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:

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

\(\Rightarrow10=x\left(KTMĐK\right)\)

\(k=\frac{3x-4}{y+15}\)(1)

Thế x = 2 ; y = 3 vào (1) ta được : \(k=\frac{3\cdot2-4}{3+15}=\frac{2}{18}=\frac{1}{9}\)

=> k = 1/9

Khi y = 12, thế y vào (1) ta được :

\(k=\frac{3x-4}{12+15}\)

<=> \(\frac{1}{9}=\frac{3x-4}{27}\)

<=> 1.27 = 9( 3x - 4 )

<=> 27 = 27x - 36

<=> 27x = 63

<=> x = 63/27 = 7/3

Vậy khi y = 12 thì x = 7/3

b: \(3x^2-2x-1=0\)

=>\(3x^2-3x+x-1=0\)

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

=>\(\left[{}\begin{matrix}x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)

a: Bạn ghi lại đề đi bạn

a: \(x^3-4x^2-x+4=0\)

=>\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)

=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)

=>\(\left(x-4\right)\left(x^2-1\right)=0\)

=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)

b: Sửa đề: \(x^3+3x^2+3x+1=0\)

=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)

=>\(\left(x+1\right)^3=0\)

=>x+1=0

=>x=-1

c: \(x^3+3x^2-4x-12=0\)

=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)

=>\(x^2\cdot\left(x+3\right)-4\left(x+3\right)=0\)

=>\(\left(x+3\right)\left(x^2-4\right)=0\)

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

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

d: \(\left(x-2\right)^2-4x+8=0\)

=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)

=>\(\left(x-2\right)^2-4\left(x-2\right)=0\)

=>\(\left(x-2\right)\left(x-2-4\right)=0\)

=>(x-2)(x-6)=0

=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

Bài 1.

(4x-12)(x³+64)=0

=> [_{x³+64=0=>x=4}^{x-12=0=>4x=12=>x=3}           olm bị lỗi nên em đừng có viết cách ra 1 quãng như kia nhé !

vậy x thuộc {3;4}

(3x-12)(x²-4)=0

=>[_{x²-4=0=>x²=4=>x=2 hoặc x=-2}^{3x-12=0=>3x=12=>x=4}

vậy x thuộc {4;2;-2}

(x+3)³:3-1=-10

(x+3)³:3=-9

(x+3)³=-9.3

=>(x+3)³=-27

=>x+3=-3

=>x=-6

(3x-1)³-2=-66

(3x-1)³=-64

(3x-1)³=-4³

=>3x-1=-4

=>3x=-3

=>x=-1

\(\left(4x-12\right)\left(x^3+64\right)=0\)

\(\Leftrightarrow4x-12=0\)

\(\Leftrightarrow4x=0+12\)

\(\Leftrightarrow4x=12\)

\(\Leftrightarrow x=12\div4\)

\(\Leftrightarrow x=3\)

\(\Leftrightarrow x^3+64=0\)

\(\Leftrightarrow x^3=0=64\)

\(\Leftrightarrow x^3=\left(-64\right)\)

\(\Leftrightarrow x^3=\left(-4\right)^3\)

\(\Leftrightarrow x=\left(-4\right)\)

\(\Rightarrow x\in\left\{-4;3\right\}\)

\(\Leftrightarrow\left(3x-12\right)\left(x^2-4\right)=0\)

\(\Leftrightarrow3x-12=0\)

\(\Leftrightarrow3x=0+12\)

\(\Leftrightarrow3x=12\)

\(\Leftrightarrow x=12\div3\)

\(x=4\)

\(\Leftrightarrow x^2-4=0\)

\(\Leftrightarrow x^2=0+4\)

\(\Leftrightarrow x^2=4\)

\(\Leftrightarrow x^2=2^2=\left(-2\right)^2\)

\(\Rightarrow x\in\left\{2;-2\right\}\)

\(\Rightarrow x\in\left\{-2;2;4\right\}\)

Các câu khác tương tự nhé !

a) (3x – 5)2 – (x +1 )2 = (3x – 5 – x – 1)(3x – 5 + x + 1)

= (2x – 6)(4x – 4) = 8(x – 1)(x – 3)

Vậy (x – 1)(x – 3) = 0 ⇒ x - 1 = 0 hoặc x - 3 = 0

⇒ x = 1hoặc x = 3