(y+1/3)+(y+1/9)+(y+1/27)+(y+1/81)=56/81
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\((3.x-2^{4} ):7^3 =2.7^{4}\)
\(=>(3.x-16):7^{3} =2.7^{4}\)
\(=>3.x-16=2.7^{4}:7^{3}\)
\(=>3.x-16=2.7^{4-3}\)
\(=>3.x-16=2.7\)
\(=>3.x-16=14\)
\(=>3.x=14+16\)
\(=>3.x=30\)
\(=>x=30:3\)
\(=>x=10\)
\(\left(3x-2^4\right):7^3=\left(2.7\right)^4\)
\(\Rightarrow\left(3x-2^4\right)=2^4.7^4.7^3\)
\(\Rightarrow3x-2^4=2^4.7^7\)
\(\Rightarrow3x=2^4.7^7+2^4\)
\(\Rightarrow3x=2^4.\left(7^7+1\right)=16\left(7^7+1\right)\)
\(\Rightarrow x=\dfrac{16\left(7^7+1\right)}{3}\)
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 :
\(3=2^2-1\)
\(15=4^2-1\)
\(35=6^2-1\)
.....
\(9999=100^2-1\)
\(\Rightarrow S=2^2+4^2+...+100^2-\left(1\right).\left(\left(100-2\right):2+1\right)\)
\(\Rightarrow S=\dfrac{100.\left(100+1\right)\left(2.100+1\right)}{6}-51\)
\(\Rightarrow S=\dfrac{100.101.201}{6}-51=338299\)
36.33-105.11+22.15
=1188-1155+330
=33+330
=363
\(36\cdot33-105\cdot11+22\cdot15\)
\(=11\cdot\left(36\cdot3-105+2\cdot15\right)\)
\(=11\cdot\left(108-105+30\right)\)
\(=11\cdot33\)
\(=33\cdot\left(10+1\right)\)
\(=330+33\)
\(=363\)
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`
\(x\in\) 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 \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< 1\)
\(\Rightarrow dpcm\)
(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 = \(\dfrac{4}{81}\)
\(\left(y+\dfrac{1}{3}\right)+\left(y+\dfrac{1}{9}\right)+\left(y+\dfrac{1}{27}\right)+\left(y+\dfrac{1}{81}\right)=\dfrac{56}{81}\)
\(\Rightarrow y+\dfrac{1}{3}+y+\dfrac{1}{9}+y+\dfrac{1}{27}+y+\dfrac{1}{81}=\dfrac{56}{81}\)
\(\Rightarrow4\times y+\dfrac{40}{81}=\dfrac{56}{81}\)
\(\Rightarrow4\times y=\dfrac{56}{81}-\dfrac{40}{81}\)
\(\Rightarrow4\times y=\dfrac{16}{81}\)
\(\Rightarrow y=\dfrac{16}{81}:4\)
\(\Rightarrow y=\dfrac{4}{81}\)