\(1+3+5+...+2x+1=625\)

\(\Rightarrow\left[\left(2x+1-1\right):2+1\right]\cdot\left(2x+1+1\right):2=625\)

\(\Rightarrow\left(2x:2+1\right)\cdot\left(2x+2\right):2=625\)

\(\Rightarrow\left(x+1\right)\cdot2\left(x+1\right):2=625\)

\(\Rightarrow\left(x+1\right)^2=625\)

\(\Rightarrow\left(x+1\right)^2=25^2\)

\(\Rightarrow\left[{}\begin{matrix}x+1=25\\x+1=-25\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=24\\x=-26\end{matrix}\right.\)

Do 1+3+5+7+......+(2x+1) = 625 nên x thuộc N

Đặt A = 1+3+5+7+......+(2x+1)

Số số hạng của A là : [(2x+1) - 1] / 2 + 1 = x+1(số hạng)

Tổng A = {[(2x+1) + 1] . (x+1)} / 2 = (2x+2)(x+1) / 2 = 2(x+1)² / 2 = (x+1)² = 625

=>x+1=25

=>x=24

Nếu thấy đúng thì k nha !!!!

tui cũng đang bí

\(2^x=16\Leftrightarrow2^x=2^4\Leftrightarrow x=4\)

\(2^x=128\Leftrightarrow2^x=2^7\Leftrightarrow x=7\)

b), c) tương tự

d) \(5^{2x}=625\Leftrightarrow5^{2x}=5^4\Leftrightarrow2x=4\Leftrightarrow x=2\)

e) tương tự d

đéo biết đâu lũ ngu đừng hỏi

giúp mình với

a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)

                 \(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)

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

           \(=>2x+\frac{3}{5}=\frac{3}{5}\)

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

                                    \(2x=0\)

                                      \(x=0:2\)

                                      \(x=0\)

b) \(\left(3x-1\right).\left(-\frac{1}{2x}+5\right)=0\)

=> \(\left(3x-1\right)=0\)hoặc \(\left(-\frac{1}{2x}+5\right)=0\)hoặc \(\left(3x-1\right)\)và\(\left(-\frac{1}{2x}+5\right)\)cùng bằng 0.

\(\orbr{\begin{cases}3x-1=0\\-\frac{1}{2x}+5=0\end{cases}}=>\orbr{\begin{cases}3x=1\\-\frac{1}{2x}=-5\end{cases}}=>\orbr{\begin{cases}x\in\varnothing\\2x=\frac{1}{5}\end{cases}}=>x=\frac{1}{5}:2=>x=\frac{1}{10}\)

a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)

\(\left(2x+\frac{3}{5}\right)^2=0+\frac{9}{25}\)

\(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)

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

\(\orbr{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)

\(\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)

b) \(\left(3x-1\right)\left(-\frac{1}{2}x+5\right)=0\)

\(\left(3x-1\right)\left(-\frac{x}{2}+5\right)=0\)

\(\left(3x-1\right)\left(5-\frac{x}{2}\right)=0\)

\(\orbr{\begin{cases}3x-1=0\\5-\frac{x}{2}=0\end{cases}}\)

\(\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}\)

\(\left(\frac{3}{5}\right)^x.\left(\frac{625}{81}\right)^3=\frac{243}{3125}\)

=>\(\left(\frac{3}{5}\right)^x.\left(\frac{5}{3}\right)^{12}=\left(\frac{3}{5}\right)^5\)

=>\(\left(\frac{3}{5}\right)^x=\left(\frac{3}{5}\right)^5:\left(\frac{5}{3}\right)^{12}\)

=>\(\left(\frac{3}{5}\right)^x=\left(\frac{3}{5}\right)^{17}\)

=>x=17

\(\left(\frac{3}{5}\right)^x.\left(\frac{625}{81}\right)^3=\frac{243}{3125}\)

\(\Rightarrow\left(\frac{3}{5}\right)^x.\left(\frac{3}{5}\right)^{12}=\left(\frac{3}{5}\right)^5\)

\(\Rightarrow\left(\frac{3}{5}\right)^x=\left(\frac{3}{5}\right)^5:\left(\frac{3}{5}\right)^{12}\)

\(\Rightarrow\left(\frac{3}{5}\right)^x=\left(\frac{5}{3}\right)^7\)

\(\Rightarrow x=-7\)