Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1) \(\left(\dfrac{-13}{17}-\dfrac{31}{52}\right)-\left(\dfrac{73}{52}-\dfrac{13}{17}+\dfrac{5}{6}\right)-\dfrac{3}{4}\)
\(=\dfrac{-13}{17}-\dfrac{31}{52}-\dfrac{73}{52}+\dfrac{13}{17}-\dfrac{5}{6}-\dfrac{3}{4}\)
\(=\left(\dfrac{-13}{17}+\dfrac{13}{17}\right)-\left(\dfrac{31}{52}+\dfrac{73}{52}\right)-\left(\dfrac{5}{6}+\dfrac{3}{4}\right)\)
\(=0-2-\dfrac{19}{12}\)
\(=-2-\dfrac{19}{12}\)
\(=\dfrac{-43}{12}\)
a, \(2^3.2^5=2^8=256\)
\(\left(-3\right)^9:\left(-3\right)^5=\left(-3\right)^4=81\)
\(\left(-6\right)^9.6^5=\left(-1\right)^9.6^9.6^5=\left(-1\right).6^{14}\\ \left(\dfrac{1}{2}\right)^5=\dfrac{1}{32}\)
b, \(\left(\dfrac{3}{5}\right)^6.\left(\dfrac{5}{3}\right)^6=\left(\dfrac{3}{5}. \dfrac{5}{3}\right)^6=1^6=1\\ \left(-\dfrac{7}{8}\right)^9:\left(\dfrac{7}{4}\right)^9=\left(-\dfrac{7}{8}:\dfrac{7}{4}\right)^9=\left(-\dfrac{1}{2}\right)^9=-\dfrac{1}{512}\\ \left(\left(-\dfrac{1}{2}\right)^2\right)^3=\left(\dfrac{1}{2}\right)^{...}\Rightarrow\left(\dfrac{1}{64}\right)=\left(\dfrac{1}{2}\right)...\Rightarrow\left(\dfrac{1}{64}\right)=\left(\dfrac{1}{2}\right)^6\)
c, \(\left(\dfrac{2}{3}\right)^8=\left(\left(\dfrac{2}{3}\right)^4\right)^{...}\Rightarrow\left(\left(\dfrac{2}{3}\right)^4\right)^2=\left(\left(\dfrac{2}{3}\right)^4\right)^{...}\Rightarrow\left(\dfrac{2}{3}\right)^8=\left(\left(\dfrac{2}{3}\right)^4\right)^2\\ \left(\dfrac{1}{3}\right)^{12}:\left(-\dfrac{3}{9}\right)^{12}=\left(\dfrac{1}{3}.\left(-3\right)\right)^{12}=\left(-1\right)^{12}=1\\ \left(\dfrac{1}{3}\right)^{12}:\left(\dfrac{1}{3}\right)^{10}=\left(\dfrac{1}{3}\right)^2=\dfrac{1}{9}\)
Cho mình hỏi từ câu C trở xuống đc ko ạ tại mắt mình yếu nên nhìn không rõ ấy
a: \(=\dfrac{4}{7}+\dfrac{3}{7}\cdot\dfrac{-2}{3}\)
\(=\dfrac{4}{7}-\dfrac{2}{7}=\dfrac{2}{7}\)
c.
$2|3x+1|-5=3$
$2|3x+1|=8$
$|3x+1|=4$
$\Rightarrow 3x+1=4$ hoặc $3x+1=-4$
$\Rightarrow x=1$ hoặc $x=\frac{-5}{3}$
f.
$3|4x+3|-5=4$
$3|4x+3|=9$
$|4x+3|=3$
$\Rightarrow 4x+3=3$ hoặc $4x+3=-3$
$\Rightarrow x=0$ hoặc $x=\frac{-6}{4}=\frac{-3}{2}$
i.
$|2x+5|-\frac{1}{3}=\frac{2}{3}$
$|2x+5|=\frac{2}{3}+\frac{1}{3}=1$
$\Rightarrow 2x+5=1$ hoặc $2x+5=-1$
$\Rightarrow x=-2$ hoặc $x=-3$
n.
$\frac{1}{5}+2\frac{1}{3}|x+0,2|=\frac{8}{15}$
$\frac{7}{3}|x+0,2|=\frac{8}{15}-\frac{1}{5}=\frac{1}{3}$
$|x+0,2|=\frac{1}{3}: \frac{7}{3}=\frac{1}{7}$
$\Rightarrow x+0,2=\frac{1}{7}$ hoặc $x+0,2=\frac{-1}{7}$
$\Rightarrow x=\frac{-2}{35}$ hoặc $x=\frac{-12}{35}$