\(5^{2.x+1}=125\)

\(\Rightarrow5^{2x}.5=125\)

\(\Rightarrow5^{2x}=25\)

\(\Rightarrow25^x=25\)

\(\Rightarrow x=1\)

=>\(5^{2x}.5=125\)

=>\(5^{2x}=25\)

=> \(25^x=25\Rightarrow x=1\)

     \(5^{2x+1=125}\)

\(\Rightarrow5^{2x+1}=5^3\)

\(\Rightarrow2x+1=3\)

\(\Rightarrow2x=2\)

\(\Rightarrow x=1\)

Chúc bạn lam bài tốt nhé !

đè sai hả kg ra kq nếu đề sai k tui đưa đề ms lm cho

\(2^3+(3x+1)^3=24\)

\(\Rightarrow8+(3x+1)^3=24\)

\(\Rightarrow(3x+1)^3=16\)

\(\Rightarrow x\in\varnothing\)

| x+1| +5 = 2

|x+1| = 2-5

| x+ 1| = -3 ( Vô Lí)

=> \(x\in\left\{\varnothing\right\}\)

\(x+\dfrac{3}{4}=\dfrac{28}{6}+\dfrac{1}{2}\\ x+\dfrac{3}{4}=\dfrac{31}{6}\\ x=\dfrac{31}{6}-\dfrac{3}{4}\\ x=\dfrac{53}{12}\)

\(x+\dfrac{3}{4}=\dfrac{28+3}{6}=\dfrac{31}{6}\Leftrightarrow x=\dfrac{31}{6}-\dfrac{3}{4}=\dfrac{62-9}{12}=\dfrac{53}{12}\)

Dễ thế mà không làm được

Ta có:

\(\frac{x}{4}-\frac{1}{y}=\frac{1}{2}\)

\(\frac{xy}{4y}-\frac{4}{4y}=\frac{1}{2}\)

\(\frac{xy-4}{4y}=\frac{1}{2}\)

      \(2.\left(xy-4\right)=4y\)

      \(2xy-8-4y=0\)

       \(2xy-2-4-4y=0\)

        \(2.\left(xy+1\right)-4.\left(y+1\right)=0\)

        \(2.\left(xy+1\right)-2.2.\left(y+1\right)=0\)

        \(2.\left[\left(xy+1\right)-2.\left(y+1\right)\right]=0\)

        \(xy+1-2y-2=0\)

        \(y.\left(x-2\right)=1\)

Ta có:1=1.1=(-1).(-1)

     Do đó ta có bảng sau:

          Vậy cặp (x;y) TM là:(3;1)(1;-1)

Theo đầu bài ta có:
\(\frac{x}{4}-\frac{1}{y}=\frac{1}{2}\)
\(\Rightarrow\frac{xy-4}{4y}=\frac{1}{2}\)
\(\Rightarrow2\left(xy-4\right)=4y\)
\(\Rightarrow xy-4=2y\)
\(\Rightarrow xy-2y=4\)
\(\Rightarrow y\left(x-2\right)=4\)
Từ đó ta có bảng sau: