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a: \(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5+3^5}\cdot\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5+2^5+2^5+2^5+2^5}=2^x\)
\(\Leftrightarrow2^x=\dfrac{4^5}{3^5}\cdot\dfrac{6^5}{2^5}=4^5=2^{10}\)
=>x=10
b: \(\left(x-1\right)^{x+4}=\left(x-1\right)^{x+2}\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow x\left(x-1\right)^{x+2}\cdot\left(x-2\right)=0\)
hay \(x\in\left\{0;1;2\right\}\)
c: \(6\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)
\(\Leftrightarrow5\cdot\left(6-x\right)^{2003}=0\)
\(\Leftrightarrow6-x=0\)
hay x=6

n.(2x-5)2=9
(2x-5)2=32
* 2x-5=3 * 2x-5=-3
2x=3+5 2x=-3+5
2x=8 2x=2
x=8:2 x=2:2
x=4 x=1
vậy x=4 hoặc x=1
o.(1-3x )3=-8
(1-3x)3=(-2)3
1-3x=-2
3x=1-(-2)
3x=3
x=3:3
x=1
vậy x=1

\(\left(\dfrac{3}{5}\right)^5.x=\left(\dfrac{9}{25}\right)^3.\left(\dfrac{3}{5}\right)^2\)
\(\Rightarrow\left(\dfrac{3}{5}\right)^3.x=\left(\dfrac{9}{25}\right)^3\)
\(\Rightarrow x=\left(\dfrac{3}{5}\right)^3\)
\(\Rightarrow x=\dfrac{27}{125}\)
Bạn sai rồi nhé như thế này mới đúng :
Lời giải :
\(\left(\dfrac{3}{5}\right)^5.x=\left(\dfrac{9}{25}\right)^3.\left(\dfrac{3}{5}\right)^2\)
\(\left(\dfrac{3}{5}\right)^2.\left(\dfrac{3}{5}\right)^3.x=\left(\dfrac{9}{25}\right)^3.\left(\dfrac{3}{5}\right)^2\)
Vì : \(\left(\dfrac{3}{5}\right)^2=\left(\dfrac{3}{5}\right)^2\)
Nên : \(\left(\dfrac{3}{5}\right)^3.x=\left(\dfrac{9}{25}\right)^3\)
\(\Rightarrow x=\left(\dfrac{9}{25}\right)^3:\left(\dfrac{3}{5}\right)^3\)
\(\Rightarrow x=\left(\dfrac{9}{25}:\dfrac{3}{5}\right)^3\)
\(\Rightarrow x=\left(\dfrac{3}{5}\right)^3\)
\(\Rightarrow x=\dfrac{3.3.3}{5.5.5}\)
\(\Rightarrow x=\dfrac{3^3}{5^3}\)
\(\Rightarrow x=\left(3:5\right)^3\)
\(\Rightarrow x=0,6^3\)
\(\Rightarrow x=0,216\)
Vậy \(x=0,216\)

\(\frac{3-x}{5-x}=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow\frac{3-x}{5-x}=\frac{9}{25}\)
\(\Rightarrow\left(3-x\right)25=9\left(5-x\right)\)
\(\Rightarrow75-25x=45-9x\)
\(\Rightarrow-25x+9x=45-75\)
\(\Rightarrow-16x=-30\)
\(\Rightarrow x=\frac{15}{8}\)

3) \(\left(x+\dfrac{1}{5}\right)^2\) + \(\dfrac{17}{25}\) = \(\dfrac{26}{25}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{26}{25}\) - \(\dfrac{17}{25}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{9}{25}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{3}{5}.\dfrac{3}{5}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\left(\dfrac{3}{5}\right)^2\)
=> \(x\) + \(\dfrac{1}{5}\) = \(\dfrac{3}{5}\)
=> \(x\) = \(\dfrac{3}{5}\) - \(\dfrac{1}{5}\)
=> \(x\) = \(\dfrac{2}{5}\)
4) -1\(\dfrac{5}{27}\) - \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-24}{27}\)
=> \(\dfrac{-32}{27}\) - \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-8}{9}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-32}{27}\) - \(\dfrac{-8}{9}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-8}{27}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-2}{3}\) . \(\dfrac{-2}{3}\) . \(\dfrac{-2}{3}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\left(\dfrac{-2}{3}\right)^3\)
=> \(3x-\dfrac{7}{9}=\dfrac{-2}{3}\)
=> \(3x=\dfrac{-2}{3}+\dfrac{7}{9}\)
=> \(3x=\dfrac{1}{9}\)
=> \(x=\dfrac{1}{9}:3\)
=> \(x=\dfrac{1}{27}\)

a) 15 + 23 = 1 + 8 = 9 = 32 ( là số chính phương )
b) 52 + 122 = 25 + 144 = 169 = 132 ( là số chính phương )
c) 26 + 62 = 64 + 36 = 100 = 1002 ( là số chính phương )
d) 13 + 23 + 33 + 43 + 53 + 63
= 1 + 8 + 27 + 64 + 125 + 216
= 441 = 212 ( là số chính phương )
a) 15 + 23=1 + 8 = 9 (là số chính phương)
b) 52 + 122= 25 + 144= 169 (là số chính phương)
c) 26 + 62= 64 + 36=100 (là số chính phương)
d) 142 – 122= 196 - 144=52 (không là số chính phương)
e) 13 + 23 + 33 + 43 + 53 + 63= 1 + 8 + 27 + 64 + 125 + 216 = 411 (là số chính phương)

So sánh :
a ) 31^11 và 17^14
31^11 < 32^11= (25)11 = 2^55
=> 31^11 < 2^55
17^14>16^14=(24)14 = 2^56
=>17^14>2^56
=>31^11 < 2^55 < 2^56 < 17^14
=>31^11 < 17^14
b ) 3^500 và 7^300
3^500 = ( 35)100 = 243100
7^300 = ( 73)100 = 343100
=> 243100 < 343100
=> 3^500 < 7^300
Tìm x :
a ) 2x . 4 = 128
=> 2x = 32
=> 2x = 25
=> x = 5
b ) 2x . 22 = ( 23)2 = 64
=> 2x = 64 : 22 = 16
=> 2x = 24
=> x = 4
Bài cuối bạn tham khảo tại : Câu hỏi của Linh Phan - Toán lớp 6 - Học toán với OnlineMath
Link : https://olm.vn/hoi-dap/detail/198524999512.html

a) \(2^x=32\)
Ta có: \(2^5=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
b) Sửa đề tí: \(9< 3^x< 81\)
\(\Rightarrow3^2< 3^x< 3^4\)
\(\Rightarrow2< x< 4\)
\(\Rightarrow x=\left\{3\right\}\)
Vậy x = 3
c) Ta có: \(25\le5^x\le125\)
\(\Rightarrow5^2\le5^x\le5^3\)
\(\Rightarrow2\le x\le3\)
\(\Rightarrow x=\left\{2;3\right\}\)
Vậy x = 2 hoặc x = 3
d) \(\left(x-2\right)^3\times5=40\)
\(\Rightarrow\left(x-2\right)^3=8\)
Mà \(8=2^3\Rightarrow\left(x-2\right)^3=2^3\)
Suy ra: x - 2 = 2
Vậy x = 4
5( x-25)= 23. 3. 5
5( x- 25)= 120
5x- 125= 120
5x= 120+ 125
5x= 245
x= 245: 5
x= 49
vậy x= 49
5(x - 25)= 23 x 3 x 5
5(x - 25)= 120
x - 25 = 120 : 5
x - 25 = 24
x = 24 + 25
x = 29