Question

ai giúp mih luôn với ạ mih cần gấp

tìm x

ý a (2x-5)^2-(5+2x)^2=0

y b 27x^3-54x^2+36x=8

ai giúp mih mih cảm ơn nhiều

Answer 1

b) \(27x^3-54x^2+36x=8\)

\(\Rightarrow27x^3-54x^2+36x-8=0\)

\(\Rightarrow\left(3x\right)^3-3.\left(3x\right)^2.2+3.3x.2^2-2^3=0\)

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

\(\Rightarrow3x-2=0\)

\(\Rightarrow3x=2\)

\(\Rightarrow x=\dfrac{2}{3}\)

Answer 2

(2x-5)^2-(5+2x)^2=0
<=>(2x-5-5-2x)(2x-5+5+2x)=0
<=>(-10).(4x)=0
<=>(-40x)=0
<=>x =0
27x^3-54x^2+36x=8
<=>27x^3-54x^2+36x-8=0
<=>(3x-2)^3=0
<=>3x-2=0
<=>3x=2
<=>x=2/3

Answer 3

a,\(\left(2x-5\right)^2-\left(5+2x\right)^2=0\)

\(\Rightarrow\left(2x-5\right)^2+\left(2x-5\right)^2=0\)

Do \(\left(2x-5\right)^2\ge0\)

\(\Rightarrow\left(2x-5\right)^2=0\)

\(\Rightarrow2x-5=0\)

\(\Rightarrow x=\dfrac{5}{2}\)

@Yukru giúp câu b với!