Thay a = 2 vào biểu thức:

54 : a + 32 - 4 ta được:

54 : 2 + 32 - 4

= 27 + 32 - 4

= 59 - 4

= 55

54 : 2 + 32 - 4

= 27 + 32 - 4

= 59 - 4

= 55

\(\boxed{\star19/08/2004}\)

\(\left(-\dfrac65\right)^{3x-1}=\left(\dfrac{36}{25}\right)^7\)

\(\Rightarrow\left(-\dfrac65\right)^{3x-1}=\left\lbrack\left(-\dfrac65\right)^2\right\rbrack^7\)

\(\Rightarrow\left(-\dfrac65\right)^{3x-1}=\left(-\dfrac65\right)^{2\cdot7}\)

\(\Rightarrow\left(-\dfrac65\right)^{3x-1}=\left(-\dfrac65\right)^{14}\)

\(\Rightarrow3x-1=14\)

\(\Rightarrow3x=15\)

\(\Rightarrow x=5\)

Vậy \(x=5\)

(-6/5) ^ 3x-1 = (36/25)^7

(-6/5) ^ 3x-1=[(-6/5)^2]^7

(-6/5) ^ 3x-1= (-6/5)^14

⇒3x-1=14

3x=14+1

3x=15

x=15:3

x=5

Vậy x=5

Ta có:

S = (1 - 3 + 3^2 - 3^3) + (3^4 - 3^5 + 3^6 - 3^7) + ... + (3^96 - 3^97 + 3^98 - 3^99)

= (1 - 3 + 9 - 27) + 3^4(1 - 3 + 9 - 27) + ... + 3^96(1 - 3 + 9 - 27)

= -20 - 20.3^4 - ... - 20.3^96

= -20(1 + 3^4 + 3^8 + ... + 3^96)

Vậy S là bội của (-20) ĐPCM

ta có: số lượng số hạng là: (99-0):1+1=100( số hạng)

<=> \(S=\left(1-3+3^2-3^3\right)+\left(3^4-3^5+3^6-3^7\right)+\cdots+\left(3^{96}-3^{97}+3^{98}-3^{99}\right)\)

= \(\left(-20\right)+3^4\left(1-3+3^2-3^3\right)+\cdots+3^{96}\left(1-3+3^2-3^3\right)\)

= \(\left(-20\right)\left(1+3^4+\cdots+3^{96}\right)\)

vì \(\left(1+3^4+\cdots+3^{96}\right)\) là một số nguyên

=> S là bội của -20(đpcm)

7 tạ 87 kg = 7 tạ + 87kg = 700kg + 87kg = 787kg

đổi 7 tạ=700kg

7 tạ 87 kg = 7 tạ + 87kg = 700 kg +87 kg = 787kg

`8^2 * 4^5 * 2^20`

`= 2*(3*2) * 2^(2*5) + 2^20`

`= 2*6 * 2^10 * 2^20`

`= 2^(6 + 10 +20)`

`=2^36`

\(\boxed{\star19/08/2004}\)

\(\dfrac{8^2\cdot4^5}{2^{20}}\)

\(=\dfrac{2^6\cdot2^{10}}{2^{20}}\)

\(=\dfrac{2^{6+10}}{2^{20}}\)

\(=\dfrac{2^{16}}{2^{20}}\)

\(=\dfrac{1}{2^{20-16}}\)

\(=\dfrac{1}{2^4}\)

\(=\dfrac{1}{16}\)

-3 + x = 2

x = 2 - (-3)

x = 5

Vậy x = 5

x = 2 + 3 = 5
vay x = 5

238 - 4 \(\times\) (\(x\) + 9) = 10

4 \(\times\) (\(x\) + 9) = 238 - 10

4 \(\times\) (\(x\) + 9) = 228

\(x\) + 9 = 228 : 4

\(x\) + 9 = 57

\(x\) = 57 - 9

\(x\) = 48

Vậy \(x\) = 48

$\textbf{a)}$

$A=2010\cdot2010=2010^2.$

$B=2009\cdot2011=(2010-1)(2010+1)=2010^2-1.$

=> $A-B=2010^2-(2010^2-1)=1>0.$

Vậy $A>B.$

$\textbf{b)}$

$A=234234\cdot233.$

$B=233233\cdot234.$

Ta có $A-B=234234\cdot233-233233\cdot234$

$=234234\cdot233-233233\cdot233-233233$

$=(234234-233233)\cdot233-233233$

$=1001\cdot233-233233$

$=233233-233233$

$=0.

Vậy $A=B.$$=0.$

Vậy $A=B.$

a)

\(A = 2010 \cdot 2010 = 201 0^{2} .\)

\(B = 2009 \cdot 2011 = \left(\right. 2010 - 1 \left.\right) \left(\right. 2010 + 1 \left.\right) = 201 0^{2} - 1.\)

=> \(A - B = 201 0^{2} - \left(\right. 201 0^{2} - 1 \left.\right) = 1 > 0.\)

Vậy \(A > B .\)

\(\textbf{b})\)

\(A = 234234 \cdot 233.\)

\(B = 233233 \cdot 234.\)

Ta có \(A - B = 234234 \cdot 233 - 233233 \cdot 234\)

\(= 234234 \cdot 233 - 233233 \cdot 233 - 233233\)

\(= \left(\right. 234234 - 233233 \left.\right) \cdot 233 - 233233\)

\(= 1001 \cdot 233 - 233233\)

\(= 233233 - 233233\)

$=0.

Vậy $A=B.$$=0.$

Vậy

\(A = B .\)

\(\frac{\left(\right. 2^{6} \left.\right)^{x}}{\left(\right. 2^{4} \left.\right)^{x}} = 2^{8}\)

\(\frac{2^{6 x}}{2^{4 x}} = 2^{8}\)

\(2^{6 x - 4 x} = 2^{8}\)

\(2^{2 x} = 2^{8}\)

\(2 x = 8\)

\(x = 4\)

`64^x : 16^x = 256`

`=> 2^(6x) : 2^(4x) = 2^8`

`=> 6x - 4x = 8`

`=> 2x = 8`

`=> x = 4`

Vậy `x = 4`