b

\(\left|6+x\right|\ge0;\left(3+y\right)^2\ge0\Rightarrow\left|6+x\right|+\left(3+y\right)^2\ge0\)

Suy ra \(\left|6+x\right|+\left(3+y\right)^2=0\)\(\Leftrightarrow\hept{\begin{cases}6+x=0\\3+y=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-6\\y=-3\end{cases}}\)

a

Ta có:\(\left|3x-12\right|=3x-12\Leftrightarrow3x-12\ge0\Leftrightarrow3x\ge12\Leftrightarrow x\ge4\)

\(\left|3x-12\right|=12-3x\Leftrightarrow3x-12< 0\Leftrightarrow3x< 12\Leftrightarrow x< 4\)

Với \(x\ge4\) ta có:

\(3x-12+4x=2x-2\)

\(\Rightarrow5x=10\)

\(\Rightarrow x=2\left(KTMĐK\right)\)

Với  \(x< 4\) ta có:

\(12-3x+4x=2x-2\)

\(\Rightarrow10=x\left(KTMĐK\right)\)

1)Tìm x

a) (x+1)(x-2)<0

=>Có 2TH:

TH1:

x+1<0=>x< -1

x-2>0=>x>2

=>Vô lí 

TH2:

x+1>0=>x> -1

x-2<0=>x<2

=> -1<x<2

Vậy x thuộc {0;1}

b) Tương tự a thôi ạ. 

c) (x-2)(3x+2)

=> Có hai TH:

TH1:

x-2<0=>x<2

3x+2<0=>3x< -2=>x< -2/3

=>x< -2/3

TH2:

x-2>0=>x>2

3x+2>0=>3x> -2=>x> -2/3

=>x>2

Vậy x< -2/3 hoặc x>2

2)Tìm x

x.x=x

<=>x²-x=0

<=>x(x-1)=0

<=>x=0 hoặc x=1

Cảm ơn nha Linh

a. x( x+ 3)= 0 

⇔ x= 0 hoặc x+ 3= 0

⇔ x= 0          x = -3

b. x( 2x− 1)+ 2( 2x− 1) =0 

⇔ ( 2x− 1)(x+ 2) =0

⇔ 2x− 1 =0 hoặc  x+ 2 =0

⇔ 2x       =1          x      = -2

⇔   x       =\(\dfrac{1}{2}\)         x      = -2

a) 2y - 12y = 0

\(\Rightarrow\) y ( 2-12) = 0

\(\Rightarrow\) y . (-10) =0 

\(\Rightarrow\) y = 0 : (-10) = 0

b) (y-7)(y-8) = 0

\(\Rightarrow\orbr{\begin{cases}y-7=0\\y-8=0\end{cases}\Rightarrow\orbr{\begin{cases}y=0+7\\y=0+8\end{cases}\Rightarrow}\orbr{\begin{cases}y=7\\y=8\end{cases}}}\)

c) x + x.2+x.3+x.4+...+x.10 = 165

\(\Rightarrow\) x ( 1+2+3+.....+8+9+10) = 165

\(\Rightarrow\)x . \(\frac{\left(1+10\right).10}{2}\)=165

\(\Rightarrow\) x . 55 = 165

\(\Rightarrow x=\frac{165}{55}=3\)

Can you k for me ,Lê Thị Kim Chi!

a) \(2y-12y=0\)

\(\Leftrightarrow-10y=0\)

\(\Leftrightarrow y=0:\left(-10\right)\)

\(\Leftrightarrow y=0\)

b) \(\left(y-7\right)\left(y-8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}y-7=0\\y-8=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}y=0+7\\y=0+8\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}y=7\\y=8\end{cases}}\)

c) \(x+x.2+x.3+......+x.10=165\)

\(\Leftrightarrow x.\left(1+2+3+.....+10\right)=165\)

\(\Leftrightarrow x.55=165\)

\(\Leftrightarrow x=165:55\)

\(\Leftrightarrow x=3\)

`@` `\text {Ans}`

`\downarrow`

`1,`

`x^2 - 9 = 0`

`<=> x^2 = 0 + 9`

`<=> x^2 = 9`

`<=> x^2 = (+-3)^2`

`<=> x = +-3`

Vậy, `S = {3; -3}`

`2,`

`25 - x^2 = 0`

`<=> x^2 = 25 - 0`

`<=> x^2 = 25`

`<=> x^2 = (+-5)^2`

`<=> x = +-5`

Vậy,` S= {5; -5}`

`3,`

`-x^2 + 36 = 0`

`<=> -x^2 = 0 - 36`

`<=> -x^2 = -36`

`<=> x^2 = 36`

`<=> x^2 = (+-6)^2`

`<=> x = +-6`

Vậy, `S= {6; -6}`

`4,`

`4x^2 - 4 = 0`

`<=> 4x^2 = 0+4`

`<=> 4x^2 = 4`

`<=> x^2 = 4 \div 4`

`<=> x^2 = 1`

`<=> x^2 = (+-1)^2`

`<=> x = +-1`

Vậy, `S= {1; -1}`

`@` `\text {Kaizuu lv uuu}`

a) \(\sqrt{x}\left(\sqrt{x}-1\right)=0\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

b) \(\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)=0\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-2=0\\\sqrt{x}+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\\sqrt{x}=-3\left(vôlí\right)\end{cases}}\)

c) \(\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)=0\Leftrightarrow\orbr{\begin{cases}\sqrt{x}+1=0\\\sqrt{x}+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=-1\left(vôlí\right)\\\sqrt{x}=-3\left(vôlí\right)\end{cases}}\)

a: \(8x\left(x-2017\right)-2x+4034=0\)

\(\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)