Có : \(\dfrac{x}{5}=\dfrac{125}{x}\)

\(\Leftrightarrow x^2=5.125\)

\(\Leftrightarrow x=\pm25\)

Có : \(\dfrac{x}{5}=\dfrac{125}{x}\)

\(\Leftrightarrow x^2=5.125\)

\(\Leftrightarrow x=\pm25\)

b) Ta có: \(-5+\left|3x-1\right|+6=\left|-4\right|\)

\(\Leftrightarrow\left|3x+1\right|+1=4\)

\(\Leftrightarrow\left|3x+1\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=3\\3x+1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{4}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{2}{3};-\dfrac{4}{3}\right\}\)

c) Ta có: \(\left(x-1\right)^2=\left(x-1\right)^4\)

\(\Leftrightarrow\left(x-1\right)^2-\left(x-1\right)^4=0\)

\(\Leftrightarrow\left(x-1\right)^4-\left(x-1\right)^2=0\)

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

\(\Leftrightarrow\left(x-1\right)^2\cdot\left(x-1-1\right)\left(x-1+1\right)=0\)

\(\Leftrightarrow x\cdot\left(x-1\right)^2\cdot\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x-1\right)^2=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=2\end{matrix}\right.\)

Vậy: \(x\in\left\{0;1;2\right\}\)

d) Ta có: \(5^{-1}\cdot25^x=125\)

\(\Leftrightarrow5^{-1}\cdot5^{2x}=5^3\)

\(\Leftrightarrow5^{2x-1}=5^3\)

\(\Leftrightarrow2x-1=3\)

\(\Leftrightarrow2x=4\)

hay x=2

Vậy: x=2

a , 5^-1 . 25^x = 125

\(\frac{1}{5}.25^x=125\)

\(25^x=125.5=625\)

\(25^x=25^2\)

x = 2

b , 5x - | 9 - 7x | = 3

Với x < \(\frac{9}{7}\)

9 - 7x < 0

| 9 - 7x | = - ( 9 - 7x ) = 7x - 9

5x - ( 7x - 9 ) = 3

5x - 7x + 9 = 3

-2x = -6

x = -6 : (-2 )

x = 3

Với x >= \(\frac{7}{9}\)

9 - 7x >= 0

| 9 - 7x | = 9 - 7x

5x - 9 - 7x = 3

-2x - 9 = 3

-2x = 3 + 9

-2x = 12

x = 12 : ( -2 ) = -6

so sánh gì thế bạn

\(x^3=-125\)

\(x^3=\left(-5\right)^3\)

\(\Rightarrow x=-5\)

x³ = -125

x³ = ( -5)³

=> x = - 5

Đây bạn :V

Ta có: \(x^3=-125\)

\(\Rightarrow x^3=\left(-5\right)^3\)

\(\Rightarrow x=-5\)

Học tốt nha :V

a, \(\left(3x-7\right)-\left(-5-7x\right)=4x+9\)

\(\Leftrightarrow3x-7+5+7x=4x+9\)

\(\Leftrightarrow3x+7x-4x=9+7-5\)

\(\Leftrightarrow\left(3+7-6\right)x=11\Rightarrow4x=11\Rightarrow x=\frac{11}{4}\notin Z\)

Vậy : \(x\in\varnothing\)

b, \(\left(3x-7\right)-\left(-5-7x\right)=5x+8\)

\(\Leftrightarrow3x-7+5+7x=5x+8\)

\(\Leftrightarrow3x+7x-5x=8+7-5\)

\(\Leftrightarrow\left(3+7-5\right)x=10\Rightarrow5x=10\Rightarrow x=10\div5=2\in Z\)

Vậy : x = 2

c, \(\left(3x-2\right)^3=125\Leftrightarrow\left(3x-2\right)^3=5^3\)

\(\Rightarrow3x-2=5\Rightarrow3x=5+2\Rightarrow3x=7\)

\(\Rightarrow x=7\div3\Rightarrow x=\frac{7}{3}\notin Z\)

Vậy \(x\in\varnothing\)

a, (3x-7)-(-5-7x)=4x+9

3x-7+5+7x=4x+9

3x+7x-4x=9+7-5

6x=11

x=11/6(thuộc Z t/m)

Vậy...

b, (3x-7)-(-5-7x)=5x+8

3x-7+5+7x=5x+8

3x+7x-5x=8+7-5

5x=10

x=2(thuộc Z t/m)

vây...

c, (3x-2)^3=125

(3x-2)^3=5^3

3x-2=5

x=7/3(thuộc Z thỏa mãn)

Vậy...

a) \(5^{-1}.25^x=125\)

\(\Rightarrow5^{-1}.5^{2x}=5^3\)

\(\Rightarrow5^{2x-1}=5^3\)

\(\Rightarrow2x-1=3\)

\(\Rightarrow2x=4\)

\(\Rightarrow x=2\)

Vậy \(x=2\)

b) \(|x+1|+|x+2|+|x+3|=4x\)

Vì \(\hept{\begin{cases}|x+1|\ge0\forall x\\|x+2|\ge0\forall x\\|x+3|\ge0\forall x\end{cases}}\)

\(\Rightarrow|x+1|+|x+2|+|x+3|\ge0\)

\(\Rightarrow4x\ge0\)

\(\Rightarrow x\ge0\)

\(\Rightarrow\hept{\begin{cases}x+1>0\\x+2>0\\x+3>0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}|x+1|=x+1\\|x+2|=x+2\\|x+3|=x+3\end{cases}}\)

\(\Rightarrow\left(x+1\right)+\left(x+2\right)+\left(x+3\right)=4x\)

\(\Rightarrow3x+6=4x\)

\(\Rightarrow x=6\)

Vậy \(x=6\)

Chọn A.

Phương trình 

Lấy logarit cơ số 3 hai vế của (*), ta được 

Suy ra 

Cách nhanh nhất có loại câu này khỏi mục "chưa ai trả lời"

2.

c) /x-2/=9+15=24

x=26

hoạc

x=-22