a, \(2^3< 2^x< 2^9.2^{-5}\)

\(2^3< 2^x< 2^4\)

cn lại mk ko bt, hình như đề hơi sai sai

a)hình như đề sai thì phải

sửa lại

\(\left(\dfrac{1}{7}-\dfrac{2}{5}\right).\dfrac{2016}{2017}+\left(\dfrac{13}{7}+\dfrac{2}{5}\right).\dfrac{2016}{2017}\)

=\(\dfrac{2016}{2017}.\left(\dfrac{1}{7}-\dfrac{2}{5}+\dfrac{13}{7}+\dfrac{2}{5}\right)\)

=\(\dfrac{2016}{2017}.2=\dfrac{4032}{2017}\)

\(\dfrac{-1}{4}x+\dfrac{2}{3}=\dfrac{5}{9}\)\(\Rightarrow\)\(\dfrac{-1}{4}x=\dfrac{5}{9}-\dfrac{2}{3}\)=\(\dfrac{-1}{9}\)

\(\Rightarrow\) x = \(\dfrac{-1}{9}:\dfrac{-1}{4}\)=\(\dfrac{4}{9}\).

\(x.\left(\dfrac{3}{5}\right)^3=\dfrac{3}{5}\)

\(\Rightarrow\)x=\(\dfrac{3}{5}:\left(\dfrac{3}{5}\right)^3=\left(\dfrac{3}{5}\right)^{-2}\)= \(2\dfrac{7}{9}\)

\(\left|x\right|\) + \(\dfrac{1}{5}=2-\left(\dfrac{2}{3}-\dfrac{3}{4}\right)\)=2 - \(\dfrac{-1}{12}\)=2\(\dfrac{1}{12}\)

\(\Rightarrow\)\(\left|x\right|\)=\(2\dfrac{1}{12}\)-\(\dfrac{1}{5}\)=\(1\dfrac{53}{60}\)

\(\Rightarrow\)x=\(\left[{}\begin{matrix}1\dfrac{53}{60}\\-1\dfrac{53}{60}\end{matrix}\right.\)

\(\left(\dfrac{-3}{4}\right)^x=\dfrac{81}{256}\)=\(\dfrac{(-3)^4}{4^4}\)=\(\left(\dfrac{-3}{4}\right)^4\)

\(\Rightarrow\) x = 4

14 Mẹo Vặt Sáng Tạo Dành Cho Học Sinh - YouTube

thử xem mẹo thứ 5 đi chứ theo mk đây là bài toán dễ lớp 6

a: \(\dfrac{x+1}{5}+\dfrac{x+1}{6}=\dfrac{x+1}{7}+\dfrac{x+1}{8}\)

\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}-\dfrac{1}{8}\right)=0\)

=>x+1=0

hay x=-1

b: \(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\)

=>x-2010=0

hay x=2010

c: \(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Leftrightarrow\dfrac{x}{\left(x+2\right)\left(x+17\right)}=\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}\)

=>x=15

b,

\(B=\frac{1}{2000.1999}-\frac{1}{1999.1998}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(\Rightarrow B=\frac{1}{1999.2000}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1998.1999}\right)\)

\(\Rightarrow B=\frac{1}{1999.2000}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1998}-\frac{1}{1999}\right)\)

\(\Rightarrow B=\frac{1}{1999.2000}-\left(1-\frac{1}{1999}\right)\)

\(\Rightarrow B=\frac{1}{1999.2000}-\frac{1998}{1999}\)

\(\Rightarrow B=\frac{1}{1999}-\frac{1}{2000}-\frac{1998}{1999}\)

\(\Rightarrow B=\left(\frac{1}{1999}-\frac{1998}{1999}\right)-\frac{1}{2000}\)

\(\Rightarrow B=\frac{-1997}{1999}-\frac{1}{2000}\)

Cảm ơn bn!Mặc dù mik chư hiểu z hết!

Đăng từng bài một thôi bạn!

1)\(\left(-\dfrac{5}{13}\right)^{2017}.\left(\dfrac{13}{5}\right)^{2016}\)

\(=\left(-\dfrac{5}{13}\right).\left(-\dfrac{5}{13}\right)^{2016}.\left(\dfrac{13}{5}\right)^{2016}\)

\(=\left(-\dfrac{5}{13}\right).\left(\dfrac{5}{13}\right)^{2016}.\left(\dfrac{13}{5}\right)^{2016}\)

\(=\left(-\dfrac{5}{13}\right).\left(\dfrac{5}{13}.\dfrac{13}{5}\right)^{2016}\)

\(=\left(-\dfrac{5}{13}\right).1^{2016}\)

\(=-\dfrac{5}{13}\)

Cám ơn bn nhìu. giúp mk mí bài kia nữa đc ko?

1/a/ \(\left(\dfrac{2}{5}-3x\right)^2=\dfrac{9}{25}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{5}-3x=\dfrac{3}{5}\\\dfrac{2}{5}-3x=-\dfrac{3}{5}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}3x=-\dfrac{1}{5}\\3x=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{15}\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy ...

b/ \(\left(\dfrac{2}{3}x-\dfrac{1}{5}\right)^5=\dfrac{1}{243}\)

\(\Leftrightarrow\dfrac{2}{3}x-\dfrac{1}{5}=\dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{2}{3}x=\dfrac{8}{15}\)

\(\Leftrightarrow x=\dfrac{4}{5}\)

Vậy .........

2/ a/

Ta có :

\(5^{222}=\left(5^2\right)^{111}=25^{111}\)

\(2^{555}=\left(2^5\right)^{111}=32^{111}\)

Vì \(25^{111}< 32^{111}\Leftrightarrow5^{222}< 2^{555}\)

b/ Ta có :

\(3^{48}=\left(3^4\right)^{12}=81^{12}\)

\(4^{36}=\left(4^3\right)^{12}=64^{12}\)

Vì \(81^{12}>64^{12}\Leftrightarrow3^{48}>4^{36}\)