6.(x + 8) - 12x = -12

=> 6x + 48 - 12x + 12 = 0

=> -6x + 60 = 0

=> -6x = -60

=> x = 10

6(x+8)-12x=-12

6x+54-12x=-12

-6x+54=-12

-6x=-66

x=11

a) \(x^{12}-y^4=\left(x^6\right)^2-\left(y^2\right)^2=\left(x^6-y^2\right)\left(x^6+y^2\right)=\left(x^3-y\right)\left(x^3+y\right)\left(x^6+y^2\right)\)

b) \(x^9+1=\left(x^3\right)^3+1=\left(x^3+1\right)\left(x^6-x^3+1\right)=\left(x+1\right)\left(x^6-x^3+1\right)^2\)

c) \(x^6-y^6=\left(x^2\right)^3-\left(y^2\right)^3=\left(x^2-y^2\right)\left(x^4-x^2y^2+y^4\right)=\left(x-y\right)\left(x+y\right)\left(x^4-x^2y^2+y^4\right)\)

d) \(x^6+1=\left(x^2\right)^3+1=\left(x^2+1\right)\left(x^4-x^2+1\right)\)

e) \(9x^6-12x^7+4x^8=x^6\left(9-12x+4x^2\right)=x^6\left(3-2x\right)^2\)

a) x - (11 - x) = -48 + (-12 + x)

x - 11 + x = -48 + (-12) + x

x + x - x = -48 + (-12) + 11

x + 0 = (-60) + 11

x = -49

b) (15 - x) + (x - 12) = 7 - (-8 + x)

15 - x + x - 12 = 7 + 8 + x

15 - x + x - 12 = 15 + x

x + x - x = 15 - 15 + 12

x + 0 = 0 + 12

x = 12

c) (x - 12) - (2x + 31) = -6 - 5

x - 12 - 2x - 31 = -1

x - 12 - 2x = -1 + 31

x - 12 - 2x = 30

x - 2x = 30 + 12

-1x = 42

=> x = -42

d) |x + 5| - (-17) = 20

=> |x + 5| + 17 = 20

=> |x + 5| = 20 - 17

=> |x + 5| = 3

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

=> \(\orbr{\begin{cases}x=3-5\\x=-3-5\end{cases}}\)

=> \(\orbr{\begin{cases}x=-2\\x=-8\end{cases}}\)

12x² - 3x = 0

=> 12x² - 3x = (2² . 3x²) - 3x = 0

=> 3x . (4x - 1) = 0

=> 3x . 4x - 3x . 1 = 0

=> 3x . 4x - 3x = 0

=> 12x - 3x = 0

=> 4x  =0

=> x = 0

x=0 chac chan

\(a,\left(x-1\right)^2-2^2=\left(x-1-2\right)\left(x-1+2\right)=\left(x-3\right)\left(x+1\right)\\ b,=\left(2x\right)^2+2.2x.3+3^2\\ =\left(2x+3\right)^2\\ c,=x^3-\left(2y\right)^3\\ =\left(x-2y\right)\left(x^2+2xy+4y^2\right)\\ d,=x^3\left(x^2-1\right)-\left(x^2-1\right)\\ =\left(x^3-1\right)\left(x^2-1\right)\\ =\left(x-1\right)\left(x^2+x+1\right)\left(x-1\right)\left(x+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\)

\(e,=-4x^2\left(x-1\right)+\left(x-1\right)\\ =\left(1-4x^2\right)\left(x-1\right)\\ =\left(1-2x\right)\left(1+2x\right)\left(x-1\right)\)

\(f,=\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1^2+1^3\\ =\left(2x+1\right)^3\)

a: A=(-x)^3+3*(-x)^2*2+3*(-x)*2^2+2^3=(-x+2)^3

=(28+2)^3=30^3=27000

b: \(C=\left(x+2y-2\right)^3=\left(20+2\cdot9-2\right)^3\)

=36^3

c: 11^3-1

=(11-1)(11^2+11+1)

=10*(121+12)

=1330

d: x^3-y^3=(x-y)^3+3xy(x-y)

=6^3+3*6*9

=216+162

=378

a) (x-2)³ - 6(x+1)² - x³ + 12 = 0 

<=> x³-6x²+12x-8-6(x²+2x+1)-x³+12=0

<=> x³-6x²+12x-8-6x²-12x-6-x³+12=0

<=> -12x²+4=0

<=> \(x=\frac{1}{\sqrt{3}},x=-\frac{1}{\sqrt{3}}\)

vậy pt có 2 nghiệm....

b) x³ - 6x² + 12x - 8 = 0 

<=> (x³-2x²)-(4x²-8x)+(4x+8)=0

<=> (x-2)(x²-4x+4)=(x-2)³=0

=> x=2 là nghiệm

c) 8x³ - 12x² + 6x - 1 = 0

<=> (2x-1)³=0

<=> x=1/2

a) \(\left(x-2\right)^3-6\left(x+1\right)^2-x^3+12=0\)

\(\Leftrightarrow x^3-6x^2+12x-8-6\left(x^2+2x+1\right)-x^3+12=0\)

\(\Leftrightarrow x^3-6x^2+12x-8-6x^2-12x-6-x^3+12=0\)

\(\Leftrightarrow-12x^2-2=0\)

\(\Leftrightarrow-2\left(6x^2+1\right)=0\)

\(\Leftrightarrow6x^2+1=0\) (vô nghiệm)

Vậy không có giá trị nào của x thỏa mãn pt

b) \(x^3-6x^2+12x-8=0\)

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

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

Vậy x=2

c) \(8x^3-12x^2+6x-1=0\)

\(\Leftrightarrow\left(2x-1\right)^3=0\)

\(\Leftrightarrow2x-1=0\Leftrightarrow x=\frac{1}{2}\)

Vậy \(=\frac{1}{2}\)

\(12\left(x+5\right)+2x=130\\\Leftrightarrow 12x+60+2x=130\\ \Leftrightarrow14x=70\\ \Leftrightarrow x=5\\ ----\\ 23\left(x-5\right)-12x=138\\ \Leftrightarrow23x-115-12x=138\\ \Leftrightarrow23x-12x=138+115\\ \Leftrightarrow11x=253\\ \Leftrightarrow x=\dfrac{253}{11}=23\\ ----\\ 360-12x+23\left(x-5\right)=278\\ \Leftrightarrow360-12x+23x-115=278\\ \Leftrightarrow-12x+23x=278+115-360\\ \Leftrightarrow11x=33\\ \Leftrightarrow x=\dfrac{33}{11}=3\)

\(6\left(x+3\right)+3\left(x-5\right)=278\\ \Leftrightarrow6x+18-3x-15=278\\ \Leftrightarrow6x-3x=278+15-18\\ \Leftrightarrow3x=275\\ \Leftrightarrow x=\dfrac{275}{3}\\ ---\\ \left(7-x\right)\left(3x-90\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}7-x=0\\3x-90=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=30\end{matrix}\right.\)

a: \(=\dfrac{3\left(x-2\right)}{\left(x-2\right)^3}=\dfrac{3}{\left(x-2\right)^2}\)

b: \(=\dfrac{x^2\left(x+2\right)}{\left(x+2\right)^3}=\dfrac{x^2}{\left(x+2\right)^2}\)

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