\(a,5^x< 0,125\\ \Leftrightarrow x< -1,292\\ b,\left(\dfrac{1}{3}\right)^{2x+1}\ge3\\ \Leftrightarrow2x+1\le-1\\ \Leftrightarrow2x\le-2\\ \Leftrightarrow x\le-1\)

c, Điều kiện: x > 0 

\(log_{0,3}x>0\\ \Leftrightarrow x>1\)

d, Điều kiện: \(x>\dfrac{3}{2}\) 

\(ln\left(x+4\right)>ln\left(2x-3\right)\\ \Rightarrow x+4>2x-3\\ \Leftrightarrow x< 7\)

Vậy \(\dfrac{3}{2}< x< 7\)

c, 28,35: 0,125 + 42,65: 0,125 + 1:0,125

= (28,35 + 42,65 + 1) : 0,125

= 72 : 0,125

= 72 x 8

= 576

a, 1,25 x 10,8 x 0,8:0,9

= (1,25 x 0,8) x (10,8 :0,9)

= 1 x 12

= 12

a) \(1=\left(2x+0,5\right)^{600}\)

\(\Rightarrow1^{600}=\left(2x+0,5\right)^{600}\)

\(\Rightarrow\left[{}\begin{matrix}2x+0,5=1\\2x+0,5=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=0,5\\2x=-1,5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0,25\\x=-0,75\end{matrix}\right.\)

b) \(\left(x-0,125\right)^2=0,25\)

\(\Rightarrow\left(x-0,125\right)^2=0,5^2\)

\(\Rightarrow\left[{}\begin{matrix}x-0,125=0,5\\x-0,125=-0,5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0,625\\x=-0,375\end{matrix}\right.\)

c) \(\left(x-3\right)^{11}=\left(x-3\right)^{41}\)

\(\Rightarrow\left(x-3\right)^{11}-\left(x-3\right)^{41}=0\)

\(\Rightarrow\left(x-3\right)^{11}\left[1-\left(x-3\right)^{30}\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x-3\right)^{11}=0\\\left(x-3\right)^{30}=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x-3=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

`@` `\text {Ans}`

`\downarrow`

`a)`

`1 = (2x + 0,5)^600`

`=> (2x+0,5)^600 = (+-1)^600`

`=> \text {TH1: } 2x + 0,5 = 1`

`=> 2x = 1 - 0,5`

`=> 2x = 0,5`

`=> x = 0,5 \div 2`

`=> x = 0,25`

`\text {TH2: } 2x + 0,5 = -1`

`=> 2x = -1 - 0,5`

`=> 2x = -1,5`

`=> x = -1,5 \div 2`

`=> x = -0,75`

Vậy, `x \in {-0,75; 0,25}.`

`b)`

`(x - 0,125)^2 = 0,25`

`=> (x - 0,125)^2 = (+-0,5)^2`

`=> `\(\left[{}\begin{matrix}x-0,125=0,5\\x-0,125=-0,5\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0,5+0,125\\x=-0,5+0,125\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0,625\\x=-0,375\end{matrix}\right.\)

Vậy, `x \in {-0,375; 0,625}.`

`c)`

`(x - 3)^11 = (x - 3)^41`

`=> (x - 3)^11 - (x - 3)^41 = 0`

`=> (x - 3)^11 * [ 1 - (x - 3)^30] = 0`

`=>`\(\left[{}\begin{matrix}\left(x-3\right)^{11}=0\\1-\left(x-3\right)^{30}=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x-3=0\\\left(x-3\right)^{30}=1\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=3\\x-3=1\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

Vậy, `x \in {3; 4}.`

a) \(\frac{-2}{3}x+\frac{1}{5}=\frac{1}{10}\)

\(\Leftrightarrow\frac{-2}{3}x=\frac{1}{10}-\frac{1}{5}\)

\(\Leftrightarrow\frac{-2}{3}x=\frac{-1}{10}\)

\(\Leftrightarrow x=\frac{-1}{10}\div\frac{-2}{3}\)

\(\Leftrightarrow x=\frac{3}{20}\)

ko có tên à

MÌNH ĐANG CẦN GẤP

Giữ lời hứa nka Nguyễn Mạnh Tân!

làm nhanh mình k cho

( 0,25x ) : 3 = 5/6 : 0,125

( 1/4x ) : 3 = 5/6 : 1/8

( 1/4x ) : 3 = 5/6 x 8

( 1/4x ) : 3 = 20/3

1/4x          = 20/3 x 3

1/4x         = 20

x              = 20 : 1/4

x              = 80

0,01 : 2,5 = ( 0,75x ) : 0,75

1/100 : 25/10 = (3/4x) : 3/4

1/100 x 10/25 = (3/4x) : 3/4

=>(3/4x) : 3/4 = 1/250

3/4x                = 1/250 x 3/4

3/4x                = 3/1000

x                     = 3/1000 : 3/4

x                     = 1/250

Bài này dễ thôi. Bạn cứ tích chéo lên là được mà.

Điều kiện : \(\begin{cases}x\ge\frac{1}{3}\\3x\in N\end{cases}\)

Từ phương trình ban đầu \(\Leftrightarrow\sqrt{2^x.2^{2.\frac{x}{3}}.\left(\frac{1}{8}\right)^{\frac{1}{3x}}}=2^2.2^{\frac{1}{3}}\)

                                     \(\Leftrightarrow2^{\frac{x}{2}}.2^{\frac{x}{3}}.2^{\frac{-1}{2x}}=2^{\frac{7}{3}}\)

                                     \(\Leftrightarrow2^{\frac{x}{2}+\frac{x}{3}-\frac{1}{2x}}=2^{\frac{7}{3}}\)

                                     \(\Leftrightarrow\frac{x}{2}+\frac{x}{3}-\frac{1}{2x}=\frac{7}{3}\)

                                     \(\Leftrightarrow5x^2-14x-3=0\)

                                      \(\Leftrightarrow\left[\begin{array}{nghiempt}x=3\\x=-\frac{1}{5}\end{array}\right.\)

Kết hợp với điều kiện ta có \(x=3\) là nghiệm của phương trình