Question

Tìm x biết

a, \(x^{10}=x^2\)

b,\(\frac{x+3}{5}=\frac{20}{x+3}\)

c, \(\left(2x-15\right)^5=\left(2x-15\right)^3\)

Answer 1

a, x¹⁰ = x²

=> x¹⁰ - x² = 0

=> x² (x⁸ - 1) = 0

=>x² = 0 hoặc x⁸ - 1 = 0

=>x = 0 hoặc x⁸ = 1

=>x=0 hoặc x=1

b, x+3/5 = 20/x+3

=> x+3 . x+3 = 5.20

=> (x+3)² = 100

=> (x+3)² = 10²

=> x+3 = 10

=> x = 7

c, làm tương tự giống phần a,

Answer 2

\(c,\left(2x-15\right)^2=\left(2x-15\right)^3\)

=> \(\left(2x-15\right)^2-\left(2x-15\right)^3=0\)

=> \(\left(2x-15\right)^2.\left[1-\left(2x-15\right)\right]=0\)

=> \(\left[{}\begin{matrix}\left(2x-15\right)^2=0\\1-\left(2x-15\right)=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}2x-15=\sqrt{0}=0\\2x-15=1-0=1\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}2x=0+15=15\\2x=1+15=16\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=15:2=\frac{15}{2}\\x=16:2=8\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{15}{2};8\right\}\)