Câu hỏi của Nguyễn Huy Trường Lưu

Question

tìm giá trị nhỏ nhất của biểu thức:
A=(2x-1)4+3
B=-(8x-$\dfrac{4}{5}$)6+1

Nguyễn Đức Trí · Accepted Answer

$A=\left(2x-1\right)^4+3$

mà $\left(2x-1\right)^4\ge0,\forall x$

$\Rightarrow A=\left(2x-1\right)^4+3\ge0+3=3$

$\Rightarrow GTNN\left(A\right)=3\left(x=\dfrac{1}{2}\right)$

$B=-\left(8x-\dfrac{4}{5}\right)^6+1$

mà $-\left(8x-\dfrac{4}{5}\right)^6\le0,\forall x$

$\Rightarrow B=-\left(8x-\dfrac{4}{5}\right)^6+1\le0+1=1$

$\Rightarrow GTLN\left(B\right)=1\left(x=\dfrac{1}{10}\right)$Câu trả lời: $A=\left(2x-1\right)^4+3$

mà $\left(2x-1\right)^4\ge0,\forall x$

$\Rightarrow A=\left(2x-1\right)^4+3\ge0+3=3$

$\Rightarrow GTNN\left(A\right)=3\left(x=\dfrac{1}{2}\right)$

$B=-\left(8x-\dfrac{4}{5}\right)^6+1$

mà $-\left(8x-\dfrac{4}{5}\right)^6\le0,\forall x$

$\Rightarrow B=-\left(8x-\dfrac{4}{5}\right)^6+1\le0+1=1$

$\Rightarrow GTLN\left(B\right)=1\left(x=\dfrac{1}{10}\right)$