(y + \(\dfrac{1}{3}\)) + ( y + \(\dfrac{1}{9}\)) + ( y + \(\dfrac{1}{27}\)) + ( y + \(\dfrac{1}{81}\)) = \(\dfrac{56}{81}\)

( y + y + y + y ) + (\(\dfrac{1}{3}\)+ \(\dfrac{1}{9}\) + \(\dfrac{1}{27}\) + \(\dfrac{1}{81}\)) = \(\dfrac{56}{81}\)

4\(y\) + ( \(\dfrac{27}{81}\) + \(\dfrac{9}{81}\) + \(\dfrac{3}{27}\) + \(\dfrac{1}{81}\) ) = \(\dfrac{56}{81}\)

4y + \(\dfrac{40}{81}\) = \(\dfrac{56}{81}\)

4y = \(\dfrac{56}{81}\) - \(\dfrac{40}{81}\)

4y = \(\dfrac{16}{81}\)

y  = \(\dfrac{16}{81}\) : 4

y = \(\dfrac{4}{81}\)

\(\left(y+\dfrac{1}{3}\right)+\left(y+\dfrac{1}{9}\right)+\left(y+\dfrac{1}{27}\right)+\left(y+\dfrac{1}{81}\right)=\dfrac{56}{81}\)

\(\Rightarrow y+\dfrac{1}{3}+y+\dfrac{1}{9}+y+\dfrac{1}{27}+y+\dfrac{1}{81}=\dfrac{56}{81}\)

\(\Rightarrow4\times y+\dfrac{40}{81}=\dfrac{56}{81}\)

\(\Rightarrow4\times y=\dfrac{56}{81}-\dfrac{40}{81}\)

\(\Rightarrow4\times y=\dfrac{16}{81}\)

\(\Rightarrow y=\dfrac{16}{81}:4\)

\(\Rightarrow y=\dfrac{4}{81}\)

\((3.x-2^{4} ):7^3 =2.7^{4}\)

\(=>(3.x-16):7^{3} =2.7^{4}\)

\(=>3.x-16=2.7^{4}:7^{3}\)

\(=>3.x-16=2.7^{4-3}\)

\(=>3.x-16=2.7\)

\(=>3.x-16=14\)

\(=>3.x=14+16\)

\(=>3.x=30\)

\(=>x=30:3\)

\(=>x=10\)

\(\left(3x-2^4\right):7^3=\left(2.7\right)^4\)

\(\Rightarrow\left(3x-2^4\right)=2^4.7^4.7^3\)

\(\Rightarrow3x-2^4=2^4.7^7\)

\(\Rightarrow3x=2^4.7^7+2^4\)

\(\Rightarrow3x=2^4.\left(7^7+1\right)=16\left(7^7+1\right)\)

\(\Rightarrow x=\dfrac{16\left(7^7+1\right)}{3}\)

Ta có:

\(5^{75}=\left(5^5\right)^{15}=3125^{15}\)

\(7^{60}=\left(7^4\right)^{15}=2401^{15}\)

Mà: \(3125^{15}>2401^{15}\)

\(\Rightarrow5^{75}>7^{60}\)

_______________

Ta có:

\(3^{39}< 3^{42}\); \(3^{42}=\left(3^6\right)^7=729^7\)

\(11^{21}=\left(11^3\right)^7=1331^7\)

Mà: \(729^7< 1331^7\)

\(\Rightarrow3^{42}< 11^{21}\)

\(\Rightarrow3^{39}< 11^{21}\)

a) \(5^{75}=\left(5^5\right)^{15}=3125^{15}\)

\(7^{60}=\left(7^4\right)^{15}=2401^{15}\)

mà \(2401^{15}< 3125^{15}\)

\(\Rightarrow5^{75}>7^{60}\)

b) \(3^{39}=\left(3^{13}\right)^3=1594323^3;11^{21}=\left(11^7\right)^3=19487171^3\)

mà \(19487171^3>1594323^3\)

\(\Rightarrow3^{39}< 7^{21}\)

Bạn xem lại AB=8cm hay AB=10cm

a) \(EB=\dfrac{AB}{2}\) (E là trung điểm AB)

\(\Rightarrow EB=\dfrac{8}{2}=4\left(cm\right)\)

b) Vì F là trung điểm EB

\(\Rightarrow EF=FB=2\left(cm\right)\)

a) Ta có: EB = AB-AE
         T/s EB =   8 - 5
         => EB =     3(cm)   (1)
b) Ta có: EF = EB-FB
         T/s EF =  3  - 2
          => EF =    1(cm)   (2)
Từ (1), (2) => Ta thấy: 3cm > 1cm
                              Hay  EB > EF

Bài 1 :

\(S=1.3+3.5+5.7+...+99.101=3+15+35+...9999\)

Ta thấy :

\(3=2^2-1\)

\(15=4^2-1\)

\(35=6^2-1\)

.....

\(9999=100^2-1\)

\(\Rightarrow S=2^2+4^2+...+100^2-\left(1\right).\left(\left(100-2\right):2+1\right)\)

\(\Rightarrow S=\dfrac{100.\left(100+1\right)\left(2.100+1\right)}{6}-51\)

\(\Rightarrow S=\dfrac{100.101.201}{6}-51=338299\)

nhanh len nhé mik đang cần gấp ai lam trước mik tích cho

36.33-105.11+22.15

=1188-1155+330

=33+330

=363

\(36\cdot33-105\cdot11+22\cdot15\)

\(=11\cdot\left(36\cdot3-105+2\cdot15\right)\)

\(=11\cdot\left(108-105+30\right)\)

\(=11\cdot33\)

\(=33\cdot\left(10+1\right)\)

\(=330+33\)

\(=363\)

Số cần tìm là 7 thỏa mãn dãy tăng dần từ 11 - 17.

a) \(\left(-17\right)\cdot32+17\cdot\left(-68\right)-17\)

\(=17\cdot\left(-32-68-1\right)\)

\(=17\cdot\left(-101\right)\)

\(=17\cdot\left(-100-1\right)\)

\(=-1700-17\)

\(=-1717\)

b) \(25\cdot\left(34-89\right)+25\cdot89\)

\(=25\cdot\left(34-89+89\right)\)

\(=25\cdot34\)

\(=850\)

c) \(-145\cdot\left(13-57\right)+57\cdot\left(10-145\right)\)

\(=-145\cdot13+145\cdot57+57\cdot10-145\cdot57\)

\(=-145\cdot\left(13+57\right)+57\cdot\left(145+10\right)\)

\(=-145\cdot70+57\cdot155\)

\(=-10150+8835\)

\(=-1315\)

`@` `\text {Ans}`

`\downarrow`

`a)`

\((- 17) . 32 + 17 . (- 68) – 17 \)

`= (-17). (32 + 68 + 1) `

`= (-17). 101`

`= -1717`

`b)`

\(25 . (34 – 89) + 25 . 89\)

`= 25. (34 - 89 + 89)`

`= 25. 34`

`= 850`

`c)`

\(– 145. (13 – 57) + 57. (10 – 145)\)

`= (-145). 13 - (-145). 57 + 57.10 + 57. (-145)`

`= (-145). (13 - 57 + 57) + 57.10`

`= (-145). 13 + 570`

`= -1885 + 570 = -1315`

\(\Rightarrow x=269-26=243\)

\(x\in\) N ; \(x^5\) + 26 = 269

             \(x^5\)         = 269 - 26

             \(x^5\)         =  243

             \(x^5\)         =   3⁵

             \(x\)           =     3

Cho: \(A=\dfrac{2}{2^2}+\dfrac{2}{3^2}+\dfrac{2}{4^2}+....+\dfrac{2}{100^2}\)

\(A=2\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\right)\)

Và cho \(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)

Mà: 

\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}\)

....

\(\dfrac{1}{100^2}< \dfrac{1}{99\cdot100}\)

Nên: \(B< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{99\cdot100}\)

\(\Rightarrow B< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(\Rightarrow B< 1-\dfrac{1}{100}\)

\(\Rightarrow B< \dfrac{99}{100}\)

Mà: \(\dfrac{99}{100}< 1\) (tử nhỏ hơn mẫu)

\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< 1\)

\(\Rightarrow A=2\cdot\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+..+\dfrac{1}{100^2}\right)< 2\) (đpcm)

\(\dfrac{2}{2^2}+\dfrac{2}{3^2}+\dfrac{2}{4^2}+...+\dfrac{2}{100^2}\)

\(=2\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\right)\)

mà \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< 1\)

\(\Rightarrow dpcm\)