a) \(\frac{1}{81}\): 3^x = \(\frac{1}{729}\)

     3^x = \(\frac{1}{81}\): \(\frac{1}{729}\)

     3^x = 9 

=> x = 2 ( vì 3² = 9 ) 

Ta có: 25^{x + 1} . 125^x . 625^{x + 2} = (5²)⁵

=> (5²)^{x + 1} . (5³)^x  . (5⁴)^{x+ 1} = 5¹⁰

=> 5^{2x + 2} . 5^3x . 5^{4x + 8} = 5¹⁰

=> 2x + 2 + 3x + 4x + 8 = 10

=> 9x + 2 + 8 = 10

=> 9x = 10 - 2 - 8

=> 9x = 0

=> x = 0 : 9

=> x = 0

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 \(x=6\)

a,x²-25-(x+5)                                                   b,mình quên mất rồi.Đợi tí nhé

(x²-25)-(x+5)=0

(x²-5²)+(x-5)=0

(x-5)(x+5)+(x-5)=0

(x-5)(x+5+1)=0

x-5=0 hoặc x+5+1=0

x=0+5 hoặc x=0-5-1

x=5     hoặc x=-6

Vậy x=5 và x=-6

Giải bpt

A)  (x^2+1)×(4x-2)≫0(lớn hơn hoặc =0)

B) (x-2)×x^2>0

Mog mn giúp ạ

E cần gấp

Thak mn

`#040911`

\(x-\dfrac{1}{6}=x+\dfrac{5}{7}\\ \Rightarrow x-x=\dfrac{1}{6}+\dfrac{5}{7}\\\Rightarrow0=\dfrac{37}{42}\left(\text{vô lý}\right)\\ \text{Vậy, x không có giá trị nào thỏa mãn.} \)

(x-5) . (x-5)=25.4

(x-5) . (x-5)=100

10.10=100

Vậy x=15

\(\frac{x-5}{25}=\frac{4}{x-5}\)

\(\Leftrightarrow\left(x-5\right)\left(x-5\right)=25.4\)

\(\left(x-5\right)^2=100\)

\(\left(x-5\right)^2=\pm10^2\)

\(\Rightarrow\hept{\begin{cases}x-5=10\\x-5=-10\end{cases}\Rightarrow\hept{\begin{cases}x=15\\x=-5\end{cases}}}\)

a/ (-8) x 25 x (-2) x 4 x (-5) x 125

= [(-8) x 125) x (25 x 4) x [(-2) x (-5)]

= (-1000) x 100 x 10

= -1000000

b/ (a + b) x (1 + X + y)

= (a + b) x 1 + (a + b) x X + (a + b) x y

= a x 1 + b x 1 + a x X + b X x + a x y + b x y

= a + b + a x X + b x X + a x y + b x y

ủng hộ mk nha !!!

\(a,(x-2)^2-25=0\\\Leftrightarrow (x-2)^2=25\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)

\(---\)

\(b,4x(x-2)+x-2=0\\\Leftrightarrow4x(x-2)+(x-2)=0\\\Leftrightarrow(x-2)(4x+1)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\4x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{1}{4}\end{matrix}\right.\)

\(---\)

\(c,4x(x-2)-x(3+4x)(?)\)

\(d,(2x-5)^2-3x(5-2x)=0\\\Leftrightarrow(2x-5)^2+3x(2x-5)=0\\\Leftrightarrow(2x-5)(2x-5+3x)=0\\\Leftrightarrow(2x-5)(5x-5)=0\)

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

\(---\)

\(e,x^2-25-(x+5)=0(sửa.đề)\\\Leftrightarrow(x^2-5^2)-(x+5)=0\\\Leftrightarrow (x-5)(x+5)-(x+5)=0\\\Leftrightarrow(x+5)(x-5-1)=0\\\Leftrightarrow(x+5)(x-6)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=6\end{matrix}\right.\)

\(---\)

\(f,5x(x-3)-x+3=0\\\Leftrightarrow5x(x-3)-(x-3)=0\\\Leftrightarrow(x-3)(5x-1)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)

\(Toru\)