.(x+7)=0

TH1: x = 0 thì x + 7 = 7

TH2: x + 7 = 0 thì x = -7

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

TH1: x + 12 = 0 thì x = -12, x - 3 = -15

TH2:x - 3 = 0 thì x = 3, x + 12 = 15

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

TH1: ( -x + 5 ) = 0 thì -x = -5 => x = 5, 3 - x = -2

TH2: ( 3 - x ) = 0 thì x = 3 => -x + 5 = 2

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

TH1: x = 0 thì 2 + x = 2, 7 - x = 7

TH2: 2 + x = 0 thì x = -2, 7 - x = 9

TH3: 7 - x = 0 thì x = 7, 2 + x = 9

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

TH1: x - 1 = 0 thì x = 1, x + 2 = 3, -x - 3 = -4

TH2: x + 2 = 0 thì x = -2, x - 1 = -3, -x - 3 = -1

TH3: -x - 3 = 0 thì x = -3, x - 1 = -4, x + 2 = -1