Question

[Kiều Thúy]

Bài1 : tìm x

b, \(x^2.\left(x^2+4\right)-x^2-4=0\)

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

bài 2 rút gọn

a, \(90.10^k-10^{k+2}+10^{k+1}\)

b,\(2,5.5^{n-3}.10+5^n-6.5^{n-1}\)

Answer 1

[lê thị thu huyền]

b) \(x^2\left(x^2+4\right)-x^2-4=0\)

.\(\Leftrightarrow x\left(x^2+4\right)-\left(x^2+4\right)=0\)

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

\(\Rightarrow x-1=0\)(vì \(x^2+4>0\))

\(\Leftrightarrow x=1\)

Answer 2

[l҉o҉n҉g҉ d҉z҉]

Ta có : (2x - 1)² - (4x² - 1) = 0

<=> (2x - 1)² - [(2x)² - 1²] = 0

<=> (2x - 1)² - (2x - 1)(2x + 1) = 0

<=> (2x - 1)[2x - 1 - (2x + 1)] = 0

<=> (2x - 1)(-2) = 0

=> 2x - 1 = 0

=> 2x = 1

=> x = \(\frac{1}{2}\)

Vậy x = \(\frac{1}{2}\)