Tìm x 5×^2=1 2x×4=128 (2x+1)^3=125 2020^2x_6=1 X^2020=x 3^x+x^2=1 5^2×5^x=625 (5^2x×5^x+1):5=125 3^x+3^x+1+3^x+2=243×39 Các anh cj giúp em vs ạ!! Em

1. $5^{x^{2}}=1 \Leftrightarrow 5^{x^{2}}=5^{0}\Leftrightarrow x^{2}=0\Leftrightarrow x=0$

2. $2x.4=128 \Leftrightarrow 2x=128:4 \Leftrightarrow 2x=32 \Leftrightarrow x=32:2 \Leftrightarrow x=16$

3. $(2x+1)^{3}=125\Leftrightarrow (2x+1)^{3}=5^{3}\Leftrightarrow 2x+1=5\Leftrightarrow 2x=4\Leftrightarrow x=2$

4. $2020^{2x-6}=1\Leftrightarrow 2020^{2x-6}=2020^{0}\Leftrightarrow \Leftrightarrow 2x-6=0 \Leftrightarrow 2x=6 \Leftrightarrow x=3$

5. $x^{2020}=x\Leftrightarrow x^{2020}-x=0\Leftrightarrow x.(x^{2019}-1)=0$
$\Leftrightarrow \left[\begin{array}{l}x=0 \\x^{2019}-1=0 \end{array}\right.\Leftrightarrow \left[\begin{array}{l}x=0 \\x^{2019}=1 \end{array}\right.\Leftrightarrow \left[\begin{array}{l}x=0 \\x=1 \end{array}\right.$

6. $3^{x+x^{2}}=1 \Leftrightarrow 3^{x+x^{2}}=3^{0} \Leftrightarrow x+x^{2}=0 \Leftrightarrow x.(1+x)=0$
$\Leftrightarrow \left[\begin{array}{l}x=0 \\1+x=0 \end{array}\right.\Leftrightarrow \left[\begin{array}{l}x=0 \\x=-1 \end{array}\right.$

7. $5^{2}.5^{x}=625 \Leftrightarrow 5^{2+x}=5^{4} \Leftrightarrow 2+x=4 \Leftrightarrow x=2$

8. $5^{2x}.5^{x+1}:5=125 \Leftrightarrow 5^{2x}.5^{x+1}=125.5\Leftrightarrow 5^{3x+1}=625\Leftrightarrow 5^{3x+1}=5^{4}\Leftrightarrow 3x+1=4\Leftrightarrow 3x=3\Leftrightarrow x=1$

9. $3^{x}+3^{x+1}+3^{x+2}=243.39$
$\Leftrightarrow 3^{x}+3^{x}.3+3^{x}.3^{2}=9477$
$\Leftrightarrow 3^{x}.(1+3+9)=3^{6}.13\Leftrightarrow 3^{x}.13=3^{6}.13\Leftrightarrow 3^{x}=3^{6}\Leftrightarrow x=6$