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

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

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

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

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

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

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

c) \(\left|x\right|=3,5\Rightarrow\left[{}\begin{matrix}x=3,5\\x=-3,5\end{matrix}\right.\)

d) \(\left|x\right|=-2,7\Rightarrow x\in\varnothing\) 

l) \(\left|x+\dfrac{3}{4}\right|-5=-2\Rightarrow\left|x+\dfrac{3}{4}\right|=3\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=3\\x+\dfrac{3}{4}=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3-\dfrac{3}{4}\\x=-3-\dfrac{3}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\\x=\dfrac{15}{4}\end{matrix}\right.\)

Đính chính câu l \(x=-\dfrac{15}{4}\) không phải \(x=\dfrac{15}{4}\)

c) x = 3,5 hoặc -3,5

d) x = -2,7

e) x = -1,27

f) x = 0

g) x = -2

h) x = 0

l) x = 9/4

   x = -15/4

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

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

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 :

   Bạn kia trả lời đúng rồi !

Hok tốt

~ nhé bạn ~

xem lại câu b

a)(x+4)^2-(x+1)(x-1)=0    

<=>x²+8x+16-(x²-1)=0

<=>x²+8x+16-x²+1=0

<=>8x+17=0

<=>8x=-17

<=>x=-17/8

c)3(x-7)+(5-x)=2^3(x-1)

<=>3x-21+5+x=8x-8

<=>3x+x-8x=-8-5+21

<=>-4x=8

<=>x=-2

l x + 1 l - 5 = 0

l x + 1 l      = 0 + 5 = 5

  x + 1 = 5

  x       = 5 - 1 = 4

x + 1 = -5

x       = -5 - 1 = - 6

Các bạn giúp mình giải với nhé! Đúng thì mình k đúng nhé. Cảm ơn các bạn nhiều lắm. Yêu cả nhà.

\(1.\left(x-5\right)^{23}.\left(y+2\right)^7=0\)

\(\Rightarrow\hept{\begin{cases}\left(x-5\right)^{23}=0\\\left(y+2\right)^7=0\end{cases}\Rightarrow\hept{\begin{cases}\left(x-5\right)^{23}=0^{23}\\\left(y+2\right)^7=0^7\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x-5=0\\y+2=0\end{cases}\Rightarrow\hept{\begin{cases}x=0+5\\y=0-2\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x=5\\y=-2\end{cases}}\)

Vậy \(\left(x;y\right)=\left(5;-2\right)\)

2. \(A=\left(x-2\right)^2+|y+3|+7\)

Ta có :

\(\hept{\begin{cases}\left(x-2\right)^2\ge0\forall x\\|y+3|\ge0\forall y\end{cases}}\)

\(\Rightarrow\left(x-2\right)^2+|y+3|\ge0\forall x;y\)

\(\Rightarrow\left(x-2\right)^2+|y+3|+7\ge7\forall x;y\)

\(\Rightarrow A\ge7\forall x;y\)

Dấu bằng xảy ra

\(\Leftrightarrow\hept{\begin{cases}\left(x-2\right)^2=0\\|y+3|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-2=0\\y+3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2\\y=-3\end{cases}}}\)

Vậy GTNN của A là 7 khi \(\left(x;y\right)=\left(2;-3\right)\)