( 1/3+1/6+1/10+1/15+1/21) : x =5/7

5/7: x = 5/7

       x = 5/7 : 5/7

       x = 1

tk giùm nhé! chúc học giỏi!

ta có :  ( 1/3 + 1/6 + 1/10 + 1/15 + 1/21 ) :x =5/7

       =>      5/7 : x = 5/7

       => x = 5/7 : 5/7

            x = 1

vậy x=1

\(-\frac{15}{12}+x+\frac{3}{7}=\frac{6}{5}x-\frac{1}{2}\)

<=>\(x-\frac{6}{5}x=-\frac{1}{2}+\frac{15}{12}-\frac{3}{7}\)

<=>\(-\frac{1}{5}x=\frac{9}{28}\)

<=>\(x=\frac{9}{28}:\left(-\frac{1}{5}\right)\)

<=>\(x=-\frac{45}{28}\)

1.

b) \(3^x+3^{x+2}=2430\)

\(\Rightarrow3^x.1+3^x.3^2=2430\)

\(\Rightarrow3^x.\left(1+3^2\right)=2430\)

\(\Rightarrow3^x.10=2430\)

\(\Rightarrow3^x=2430:10\)

\(\Rightarrow3^x=243\)

\(\Rightarrow3^x=3^5\)

\(\Rightarrow x=5\)

Vậy \(x=5.\)

c) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)

\(\Rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)

\(\Rightarrow\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=15\\2x-15=\pm1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=15:2\\2x-15=1\\2x-15=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{15}{2}\\2x=16\\2x=14\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{15}{2}\\x=8\\x=7\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{15}{2};8;7\right\}.\)

Chúc bạn học tốt!

\(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)

\(\Rightarrow x\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)+1=0\)

\(\Rightarrow x.0+1=0\)

=> 1=0 ( Vô lý )

Vậy \(x\in\varnothing\)

\(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)

\(\Rightarrow x.\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)=1\)

\(\Rightarrow x.0=1\Rightarrow x=0\)

Vậy x=0