Ta có

5^x+2+5^x+1+5^x=5^x.5²+5^x.5+5^x

                       =31.5^x

Đề bài bạn thiếu tổng của 5^x+2+5^x+1+5^x nhé , đến đây bạn xem lại để r tự làm tiếp đi

\(\frac{1}{2}2^x+2^{x+2}=2^8+2^5\)

\(\Leftrightarrow2^{x-1}+2^{x+2}=2^5\left(1+2^3\right)\)

\(\Leftrightarrow2^{x+1}\left(1+2^3\right)=2^5\left(1+2^3\right)\)

\(\Leftrightarrow2^{x-1}=2^5\)

\(\Leftrightarrow x-1=5\)

\(\Leftrightarrow x=5+1=6\)

Vậy x=6

Bạn ơi,cho mk hỏi tại sao từ chỗ \(2^{x-1}+2^{x+2}\)bạn lại suy ra được \(2^{x+1}\left(1+2^3\right)\)vậy bạn?

Giải:

\(x-5\sqrt{x}\) = 0 (\(x\) ≥ 0)

\(\sqrt{x}\) .(\(\sqrt{x}\) - 5) = 0

\(\left[\begin{array}{l}\sqrt{x}=0\\ \sqrt{x}-5=0\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ \sqrt{x}=5\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ x=25\end{array}\right.\)

Vậy \(x\in\) {0; 25}

\(x^5\) = 2\(x^7\)

\(x^5\) - 2\(x^7\) = 0

\(x^5\).(1 - 2\(x^2\)) = 0

\(\left[\begin{array}{l}x^5=0\\ 1-2x^2=0\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ 2x^2=1\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ x^2=\frac12\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ x=\pm\sqrt{\frac12}\end{array}\right.\)

Vậy \(x\) ∈ {- \(\sqrt{\frac12}\); 0; \(\sqrt{\frac12}\)}

Giải:

\(x-5\sqrt{x}\) = 0 (\(x\) ≥ 0)

\(\sqrt{x}\) .(\(\sqrt{x}\) - 5) = 0

\(\left[\begin{array}{l}\sqrt{x}=0\\ \sqrt{x}-5=0\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ \sqrt{x}=5\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ x=25\end{array}\right.\)

Vậy \(x\in\) {0; 25}

\(x^5\) = 2\(x^7\)

\(x^5\) - 2\(x^7\) = 0

\(x^5\).(1 - 2\(x^2\)) = 0

\(\left[\begin{array}{l}x^5=0\\ 1-2x^2=0\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ 2x^2=1\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ x^2=\frac12\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ x=-\frac{1}{\sqrt2}\\ x=\frac{1}{\sqrt2}\end{array}\right.\)

Vậy \(x\) \(\in\) {- \(\frac{1}{\sqrt2}\); 0; \(\frac{1}{\sqrt2}\)}

a) 2^3x+2 = 4^x+5 = (2²)^x+5 = 2^2x+10

=> 3x + 2 = 2x + 10

=> 3x - 2x = 10 - 2

x = 8

b) 3^-1.3^x + 9.3^x = 28

3^x. ( 3^-1 + 9) = 28

3^x. 28/3 = 28

3^x = 3 = 3¹

=> x = 1

     \(2^{3x+2}=4^{x+5}\)

\(\Rightarrow2^{3x+2}=\left(2^2\right)^{x+5}\)

\(\Rightarrow3x+2=2\left(x+5\right)\)

\(\Rightarrow3x+2=2x+10\)

\(\Rightarrow3x-2x=10-2\Rightarrow x=8\)

      \(3^{-1}.3^x+9.3^x=28\)

\(\Rightarrow\frac{1}{3}.3^x+9.3^x=28\)

\(\Rightarrow3^x.\left(9+\frac{1}{3}\right)=28\)

\(\Rightarrow3^x.\frac{28}{3}=28\)

\(\Rightarrow3^x=3\Rightarrow x=1\)

Chúc bạn học tốt.

\(x+3,5-\frac{4}{7}=\frac{3}{2}-5\frac{1}{8}\)

\(x+3,5-\frac{4}{7}=-\frac{29}{8}\)

\(x+3,5=-\frac{29}{8}+\frac{4}{7}\)

\(x+3,5=-\frac{117}{56}\)

      \(\Leftrightarrow x=-\frac{367}{56}\)

a)\(\frac{x+3}{x+5}=7\Leftrightarrow x+3=7\left(x+5\right)\)

\(\Leftrightarrow x+3=7x+35\)

\(\Leftrightarrow-6x=32\)

\(\Leftrightarrow x=-\frac{16}{3}\)

b)\(\frac{2x-1}{3x+5}=-\frac{2}{3}\)

\(\Leftrightarrow3\left(2x-1\right)=-2\left(3x+5\right)\)

\(\Leftrightarrow6x-3=-6x-10\)

\(\Leftrightarrow12x=-7\)

\(\Leftrightarrow x=-\frac{7}{12}\)

c)\(\frac{x+1}{4}=\frac{9}{x+1}\Leftrightarrow\left(x+1\right)^2=36\)

\(\Leftrightarrow\left(x+1\right)^2=6^2\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=6\\x+1=-6\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-7\end{cases}}}\)

d)\(\frac{6x-1}{2x+3}=\frac{3x}{x+2}\)

\(\Leftrightarrow\left(6x-1\right)\left(x+2\right)=3x\left(2x+3\right)\)

\(\Leftrightarrow6x^2+12x-x-2=6x^2+9x\)

\(\Leftrightarrow2x=2\Leftrightarrow x=1\)