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

<=>  \(\orbr{\begin{cases}2x+1=25\\2x+1=-25\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=12\\x=-13\end{cases}}\)

Vậy...

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

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

\(\Rightarrow2x+1=25\)

\(\Leftrightarrow x=12\)

P/s tham khảo nha

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

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

\(\Rightarrow\orbr{\begin{cases}2x+1=25\\2x+1=-25\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=24\\2x=-26\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=12\\x=-13\end{cases}}}\)

Vậy \(\orbr{\begin{cases}x=12\\x=-13\end{cases}}\)

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

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

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

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

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

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

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

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

\(2x+1=\sqrt{625}\Leftrightarrow x=\frac{25-1}{2}\Leftrightarrow x=12\)

S={12}

Ta có

(2x+1)²=625

(2x+1)²=25²

=>2x+1=|25|

2x+1=25 hoặc 2x+1=-25

2x=24 hoặc 2x=-26

x=12 hoặc x=-13

Ta có

625=25²;(-25)²

^{Mà x là số tự nhiên} 

=>(2x+1)=25

=>x=12

x=312

cho li ke nhé

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 !!!!

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

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

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

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

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

a: =>2x+5=4

=>2x=-1

hay x=-1/2

b: \(\Leftrightarrow\left(3x-4\right)^2\cdot\left[\left(3x-4\right)^2-1\right]=0\)

=>(3x-4)(3x-5)(3x-3)=0

hay \(x\in\left\{1;\dfrac{4}{3};\dfrac{5}{3}\right\}\)

c: \(\Leftrightarrow3^{x+1}=3^{2x}\)

=>2x=x+1

=>x=1

d: \(\Leftrightarrow2^{2x+3}=2^{2x-10}\)

=>2x+3=2x-10

=>0x=-13(vô lý)