Ta có: \(F=5+5^3+5^5+\cdots+5^{101}\)

=>\(25F=5^3+5^5+5^7+\cdots+5^{103}\)

=>\(25F-F=5^3+5^5+5^7+\cdots+5^{103}-5-5^3-5^5-\cdots-5^{101}\)

=>\(24F=5^{103}-5\)

=>\(F=\frac{5^{103}-5}{24}\)

Ta có: \(5^{103}+1>5^{103}-5\)

=>\(\frac{5^{103}+1}{24}>\frac{5^{103}-5}{24}\)

=>E>F

Bài 8:

a: \(5^3=125;3^5=243\)

mà 125<243

nên \(5^3<3^5\)

b: \(7\cdot2^{13}<8\cdot2^{13}=2^3\cdot2^{13}=2^{16}\)

c: \(27^5=\left(3^3\right)^5=3^{3\cdot5}=3^{15}\)

\(243^3=\left(3^5\right)^3=3^{5\cdot3}=3^{15}\)

Do đó: \(27^5=243^5\)

d: \(625^5=\left(5^4\right)^5=5^{4\cdot5}=5^{20}\)

\(125^7=\left(5^3\right)^7=5^{3\cdot7}=5^{21}\)

mà 20<21

nên \(625^5<125^7\)

Bài 9:

a: \(3^{x}\cdot5=135\)

=>\(3^{x}=\frac{135}{5}=27=3^3\)

=>x=3(nhận)

b: \(\left(x-3\right)^3=\left(x-3\right)^2\)

=>\(\left(x-3\right)^3-\left(x-3\right)^2=0\)

=>\(\left(x-3\right)^2\cdot\left\lbrack\left(x-3\right)-1\right\rbrack=0\)

=>\(\left(x-3\right)^2\cdot\left(x-4\right)=0\)

=>\(\left[\begin{array}{l}x-3=0\\ x-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\left(nhận\right)\\ x=4\left(nhận\right)\end{array}\right.\)

c: \(\left(2x-1\right)^4=81\)

=>\(\left[\begin{array}{l}2x-1=3\\ 2x-1=-3\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=4\\ 2x=-2\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\left(nhận\right)\\ x=-1\left(loại\right)\end{array}\right.\)

d: \(\left(5x+1\right)^2=3^2\cdot5+76\)

=>\(\left(5x+1\right)^2=9\cdot5+76=45+76=121\)

=>\(\left[\begin{array}{l}5x+1=11\\ 5x+1=-11\end{array}\right.\Rightarrow\left[\begin{array}{l}5x=10\\ 5x=-12\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\left(nhận\right)\\ x=-\frac{12}{5}\left(loại\right)\end{array}\right.\)

e: \(5+2^{x-3}=29-\left\lbrack4^2-\left(3^2-1\right)\right\rbrack\)

=>\(2^{x-3}+5=29-\left\lbrack16-9+1\right\rbrack\)

=>\(2^{x-3}+5=29-8=21\)

=>\(2^{x-3}=16=2^4\)

=>x-3=4

=>x=4+3=7(nhận)

f: \(3+2^{x-1}=24-\left\lbrack4^2-\left(2^2-1\right)\right\rbrack\)

=>\(2^{x-1}+3=24-\left\lbrack16-4+1\right\rbrack=24-13=11\)

=>\(2^{x-1}=11-3=8=2^3\)

=>x-1=3

=>x=4(nhận)

Bài 6:

a: \(5\cdot5\cdot5\cdot5\cdot5\cdot5=5^6\)

b: \(27\cdot14\cdot7\cdot2=27\cdot14\cdot14=3^3\cdot14^2\)

c: \(x\cdot x\cdot x\cdot y=x^3\cdot y\)

d: \(5^3\cdot5^4=5^{3+4}=5^7\)

e: \(7^8:7^2=7^{8-2}=7^6\)

f: \(42^7:6^7\cdot49=7^7\cdot49=7^7\cdot7^2=7^{7+2}=7^9\)

Ta có: \(10A=\frac{10^{21}-60}{10^{21}-6}=\frac{10^{21}-6-54}{10^{21}-6}=1-\frac{54}{10^{21}-6}\)

\(10B=\frac{10^{22}-60}{10^{22}-6}=\frac{10^{22}-6-54}{10^{22}-6}=1-\frac{54}{10^{22}-6}\)

Ta có: \(10^{21}-6<10^{22}-6\)

=>\(\frac{54}{10^{21}-6}>\frac{54}{10^{22}-6}\)

=>\(-\frac{54}{10^{21}-6}<-\frac{54}{10^{22}-6}\)

=>\(-\frac{54}{10^{21}-6}+1<-\frac{54}{10^{22}-6}+1\)

=>10A<10B

=>A<B

a) diện tích △ ADG là:

20 x 9 : 2 = 90 (cm2)

diện tích △ ABE là:

14 x 8 : 2 = 56 (cm2)

diện tích hình chữ nhật ABCD là:

20 x 14 = 280 (cm2)

diện tích tứ giác AECG là:

280 - 56 - 90 = 134 (cm2)

b) tỉ số diện tích △ ABE và diện tích △ ADG là:

\(\frac{56}{90}=\frac{28}{45}\)

1: 2⋮x

mà x là số tự nhiên

nên x∈{1;2}

2: 2⋮x+1

=>x+1∈{1;-1;2;-2}

=>x∈{0;-2;1;-3}

mà x>=0

nên x∈{0;1}

3: 2⋮x+2

mà x+2>=2(Do x là số tự nhiên)

nên x+2=2

=>x=0

4: 2⋮x-1

=>x-1∈{1;-1;2;-2}

=>x∈{2;0;3;-1}

mà x>=0

nên x∈{0;2;3}

5: 2⋮x-2

=>x-2∈{1;-1;2;-2}

=>x∈{3;1;4;0}

6: 2⋮2-x

=>2⋮x-2

=>x-2∈{1;-1;2;-2}

=>x∈{3;1;4;0}

Bài 1:

2 ⋮ \(x\)(\(x\) ∈ N*)

2 ⋮ \(x\)

⇒ \(x\) ∈ Ư(2) = {-2; -1; 1; 2}

Vì \(x\) ∈ N* nên \(x\) ∈ {1; 2}

Vậy \(x\) ∈ {1; 2}

bài 3:

a: \(C=5+5^2+5^3+\cdots+5^{20}\)

\(=5\left(1+5+5^2+\cdots+5^{19}\right)\) ⋮5

b: \(C=5+5^2+5^3+\cdots+5^{20}\)

\(=\left(5+5^2\right)+\left(5^3+5^4\right)+\cdots+\left(5^{19}+5^{20}\right)\)

\(=5\left(1+5\right)+5^3\left(1+5\right)+\cdots+5^{19}\left(1+5\right)\)

\(=6\left(5+5^3+\cdots+5^{19}\right)\) ⋮6

c: \(C=5+5^2+5^3+\cdots+5^{20}\)

\(=\left(5+5^2+5^3+5^4\right)+\left(5^5+5^6+5^7+5^8\right)+\cdots+\left(5^{17}+5^{18}+5^{19}+5^{20}\right)\)

\(=5\left(1+5+5^2+5^3\right)+5^5\left(1+5+5^2+5^3\right)+\cdots+5^{17}\left(1+5+5^2+5^3\right)\)

\(=\left(1+5+5^2+5^3\right)\left(5+5^5+\cdots+5^{17}\right)=156\cdot\left(5+5^5+\cdots+5^{17}\right)\)

\(=13\cdot12\cdot\left(5+5^5+\cdots+5^{17}\right)\) ⋮13

Bài 2:

a: \(B=3+3^2+3^3+\cdots+3^{120}\)

\(=3\left(1+3+3^2+3^3+\cdots+3^{119}\right)\) ⋮3

b: \(B=3+3^2+3^3+\cdots+3^{120}\)

\(=\left(3+3^2\right)+\left(3^3+3^4\right)+\cdots+\left(3^{119}+3^{120}\right)\)

\(=3\left(1+3\right)+3^3\left(1+3\right)+\cdots+3^{119}\left(1+3\right)\)

\(=4\left(3+3^3+\cdots+3^{119}\right)\) ⋮4

c: \(B=3+3^2+3^3+\cdots+3^{120}\)

\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+\cdots+\left(3^{118}+3^{119}+3^{120}\right)\)

\(=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+\cdots+3^{118}\left(1+3+3^2\right)\)

\(=13\left(3+3^4+\cdots+3^{118}\right)\) ⋮13

Bài 1:

a: \(A=2+2^2+2^3+\ldots+2^{20}\)

\(=2\left(1+2+2^2+\cdots+2^{19}\right)\) ⋮2

b: \(A=2+2^2+2^3+\ldots+2^{20}\)

\(=\left(2+2^2\right)+\left(2^3+2^4\right)+\cdots+\left(2^{19}+2^{20}\right)\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+\cdots+2^{19}\left(1+2\right)\)

\(=3\left(2+2^3+\cdots+2^{19}\right)\) ⋮3

c: \(A=2+2^2+2^3+\ldots+2^{20}\)

\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+\cdots+\left(2^{17}+2^{18}+2^{19}+2^{20}\right)\)

\(=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+\cdots+2^{17}\left(1+2+2^2+2^3\right)\)

\(=15\left(2+2^5+\ldots+2^{17}\right)=5\cdot3\cdot\left(2+2^5+\cdots+2^{17}\right)\) ⋮5

Bài 1:

a; A = 2 + \(2^2\) + 2\(^3\) + ... + 2\(^{20}\)

A = 2 x (1+ 2+ 2\(^2\) + ... + 2\(^{19}\))

A ⋮ 2(đpcm)

b; A = 2 + \(2^2\) + 2\(^3\) + ... + 2\(^{20}\)

Xét dãy số: 1; 2;...; 20 đây là dãy số cách đều với khoảng cách là:

2 - 1 = 1

Số số hạng của dãy số trên là:

(20 - 1) : 1+ 1 = 20(số)

Vì 20 : 2 = 10

Vậy nhóm hai số hạng liên tiếp của A vào nhau khi đó ta có:

A = (2+ 2\(^2\)) + (2\(^3\) + 2\(^4\)) + ... + (2\(^{19}+\) 2\(^{20}\))

A = 2.(1 + 2) + 2\(^3\).(1+ 2) + ... + 2\(^{19}\) .(1 + 2)

A = 2.3 + 2\(^3\).3 + ... + 2\(^{19}\).3

A = 3.(2+ 2\(^3\) + ... + 2\(^{19}\))

A ⋮ 3 (đpcm)

c; A = 2 + \(2^2\) + 2\(^3\) + ... + 2\(^{20}\)

Xét dãy số: 1; 2; 3;...; 20

Dãy số trên có 20 số hạng:

Vì 20 : 4 = 5

Vậy nhóm 4 hạng tử của A thành một nhóm khi đó:

A = (2+ 2\(^2\) + 2\(^3\) + 2\(^4\)) + ... + (2\(^{17}+2^{18}+2^{19}+2^{20}\))

A = 2.(1 + 2 + 2\(^2\) + 2\(^3\)) + ... + 2\(^{17}\).(1 + 2 + 2\(^2\) + 2\(^3\))

A = (1+ 2 +2\(^2\) + 2\(^3\)).(2+ ...+ 2\(^{17}\))

A = (1 + 2 + 4 + 8).(2+ ...+ 2\(^{17}\))

A = (3+ 4 + 8).(2+ ...+ 2\(^{17}\))

A = (7 + 8)(2+ ...+ 2\(^{17}\))

A = 15.(2+ ...+ 2\(^{17}\))

A ⋮ 5(đpcm)

Giúp mình với mng ơi!!

1: \(1026-\left\lbrack\left(3^4+1\right):41\right\rbrack\)

\(=1026-82:41\)

=1026-2

=1024

\(2^{11}:\left\lbrace1026-\left\lbrack\left(3^4+1\right):41\right\rbrack\right\rbrace\)

\(=2^{11}:2^{10}=2\)

2: \(250:\left\lbrace1500:\left\lbrack4\cdot5^3-2^3\cdot25\right\rbrack\right\rbrace\)

\(=250:\left\lbrace1500:\left\lbrack4\cdot125-8\cdot25\right\rbrack\right\rbrace\)

\(=250:\left\lbrace1500:\left\lbrack500-200\right\rbrack\right\rbrace=250:\frac{1500}{3}=250:500=0,5\)

3: \(12+3\cdot\left\lbrace90:\left\lbrack39-\left(2^3-5\right)^2\right\rbrack\right\rbrace\)

\(=12+3\cdot\left\lbrace90:\left\lbrack39-\left(8-5\right)^2\right\rbrack\right\rbrace\)

\(=12+3\cdot\left\lbrace90:\left\lbrack39-3^2\right\rbrack\right\rbrace\)

\(=12+3\cdot\left\lbrace90:\left\lbrack39-9\right\rbrack\right\rbrace\)

\(=12+3\cdot\left\lbrace90:30\right\rbrace=12+3\cdot3=21\)

4: \(24:\left\lbrace390:\left\lbrack500-\left(5^3+49\cdot5\right)\right\rbrack\right\rbrace\)

\(=24:\left\lbrace390:\left\lbrack500-\left(125+245\right)\right\rbrack\right\rbrace\)

\(=24:\left\lbrace390:\left\lbrack500-125-245\right\rbrack\right\rbrace\)

\(=24:\left\lbrace390:\left\lbrack375-245\right\rbrack\right\rbrace\)

\(=24:\left\lbrace390:130\right\rbrace=\frac{24}{3}=8\)

5: \(117:\left\lbrace\left\lbrack79-3\cdot\left(3^3-17\right)\right\rbrack:7+2\right\rbrace\)

\(=117:\left\lbrace\left\lbrack79-3\cdot\left(27-17\right)\right\rbrack:7+2\right\rbrace\)

\(=117:\left\lbrace\left\lbrack79-3\cdot10\right\rbrack:7+2\right\rbrace\)

\(=117:\left\lbrace49:7+2\right\rbrace=\frac{117}{9}=13\)

6: \(514-4\cdot\left\lbrace\left\lbrack40+8\left(6-3\right)^2\right\rbrack-12\right\rbrace\)

\(=514-4\cdot\left\lbrace\left\lbrack40+8\cdot3^2\right\rbrack-12\right\rbrace\)

\(=514-4\cdot\left\lbrace\left\lbrack40+8\cdot9\right\rbrack-12\right\rbrace\)

\(=514-4\cdot\left\lbrace112-12\right\rbrace\)

\(=514-4\cdot100=514-400=114\)

7: \(25\cdot\left\lbrace32:\left\lbrack\left(12-4\right)+4\cdot\left(16:2^3\right)\right\rbrack\right\rbrace\)

\(=25\cdot\left\lbrace32:\left\lbrack8+4\cdot2\right\rbrack\right\rbrace\)

\(=25\cdot\left\lbrace32:16\right\rbrace=25\cdot2=50\)

8: \(30:\left\lbrace175:\left\lbrack355-\left(135+37\cdot5\right)\right\rbrack\right\rbrace\)

\(=30:\left\lbrace175:\left\lbrack355-\left(135+185\right)\right\rbrack\right\rbrace\)

\(=30:\left\lbrace175:\left\lbrack355-320\right\rbrack\right\rbrace=30:\left\lbrace175:35\right\rbrace=\frac{30}{5}=6\)

9: \(32:\left\lbrace160:\left\lbrack300-\left(175+21\cdot5\right)\right\rbrack\right\rbrace\)

\(=32:\left\lbrace160:\left\lbrack300-\left(175+105\right)\right\rbrack\right\rbrace\)

\(=32:\left\lbrace160:\left\lbrack300-280\right\rbrack\right\rbrace\)

\(=32:\left\lbrace160:20\right\rbrace=\frac{32}{8}=4\)

10: \(750:\left\lbrace130-\left\lbrack\left(5\cdot14-65\right)^3+3\right\rbrack\right\rbrace\)

\(=750:\left\lbrace130-\left\lbrack\left(70-65\right)^3+3\right\rbrack\right\rbrace\)

\(=750:\left\lbrace130-\left\lbrack5^3+3\right\rbrack\right\rbrace\)

\(=750:\left\lbrace130-128\right\rbrace=750:2=375\)

Ta có: \(\frac{A}{10^{10}}=\frac{10^{20}-6}{10^{20}-6\cdot10^{10}}=\frac{10^{20}-6\cdot10^{10}+6\left(10^{10}-1\right)}{10^{20}-6\cdot10^{10}}=1+\frac{6\left(10^{10}-1\right)}{10^{20}-6\cdot10^{10}}\)

\(\frac{B}{10^{10}}=\frac{10^{21}-6}{10^{21}-6\cdot10^{10}}=\frac{10^{21}-6\cdot10^{10}+6\left(10^{10}-1\right)}{10^{21}-6\cdot10^{10}}=1+\frac{6\left(10^{10}-1\right)}{10^{21}-6\cdot10^{10}}\)

Ta có: \(10^{20}<10^{21}\)

=>\(10^{20}-6\cdot10^{10}<10^{21}-6\cdot10^{10}\)

=>\(\frac{6\left(10^{10}-1\right)}{10^{20}-6\cdot10^{10}}>\frac{6\left(10^{10}-1\right)}{10^{21}-6\cdot10^{10}}\)

=>\(\frac{6\left(10^{10}-1\right)}{10^{20}-6\cdot10^{10}}+1>\frac{6\left(10^{10}-1\right)}{10^{21}-6\cdot10^{10}}+1\)

=>\(\frac{A}{10^{10}}>\frac{B}{10^{10}}\)

=>A>B