Lời giải:
a. $2^3.8=2^3.2^3=2^6$

b. $5^2.25=5^2.5^2=5^4$

c. $27:3^2=3^3:3^2=3^1$

d. $4^2.16=4^2.4^2=4^4$
e. $5^3.5^6=5^9$

f. $3^4.3=3^5$

g.$3^5.4^5=(3.4)^5=12^5$
h. $8^5.2^3=(2^3)^5.2^3=2^{15}.2^3=2^{18}$

i. $a^3.a^5=a^8$

j. $x^7.x.x^4=x^{7+1+4}=x^{12}$

k. $5^6:5^3=5^3$
l. $3^{15}:3^3=3^{15-3}=3^{12}$

m. $4^6:4^6=4^0=1$
n. $9^8:3^2=(3^2)^8:3^2=3^{16}:3^2=3^{14}$

o. $a:a=a^0=1$
p. $5^8.5.5^2=5^{8+1+2}=5^{11}$
q. $4.4^3=4^4$
r. $3.3^4=3^5$

s. $36.6^5=6^2.6^5=6^7$

t. $2^5.2^3=2^8$

u. $3^{10}:3^3=3^7$

v. $2^{10}:2^3=2^7$

w. $5^8:25=5^8:5^2=5^6$
x. $16:2^3=2^4:2^3=2^1$
y. $4.2^3=2^2.2^3=2^5$
z. $2^{10}:4=2^{10}:2^2=2^8$

Đề bài:1+3+5+...+199

Tổng trên có số số hạng là: (199-1):2+1=100 (số hạng)

Kết quả của tổng trên là: (199+1)x100:2=10000

 Vậy 1+3+5+...+199=10000

\(1+3+5+...+199=\left(\left(199-1\right):2+1\right)\left(199+1\right):2=100.200:2=10000\)

`@` `\text {Ans}`

`\downarrow`

`j)`

\(x^{17}\div x^{12}=x^{17-12}=x^5\)

`k)`

\(x^8\div x^5=x^{8-5}=x^3\)

`r)`

\(a^5\div a^5=a^{5-5}=a^0=1\)

`l)`

\(x^4\div x=x^{4-1}=x^3\)

`m)`

\(x^7\div x^6=x^{7-6}=x\)

`n)`

\(x^9\div x^9=x^{9-9}=x^0=1\)

`o)`

\(a^{12}\div a^5=a^{12-5}=a^7\)

`p)`

\(a^8\div a^6=a^{8-6}=a^2\)

`q)`

\(a^{10}\div a^7=a^{10-7}=a^3\)

`r(2),`

\(1024\div4=2^{10}\div2^2=2^8\)

`t)`

\(512\div2^3=2^9\div2^3=2^6\)

giúp mình với, mình đang vội 

777

a) \(\dfrac{x}{3}=\dfrac{2}{3}+\dfrac{-1}{7}\)

\(\Rightarrow\dfrac{x}{3}=\dfrac{11}{21}\)

\(\Rightarrow x=\dfrac{3\cdot11}{21}\)

\(\Rightarrow x=\dfrac{33}{21}\)

\(\Rightarrow x=\dfrac{11}{7}\)

b) \(\dfrac{x}{5}=\dfrac{5}{6}+\dfrac{-19}{30}\)

\(\Rightarrow\dfrac{x}{5}=\dfrac{1}{5}\)

\(\Rightarrow x=\dfrac{5\cdot1}{5}\)

\(\Rightarrow x=1\)

`@` `\text {Ans}`

`\downarrow`

\(x+\dfrac{1}{6}=\dfrac{-3}{8}\)

`=>`\(x=\dfrac{-3}{8}-\dfrac{1}{6}\)

`=>`\(x=-\dfrac{13}{24}\)

Vậy, `x =`\(-\dfrac{13}{24}\)

\(x+\dfrac{1}{6}=-\dfrac{3}{8}\)

\(\Rightarrow x=\dfrac{-3}{8}-\dfrac{1}{6}\)

\(\Rightarrow x=-\dfrac{13}{24}\)

`@` `\text {Ans}`

`\downarrow`

`a)`

\(A=\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}\)

`=`\(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{8}-\dfrac{1}{9}\)

`=`\(\dfrac{1}{3}-\left(\dfrac{1}{4}-\dfrac{1}{4}\right)-\left(\dfrac{1}{5}-\dfrac{1}{5}\right)-...-\dfrac{1}{9}\)

`=`\(\dfrac{1}{3}-\dfrac{1}{9}\)

`=`\(\dfrac{2}{9}\)

Vậy, \(A=\dfrac{2}{9}\)

`b)`

\(B=\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+...+\dfrac{1}{23\cdot24}+\dfrac{1}{24\cdot25}\)

`=`\(\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}\)

`=`\(\dfrac{1}{5}-\left(\dfrac{1}{6}-\dfrac{1}{6}\right)-\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-...-\dfrac{1}{25}\)

`=`\(\dfrac{1}{5}-\dfrac{1}{25}=\dfrac{4}{25}\)

Vậy, \(B=\dfrac{4}{25}\)

`c)`

\(C=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{98\cdot99}+\dfrac{1}{99\cdot100}\)

`=`\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

`=`\(1-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-...-\dfrac{1}{100}\)

`=`\(1-\dfrac{1}{100}=\dfrac{99}{100}\)

Vậy, \(C=\dfrac{99}{100}\)

a) \(2^5\cdot2^7\)

\(=2^{5+7}\)

\(=2^{12}\)

b) \(2^3\cdot2^2\)

\(=2^{3+2}\)

\(=2^5\)

c) \(2^4\cdot2^3\cdot2^5\)

\(=2^{4+3+5}\)

\(=2^{12}\)

d) \(2^2\cdot2^4\cdot2^6\cdot2\)

\(=2^{2+4+6+1}\)

\(=2^{13}\)

e) \(2\cdot2^3\cdot2^7\cdot2^4\)

\(=2^{1+3+7+4}\)

\(=2^{15}\)

f) \(3^8\cdot3^7\)

\(=3^{8+7}\)

\(=3^{15}\)

g) \(3^2\cdot3\)

\(=3^{2+1}\)

\(=3^3\)

h) \(3^4\cdot3^2\cdot3\)

\(=3^{4+2+1}\)

\(=3^7\)

I) \(3\cdot3^5\cdot3^4\cdot3^2\)

\(=3^{1+5+4+2}\)

\(=3^{12}\)

`@` `\text {Ans}`

`\downarrow`

`A.`

`100-99+98-97+96-95+...+4-3+2-1 ?`

Ta có:

Số phần tử của bt trên là: `(100 - 1) \div 1 + 1 = 100 (\text {phần tử})`

Mà mỗi phần tử ghép với nhau thành `1` cặp

`=>` `100 \div 2 = 50 (\text {cặp})`

`100-99+98-97+96-95+...+4-3+2-1 `

`= (100 - 99) + (98 - 97) + ... + (4-3) + (2-1)`

`= 1+1+1 + ... + 1 + 1`

Mà bt trên có `50` cặp

`=>` Có `50` số `1`

`=>` Giá trị của bt trên là `50`

`B.`

`100-98+96-94+...+4-2`

Ta có:

Số phần tử của bt trên là: `(100 - 2) \div 2 + 1 = 50 (\text {phần tử})`

Mỗi phần tử ghép với nhau thành `1` cặp

`=> 50 \div 2 = 25 (\text {cặp})`

`100-98+96-94+...+4-2`

`= (100 - 98) + (96 - 94) + ... + (4 -2)`

`= 2 + 2 + ... + 2`

Mà bt trên có `25` cặp

`=>` Giá trị của bt trên là: `2 \times 25 = 50.`

d) 5.4³ + 2.3 - 81.2 + 7

= 5.64 + 6 - 162 + 7

= 320 + 6 - 162 + 7

= 326 - 162 + 7

= 164 + 7

= 171

e) [(33 - 3) : 3]³⁺³

= (30 : 3)⁶

= 10⁶

= 1000000

g) 2⁵ + 2.{12 + 2.[3.(5 - 2) + 1] + 1} + 1

= 32 + 2.[12.(3.3 + 1) + 1] + 1

= 32 + 2.(12.10 + 1) + 1

= 32 + 2.121 + 1

= 32 + 242 + 1

= 275