i) \(2345-1000\div\left[19-2\left(21-18\right)^2\right]\)

\(=\)\(2345-1000\div\left[19-2.3^2\right]\)

\(=\)\(2345-1000\div\left[19-2.9\right]\)

\(=\)\(2345-1000\div\left[19-18\right]\)

\(=\)\(2345-1000\div1\)

\(=\)\(2345-1000\)

\(=\)\(1345\)

j) \(128-\left[68+8\left(37-35\right)^2\right]\div4\)

\(=\)\(128-\left[68+8.2^2\right]\div4\)

\(=\)\(128-\left[68+8.4\right]\div4\)

\(=\)\(128-\left[68+32\right]\div4\)

\(=\)\(128-100\div4\)

\(=\)\(128-25\)

\(=\)\(3\)

k) \(568-\left\{5\left[143-\left(4-1\right)^2\right]+10\right\}\div10\)

\(=\)\(568-\left\{5\left[143-3^2\right]+10\right\}\div10\)

\(=\)\(568-\left\{5\left[143-9\right]+10\right\}\div10\)

\(=\)\(568-\left\{5.134+10\right\}\div10\)

\(=\)\(568-\left\{670+10\right\}\div10\)

\(=\)\(568-680\div10\)

\(=\)\(568-68\)

\(=\)\(500\)

a) \(107-\left\{38+\left[7.3^2-24\div6+\left(9-7\right)^3\right]\right\}\div15\)

\(=\)\(107-\left\{38+\left[7.3^2-24\div6+2^3\right]\right\}\div15\)

\(=\)\(107-\left\{38+\left[7.9-4+8\right]\right\}\div15\)

\(=\)\(107-\left\{38+\left[63-4+8\right]\right\}\div15\)

\(=\)\(107-\left\{38+67\right\}\div15\)

\(=\)\(107-105\div15\)

\(=\)\(107-7\)

\(=\)\(7\)

b) \(307-\left[\left(180-160\right)\div2^2+9\right]\div2\)

\(=\)\(307-\left[20\div4+9\right]\div2\)

\(=\)\(307-\left[5+9\right]\div2\)

\(=\)\(307-14\div2\)

\(=\)\(307-7\)

\(=\)\(300\)

c) \(205-\left[1200-\left(4^2-2.3\right)^3\right]\div40\)

\(=\)\(205-\left[1200-\left(16-6\right)^3\right]\div40\)

\(=\)\(205-\left[1200-10^3\right]\div40\)

\(=\)\(205-\left[1200-1000\right]\div40\)

\(=\)\(205-200\div40\)

\(=\)\(205-5\)

\(=\)\(200\)

a/\(A=\left(3+5\right)^2=8^2=64\)

\(B=3^2+5^2=9+25=34\)

\(\Rightarrow A>B\)

b/ \(C=\left(3+5\right)^3=8^3=512\)

\(D=3^3+5^3=27+125=152\)

\(\Rightarrow C>D\)

a/ A= (3+5)² = 8² = 64

   B = 3² + 5² = 9 + 25 = 34

vì 64>34 => A > B

b/ C = (3+5)³ = 8³ = 512

    D = 3³ + 5³ = 27 + 125 = 152

Vì 512>152 => C > D

a/ A = 8²

A = 64

B = 9 + 25, B = 34

b/ C = 8³, CC = 512

D = 27 + 125

D = 152

a. S = 1 + 2 + 2^2 + 2^3 + ... + 2^8 + 2^9

Ta có: 2 = 1 . 2

           2^2 = 2 . 2

           2^3 = 2^2 . 2

           .....

=>       1 + 2 + 2^2 + ... + 2^8 + (2^8 . 2)

=>       1 + 2 + 2^2 + ... + (2^8 . 3)

=>       1 + 2 + 2^2 + ... + 2^7 + (2^7 .6)

=>       1 + 2 + 2^2 + ... + (2^7 . 7)

=>        .....

=>        1 + 2 . 311

A= 1+2+2²+2³+.......+2⁹⁸+2⁹⁹     

A= (1+2)+(2²+2³)+.......+(2⁹⁸+2⁹⁹ )

A=3+2².(1+2)+...+2⁹⁸.(1+2)

A=   3+2².3+...+2⁹⁸.3 

A=3.(2²+...+2⁹⁸)

Vid 3 chia hết cho 3 nên A chia hết cho 3

Đơn giản như đang giỡn

HT

giúp mình với

Đuôi thì chẳng thấy, mà có hai đầu

a: \(2^3-5^3:5^2+12\cdot2^2\)

\(=8-5+12\cdot4\)

=3+48

=51

b: \(142-\left\lbrack50-\left(2^3\cdot10-2^3\cdot5\right)\right\rbrack\)

\(=142-50+\left(2^3\cdot10-2^3\cdot5\right)\)

\(=92+2^3\left(10-5\right)\)

\(=92+8\cdot5\)

=92+40

=132

=21 .[ 2232 : 8 - 180 ] + 21

=21.99 +21

=2100

\(21\cdot\left[\left(1245+987\right):2^3-15\cdot12\right]+21\)

\(=21\left(279-180+1\right)\)

\(=21\cdot100=2100\)