Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{4}{5}-|x-\frac{1}{6}|=\frac{2}{3}\)
\(\Rightarrow|x-\frac{1}{6}|=\frac{2}{15}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{6}=\frac{2}{15}\\x-\frac{1}{6}=-\frac{2}{15}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{10}\\x=\frac{1}{30}\end{cases}}\)
Vậy.....
$\textbf{a)}$
$y=\dfrac{\left(\dfrac49\right)^2\cdot\left(-\dfrac9{16}\right)\cdot(-1)^{19}}{\left(\dfrac4{25}\right)^2\cdot\left(-\dfrac{25}{144}\right)^2\cdot\left(-\dfrac{49}{144}\right)^2}$
$=\dfrac{\dfrac{16}{81}\cdot\left(-\dfrac9{16}\right)\cdot(-1)}{\dfrac{16}{625}\cdot\dfrac{625}{144^2}\cdot\dfrac{49^2}{144^2}}$
$=\dfrac{\dfrac19}{\dfrac{49^2}{144^4}}$
$=\dfrac19\cdot\dfrac{144^4}{49^2}$
$=\dfrac19\cdot\dfrac{(2^4\cdot3^2)^4}{7^4}$
$=\dfrac{2^{16}\cdot3^6}{7^4}$
$=\dfrac{47775744}{2401}.$
$\textbf{b)}$
$B=\dfrac{3^8\cdot20^5}{6^8\cdot10^2}$
$=\dfrac{3^8\cdot(2^2\cdot5)^5}{(2\cdot3)^8\cdot(2\cdot5)^2}$
$=\dfrac{3^8\cdot2^{10}\cdot5^5}{2^8\cdot3^8\cdot2^2\cdot5^2}$
$=5^3$
$=125.$
Ta có: |2x - 1| = |1 - 2x|
Lại có: \(\left|2x+3\right|+\left|1-2x\right|\ge\left|2x+3+1-2x\right|=\left|4\right|=4\)
Mà \(\left|2x+3\right|+\left|1-2x\right|=\frac{8}{3\left(x+1\right)^2+2}\)
\(\Rightarrow\frac{8}{3\left(x+1\right)^2+2}=4\)\(\Rightarrow3\left(x+1\right)^2+2=8\div4\)\(\Rightarrow3\left(x+1\right)^2+2=2\)\(\Rightarrow3\left(x+1\right)^2=2-2=0\)\(\Rightarrow\left(x+1\right)^2=0\)\(\Rightarrow x+1=0\)\(\Rightarrow x=-1\)
Sửa bài:
\(\left|2x+3\right|+\left|2x-1\right|=\left|2x+3\right|+\left|1-2x\right|\ge\left|2x+3+1-2x\right|=4\) với mọi x
\(\frac{8}{3\left(x+1\right)^2+2}\le\frac{8}{3.0+2}=4\)với mọi x
=> \(\left|2x+3\right|+\left|2x-1\right|\ge\frac{8}{3\left(x+1\right)^2+2}\)với mọi x
=> \(\left|2x+3\right|+\left|2x-1\right|=\frac{8}{3\left(x+1\right)^2+2}\)
<=> \(\hept{\begin{cases}\left(2x+3\right)\left(1-2x\right)\ge0\\\left(x+1\right)^2=0\end{cases}\Leftrightarrow}x=-1\)
Vậy S = { -1 }
dau[ va ] i