a)=-18790

b)=-1008

Bài 1 : 

A = 1 + 2 + 2² + ... + 2¹¹

A = ( 1 + 2 ) + ( 2² + 2³ ) + ... + ( 2¹⁰ + 2¹¹ )

A = 3 + 2²(1+2) + ... + 2¹⁰(1+2)

A = 1.3 + 2².3 + ... + 2¹⁰.3

A = 3.(1+2²+...+2¹⁰) chia hết cho 3

Bài 2 :

2.5² + 3:71⁰ - 54:3³

= 2.25 + 3:1 - 54:27

= 50 + 3 - 2

= 49

Bài 3 :

a) ( 2x - 6 ) . 4⁷ = 4⁹

2x - 6 = 4² = 16

2x = 16

=> x = 8

b) ( 27x + 6 ) : 3 - 11 = 9

( 27x + 6 ) : 3 = 20

27x + 6 = 60

27x = 54

=> x = 2

c) 740 : ( x + 10 ) = 10² - 2.13

740 : ( x + 10 ) = 74

x + 10 = 10

=> x = 0

d) ( 15 - 6x ) . 3⁵ = 3⁶

15 - 6x = 3

6x = 12

=> x = 2

Bài 4 :

Ta có : ab + ba = ( 10a + b ) + ( 10b + a ) = ( 10a + a ) + ( 10b + b ) = 11a + 11a = 11.(a+b) chia hết cho 11 

Bài 1 : 

A = 1 + 2 + 2² + ... + 2¹¹

A = ( 1 + 2 ) + ( 2² + 2³ ) + ... + ( 2¹⁰ + 2¹¹ )

A = 3 + 2²(1+2) + ... + 2¹⁰(1+2)

A = 1.3 + 2².3 + ... + 2¹⁰.3A = 3.(1+2²+...+2¹⁰) chia hết cho 3

Bài 2 :

2.5² + 3:71⁰ - 54:3³

= 2.25 + 3:1 - 54:27

= 50 + 3 - 2= 49

Bài 3 :

a) ( 2x - 6 ) . 4⁷ = 4⁹

2x - 6 = 4² = 16

2x = 16

=> x = 8

b) ( 27x + 6 ) : 3 - 11 = 9

( 27x + 6 ) : 3 = 20

27x + 6 = 60

27x = 54

=> x = 2

c) 740 : ( x + 10 ) = 10² - 2.13

740 : ( x + 10 ) = 74

x + 10 = 10

=> x = 0

d) ( 15 - 6x ) . 3⁵ = 3⁶

15 - 6x = 3

6x = 12

=> x = 2

Bài 4 :

Ta có : ab + ba = ( 10a + b ) + ( 10b + a ) = ( 10a + a ) + ( 10b + b ) = 11a + 11a = 11.(a+b) chia hết cho 11 

Bài 1:

6) 3x + 2³ = 17 + 3²

3x + 8 = 17 + 9

3x + 8 = 26

3x = 26 - 8

3x = 18

x = 18 : 3

x = 6

Vậy x = 6

Bài 2:

3) 145 - (125 + x) = 12

125 + x = 145 - 12

125 + x = 133

x = 133 - 125

x = 8

Vậy x = 8

6) 3³ - (x - 5) = 2²

27 - (x - 5) = 4

x - 5 = 27 - 4

x - 5 = 23

x = 23 + 5

x = 28

Vậy x = 28

9) (x + 7) - 15⁰ = 202 - 19

(x + 7) - 1 = 189

x + 7 = 189 + 1

x + 7 = 190

x = 190 - 7

x - 183

Vậy x = 183

B = 1 + 2\(^2\) + 2\(^4\) + ... + 2\(^{2024}\) + 2\(^{2026}\)

2\(^2\) B = 2\(^2\) + 2\(^4\) + ...+ \(2^{2026}\) + 2\(^{2028}\)

4B - B = 2\(^2\) + 2\(^4\) + ...+ \(2^{2026}\) + 2\(^{2028}\) - 1 - 2\(^2\) - 2\(^4\) - ... - 2\(^{2024}\) - 2\(^{2026}\)

3B = (2\(^2\) - 2\(^2\)) + (2\(^4\) - 2\(^4\)) +...+ (2\(^{2026}\) - \(2^{2026}\)) + (2\(^{2028}\) - 1)

3B = 0 + 0 +... + 0 + 2\(^{2028}\) - 1

3B = 2\(^{2028}\) - 1

B = \(\frac{2^{2028}-1}{3}\)

Ta có: \(B=1+2^2+2^4+\cdots+2^{2024}+2^{2026}\)

=>\(4B=2^2+2^4+2^6+\ldots+2^{2026}+2^{2028}\)

=>\(4B-B=2^2+2^4+2^6+\cdots+2^{2026}+2^{2028}-1-2^2-\cdots-2^{2026}\)

=>\(3B=2^{2028}-1\)

=>\(B=\frac{2^{2028}-1}{3}\)

đổi hết về cùng cơ số rồi tính nha, cùng cơ số r thì tính nhu bth thôi

3^6 : 9^3 = 3^6 : ( 3^3.3^3)

=3^6 : 3^6

= 1

mấy câu sau cứ đổi tương tự thôi

Tìm x:
b) 1/3.x+2/5.(x-1)=0

\(<=> \dfrac{1}{3} .x +\dfrac{2}{5}x - \dfrac{2}{5} =0\)

\(<=> \dfrac{11}{15}x = \dfrac{2}{5}\)

\(<=> x= \dfrac{6}{11}\)

Vậy \( x= \dfrac{6}{11}\)
c) (2x-3).(6-2x)=0
\(<=> \begin{cases} 2x-3=0 \\ 6-2x=0 \end{cases}\) \(<=> \begin{cases} 2x=3 \\ -2x=-6 \end{cases}\) \(<=>\begin{cases} x=\dfrac{3}{2} \\ x=3 \end{cases}\)

Vậy \(x=( \dfrac{3}{2} ; 3)\)
d) -2/3-1/3.(2x-5)= 3/2

\(<=> 2x-5= \dfrac{5}{2}\)

\(<=> 2x= \dfrac{15}{2}\)

\(<=> x= \dfrac{15}{4}\)

Vậy \(x= \dfrac{15}{4}\)
f) 1/3.x-1/2=4 và 1/2 (Hỗn số ý '^')

\(<=> \dfrac{1}{3} x -\dfrac{1}{2} = \dfrac{9}{2}\)

\(<=> \dfrac{1}{3}x =5\)

\(<=> x= 15\)

Vậy \(x= 15\)

Cảm ơn cậu nhiều nhé!

Tìm x εIN biết
a) 390 - (x-8) = 168:13
b) (x-140) : 7 = 27 - 24
c) x- 6 :2 - ( 48 - 24 ) :2 :6 - 3 = 0
d) x+5.2-(32+16.3:6-15)=0

b) \(\left(x-140\right):7=27-24\)

\(\left(x-140\right):7=3\)

\(x-140=21\)

\(x=161\)

      vay   \(x=161\)

c) \(x-6:2-\left(48-24\right):2:6-3=0\)

\(x-3-24:2:6-3=0\)

\(x-3-2-3=0\)

\(x-8=0\)

\(x=8\)

vay \(x=8\)

d) \(x+5.2-\left(32+16.3:6-15\right)=0\)

\(x+10-\left(32+8-15\right)=0\)

\(x+10-25=0\)

\(x-15=0\)

\(x=15\)

vay \(x=15\)

a) \(390-\left(x-8\right)=168:13\)

\(390-x+8=\frac{168}{13}\)

\(x+8=390-\frac{168}{13}\)

\(x+8=\frac{5070}{13}-\frac{168}{13}\)

\(x+8=\frac{4902}{13}\)

\(x=\frac{4902}{13}-8\)

\(x=\frac{4798}{13}\)

vay \(x=\frac{4798}{13}\)

a ) + ) Xét x \(\ge\) 3 => |x - 3| = x - 3

=> |x - 3| + x - 3 = x - 3 + x - 3 = 0

<=> 2x - 6 = 0

<=> 2x = 6

=> x = 3

+ ) Xét x < 3 => |x - 3| = 3 - x

=> |x - 3| + x - 3 = 3 - x + x - 3 = 0

<=> 0 = 0 ( loại )

Vậy x = 3

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