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Đặt \(A=5+5^3+5^5+....+5^{47}+5^{49}\)
\(\Rightarrow5^2A=5^3+5^5+5^7+.....+5^{49}+5^{51}\)
\(\Rightarrow5^2A-A=\left(5^3+5^5+5^7+....+5^{49}+5^{51}\right)-\left(3+3^3+3^5+....+5^{47}+5^{49}\right)\)
\(\Rightarrow24A=5^{51}-5\)
\(\Rightarrow A=\dfrac{5^{51}-5}{24}\)
Vậy ............................................................
1)a) \(\left(3x-7\right)^5=32\Rightarrow\left(3x-7\right)^5=2^5\)
\(\Rightarrow3x-7=2\Rightarrow3x=9\Rightarrow x=3\)
Vậy \(x=3\)
b) \(\left(4x-1\right)^3=-27.125\)
\(\Rightarrow\left(4x-1\right)^3=-3^3.5^3=-15^3\)
\(\Rightarrow4x-1=-15\Rightarrow4x=-14\Rightarrow x=-3,5\)
Vậy \(x=-3,5\)
c) \(3^{4x+4}=81^{x+3}\Rightarrow3^{4x+4}=3^{4x+12}\)
\(\Rightarrow4x+4=4x+12\)
\(\Rightarrow4x=4x+8\)
\(\Rightarrow x\in\varnothing\)
d) \(\left(x-5\right)^7=\left(x-5\right)^9\)
\(\Rightarrow\left(x-5\right)^7-\left(x-5\right)^9=0\)
\(\Rightarrow\left(x-5\right)^7.\left[1-\left(x-5\right)^2\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^7=0\\1-\left(x-5\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\\left(x-5\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x-5=-1\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)
nên các bn giúp mk nha,còn nhìu ý nữa cơ 
mai hoc rùi
Bài 1: Tính:
a) 27 : 22 + 54 : 53. 24 - 3. 25
= 25 + 5 . 24 - 3 . 25
= 32 + 5 . 16 - 3 . 32
= 32 + 80 - 96
= 112 - 96
= 16
b) ( 37 . 35) : 310+ 5 . 24 - 73 : 7
= 312 : 310 + 5 . 24 - 72
= 32 + 5 . 24 - 72
= 9 + 5 . 16 - 49
= 9 + 80 - 49
= 89 - 49
= 40
Bài 2: Tính hợp lí:
a) ( 62007 - 62006 ) : 62006
= 62007 : 62006 - 62006 : 62006
= 6 - 1
= 5
b) ( 112003 + 112002 ) : 112002
= 11 + 1
= 12
c) 320 : ( x3 - 24 ) + 24 = 32
320 : ( x3 - 24 ) = 32 - 24 = 8
x3 - 24 = 320 : 8
x3 - 24 = 40 + 24
x3 = 64
x3 = 43 = 4
d) 130 - ( 100 + x ) = 25
( 100 + x ) = 103 - 25
100 + x = 105 - 100
x = 5
Bn ơi đừng tự ti như vậy nha !!! Mỗi người đều có một khuyết điểm mà, tri thức luôn rộng lớn bao la. Hãy làm việc đó bằng cách bn tự làm những bài kia nha.
Chúc bn hc tốt môn toán :))
2)
a) \(\left(6^{2007}-6^{2006}\right):6^{2006}\)
\(=\left(6^{2006}.6-6^{2006}.1\right):6^{2006}\)
\(=\left[6^{2006}.\left(6-1\right)\right]:6^{2006}\)
\(=6^{2006}:6^{2006}.5\)
\(=5\)
b) \(\left(11^{2003}+11^{2002}\right):11^{2002}\)
\(=\left(11^{2002}.11+11^{2002}.1\right):11^{2002}\)
\(=\left[11^{2002}.\left(11+1\right)\right]:11^{2002}\)
\(=11^{2002}:11^{2002}.12\)
\(=12\)
c) \(130:\left(x^3-24\right)+24=32\)
\(\Leftrightarrow130:\left(x^3-24\right)=32-24\)
\(\Leftrightarrow130:\left(x^3-24\right)=8\)
\(\Leftrightarrow x^3-24=\dfrac{65}{4}\)
\(\Leftrightarrow x^3=\dfrac{65}{4}+24\)
\(\Leftrightarrow x^3=\dfrac{161}{4}\)
\(\Leftrightarrow x=\sqrt[3]{\dfrac{161}{4}}\)
Vậy \(x=\sqrt[3]{\dfrac{161}{4}}\)
d) \(130-\left(100+x\right)=25\)
\(\Leftrightarrow100+x=130-25\)
\(\Leftrightarrow100+x=105\)
\(\Leftrightarrow x=105-100\)
\(\Leftrightarrow x=5\)
Vậy \(x=5\)
\(a.3^{500}=\left(3^5\right)^{100}=125^{100}\)
\(7^{300}=\left(7^3\right)^{100}=343^{100}\)
\(V\text{ì}\)\(125^{100}< 343^{100}=>3^{500}< 7^{300}\)
\(99^{20}=\left(9^2\right)^{10}=81^{10}\)
Vì 8110 < 999910 => 9920 < 999910
!!!
Ta có :
\(B=3+3^3+3^5+..............+3^{1991}\)
\(\Leftrightarrow B=\left(3+3^3+3^5\right)+\left(3^7+3^9+3^{11}\right)+...............+\left(3^{1987}+3^{1989}+3^{1991}\right)\)
\(\Leftrightarrow B=1\left(3+3^3+3^5\right)+..............+3^{1987}\left(3+3^3+3^5\right)\)
\(\Leftrightarrow B=273+.............+3^{1987}.273\)
\(\Leftrightarrow B=273\left(1+..........+3^{1987}\right)\)
Mà \(273⋮13\)
\(\Leftrightarrow B⋮13\Leftrightarrowđpcm\)
Lại có :
\(B=3+3^3+3^5+..............+3^{1991}\)
\(\Leftrightarrow B=\left(3+3^3+3^5+3^7\right)+..........\left(3^{1985}+3^{1987}+3^{1989}+3^{1991}\right)\)
\(\Leftrightarrow B=1\left(3+3^3+3^5+3^7\right)+..........+3^{1985}\left(3+3^3+3^5+3^7\right)\)
\(\Leftrightarrow B=2460+..............+3^{1985}.2460\)
\(\Leftrightarrow B=2460\left(1+............+3^{1985}\right)\)
Mà \(2460⋮41\)
\(\Leftrightarrow B⋮41\rightarrowđpcm\)
Bài 1:
a, 2x-15=17
=>2x=21
=>x=1
b)(7x-11)3=25*52+200
=>(7x-11)3=32*25+200
=>(7x-11)3=800+200
=>(7x-11)3=103
=>7x-11=10
=>7x=21
=>x=3
Bài 2:
a,523 và 6.522
6.522=(5+1).522=523+5>523
=>523<6.522
b,c tương tự
1. So sánh:
a) 230 và 320
Ta có :
230 = 23.10 = (23)10 = 810
320 = 32.10 = (32)10 = 910
Vì : 810 < 910
=> 230 < 320
b) 1020 và 2010
Ta có :
1020 = 102.10 = (102)10 = 10010
Vì 10010 > 2010
=> 1020 > 2010
1) So sánh :
a)\(^{2^{30}}\) và \(^{3^{20}}\)
\(^{2^{30}}\)= \(^{2^3}\).\(^{2^3}\).\(^{2^3}\).......\(^{2^3}\)
10 thừa số
=8.8.8.......8
10 thừa số
=\(^{8^{10}}\)
\(^{3^{20}}\)=\(^{3^2}\).\(^{3^2}\).\(^{3^2}\)......\(^{3^2}\)
10 thừa số
=9.9.9.....9
10 thừa số
=\(^{9^{10}}\)
Vì \(^{8^{10}}\)<\(^{9^{10}}\)\(\Rightarrow\) \(^{2^{30}}\)<\(^{3^{20}}\)
b) \(^{10^{20}}\) và\(^{20^{10}}\)
\(^{10^{20}}\)=\(^{10^2}\).\(^{10^2}\).\(^{10^2}\).......\(^{10^2}\)
10 thừa số
=100.100.100....100
10 thừa số
=\(^{100^{10}}\)
Vì \(^{100^{10}}\)>\(^{20^{10}}\)\(\Rightarrow\)\(^{10^{20}}\)>\(^{20^{10}}\)
a) \(100:\left\{250:\left[450-\left(4.5^3-2^2.25\right)\right]\right\}\)
\(=100:\left\{250:\left[450-\left(4.125-4.25\right)\right]\right\}\)
\(=100:\left\{250:\left[450-\left(500-100\right)\right]\right\}\)
\(=100:\left[250:\left(450-400\right)\right]\)
\(=100:\left(250:50\right)\)
\(=100:5\)
\(=20\)
b) \(109.5^2-3^2.25\)
\(=109.25-9.25\)
\(=25\left(109-9\right)\)
\(=25.100\)
\(=2500\)
c) \(\left[5^2.6-20.\left(37-2^5\right)\right]:10-20\)
\(=\left[5^2.6-20.\left(37-32\right)\right]:10-20\)
\(=\left(5^2.6-20.5\right):10-20\)
\(=\left(25.6-20.5\right):10-20\)
\(=\left(150-100\right):10-20\)
\(=50:10-20\)
\(=5-20\)
\(=-15\)
mk chỉ bt lm b2 thoy
a,-12(x-5)+7(3-x)=5
\(\Leftrightarrow\)-12x+60+21-7x-5=0
\(\Leftrightarrow\)-19x+76=0
\(\Leftrightarrow\)-19x=-76
\(\Leftrightarrow\)x=4
Vậy...
b,30(x+2)-6(x-5)-24x=100
\(\Leftrightarrow\)30x+60-6x+30-24x-100=0
\(\Leftrightarrow\)ko có gtrị nào của x
2e)Đặt \(A=1+3+3^2+...+3^{200}\)
\(\Rightarrow3A=3+3^2+3^3+...+3^{201}\)
\(\Rightarrow2A=3^{201}-1\)
\(\Rightarrow A=\frac{3^{201}-1}{2}\)
\(\Rightarrow A< 3^{201}\)
Hay \(1+3+3^2+...+3^{200}< 3^{201}\)