Question

9 Tìm x biết:

a) (2x-3)²​=16 ; b) (3x-2)⁵​=-243 ; c) (2x-5)²⁰⁰⁰+(3y+4)²⁰⁰²​\(\le\)0

Answer 1

a) Thôi dễ rồi

b) \(\left(3x-2\right)^5=-243\)

\(\Rightarrow\left(3x-2\right)^5=\left(-3\right)^5\)

\(\Rightarrow3x-2=-3\Rightarrow x=\dfrac{-1}{3}\)

c) Vì \(\left(2x-5\right)^{2000}\ge0\forall x;\left(3y+4\right)^{2002}\ge0\forall y\)

\(\Rightarrow\left(2x-5\right)^{2000}+\left(3y+4\right)^{2002}\ge0\forall x,y\)

Mà theo bài ra \(\left(2x-5\right)^{2000}+\left(3y+4\right)^{2002}\le0\)

\(\Rightarrow\left(2x-5\right)^{2000}+\left(3y+4\right)^{2002}=0\)

\(\Rightarrow\left\{{}\begin{matrix}2x-5=0\\3y+4=0\end{matrix}\right........\)