Question

tìm x biết: (4x-12)(x^3+64)=0

(3x-12)(x^2-4)=0

(x+3)^3:3-1=-10

(3x-1)^3-2=-66

m.n giúp mik vs tối mik phải nộp r ak

 

Answer 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

Answer 2

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

\(\Leftrightarrow4x-12=0\)

\(\Leftrightarrow4x=0+12\)

\(\Leftrightarrow4x=12\)

\(\Leftrightarrow x=12\div4\)

\(\Leftrightarrow x=3\)

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

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

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

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

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

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

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

\(\Leftrightarrow3x-12=0\)

\(\Leftrightarrow3x=0+12\)

\(\Leftrightarrow3x=12\)

\(\Leftrightarrow x=12\div3\)

\(x=4\)

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

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

\(\Leftrightarrow x^2=4\)

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

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

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

Các câu khác tương tự nhé !