mk mới học lớp 6

bạn có thể trả lời giúp mình câu hỏi này ko?

a) 5^x = 125

5^x = 5³

=> x = 3

b) x³ = 64

x³ = 4³

=> x = 4

c) ( x - 1 ) ² = 25

( x - 1 ) ² = 5²

=> x - 1 = 5

=> x = 5 + 1

=> x = 6

d) 5^x + 5^x+2 = 130

5^x . 1 + 5^x . 5² = 130

5^x . ( 1 + 5² ) = 130

5^x . 26 = 130

5^x = 130 : 26

5^x = 5

=> x = 1

\(a.\)\(5^x=125\)

\(5^x=5^3\Rightarrow x=3\)

\(b.\)\(x^3=64\)

\(x^3=4^3\Rightarrow x=4\)

\(c.\)\(\left(x-1\right)^2=25\)

\(x^2=5^2\)mà \(\left(x-1\right)\)nên \(x=5+1=6\)

P/s: Ý d thiếu nhé bạn 

a)5^x=125

5^x=5³

=>x=3

b)tương tụ câu a

c)(x-1)²=25

(x-1)²=5²

=>x-1=5

x=5+1

x=6

d)thiếu đề!

a) 5³⁶ = 5¹² (5³)¹² = 125¹²; 11²⁴ = 11^2.12 = (11²)¹² = 121¹²

a: \(4^5\cdot8^7=2^{10}\cdot2^{21}=2^{31}\)

b: \(125^5\cdot25^3=5^{15}\cdot5^6=5^{21}\)

a) \(4^5\)x \(8^7\)

= \(\left(2^2\right)^5\) x \(\left(2^3\right)^7\)

= \(2^{10}\) x \(2^{21}\)

= \(2^{31}\)

b) \(125^5\) x \(25^3\)

= \(\left(5^3\right)^5\) x \(\left(5^2\right)^3\)

= \(5^{15}\) x \(5^6\)

= \(5^{21}\)

Bài 2: 

a: \(\Leftrightarrow\left(x-5\right)\left(x+5\right)-\left(x+5\right)=0\)

=>(x+5)(x-6)=0

=>x=-5 hoặc x=6

b: \(\Leftrightarrow4x^2-4x+1-4x^2+1=0\)

=>-4x+2=0

hay x=1/2

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

=>x=1 hoặc x=-1

Bạn có thể tham khảo lời giải ở đây nhé:

https://diendantoanhoc.org/topic/81694-t%C3%ADnh-t%E1%BB%95ng-s-13-23-33-n3/

mk not biết

( 3x - 1 )^2 = 25 = (+-5)^2

+) 3x - 1 = 5

3x = 6

x = 2

+) 3x - 1 = -5

3x = -4

x = -4/3

Vậy,.........

Các câu còn lại tương tự

\(\left(9^{30}-27^{19}\right):3^{57}+\left(125^9-25^{12}\right):5^{24}\)

\(=\left(3^{60}-3^{57}\right):3^{57}+\left(5^{27}-5^{24}\right):5^{24}\)

\(=3^{57}\left(3^3-1\right):3^{57}+5^{24}\left(5^3-1\right):5^{24}\)

\(=3^3-1+5^3-1\)

\(=27-1+125-1\)

\(=150\)

2 )

\(x^2-25-\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-5\right)-\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-5-1\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-6\right)=0\)

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

Vậy ...

b )

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

\(\Leftrightarrow4x^2-4x+1-4x^2+1=0\)

\(\Leftrightarrow2-4x=0\)

\(\Leftrightarrow4x=2\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy ...

c )

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

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

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

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

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

Vậy ...