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

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

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

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

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

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

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

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

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

giúp mình với

a: \(5^{x}\cdot\left(5^3\right)^2=625\)

=>\(5^{x}=\frac{5^4}{5^6}=5^{-2}\)

=>x=-2

b: \(\left(\frac{12}{15}\right)^{x}=\left(\frac53\right)^{-5}-\left(-\frac35\right)^4\)

=>\(\left(\frac45\right)^{x}=\left(\frac35\right)^5-\left(\frac35\right)^4=\left(\frac35\right)^4\cdot\left(\frac35-1\right)=\left(\frac35\right)^4\cdot\frac{-2}{5}=\frac{-2\cdot3^4}{5^5}\)

=>\(x=\log_{0,8}\left(-2\cdot\frac{3^4}{5^5}\right)\)

c: \(\left(-\frac34\right)^{3x-1}=\frac{256}{81}\)

=>\(\left(-\frac34\right)^{3x-1}=\left(-\frac34\right)^{-4}\)

=>3x-1=-4

=>3x=-3

=>x=-1

d: \(172x^2-7^9:98^3=2^{-3}\)

=>\(172x^2=\frac18+\frac{7^9}{7^6\cdot2^3}=\frac18+\frac{7^3}{2^3}=\frac{1+343}{8}=\frac{344}{8}\)

=>\(x^2=\frac{344}{8}:172=\frac{344}{8\cdot172}=\frac28=\frac14\)

=>\(\left[\begin{array}{l}x=\frac12\\ x=-\frac12\end{array}\right.\)

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: =>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)