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

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

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

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

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

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}

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

Câu c:

C = \(9^{2n+1}\) + 1

CM C ⋮ 10

Giải:

9 ≡ -1 (mod 10)

\(9^{2n+1}\) ≡ -1\(^{2n+1}\) (mod 10)

9\(^{2n+1}\) ≡ -1 (mod 10)

1 ≡ 1 (mod 10)

Cộng vế với vế ta có:

9\(^{2n+1}\) + 1 ≡ (-1) + 1 (mod 10)

9\(^{2n+1}\) + 1 ≡ 0 (mod 10)

C = 9\(^{2n+1}\) + 1 ⋮ 10 (đpcm)

\(n^2+n=n\left(n+1\right)\) là tích của hai số tự nhiên liên tiếp

=>\(n^2+n\) chỉ có thể có tận cùng là 0;2;6

=>\(n^2+n+1\) sẽ có tận cùng là 1;3;7

mà \(1995^{2000}\) có chữ số tận cùng là 5

nên \(n^2+n+1\) sẽ không chia hết cho \(1995^{2000}\)

bài 14:

\(a.\left(x-1\right)\cdot100=0\)

\(x-1=0\Rightarrow x=1\)

\(b.200-11x=24\)

\(11x=200-24\)

\(11x=176\)

\(x=\frac{176}{11}=16\)

\(c.165:\left(2x+1\right)=15\) (đkxđ: x khác \(-\frac12)\)

\(2x+1=\frac{165}{15}=11\)

\(2x=11-1=10\)

\(x=\frac{10}{2}=5\)

\(d.375:\left(45-4x\right)=15\) (đkxđ: \(x\ne\frac{45}{4})\)

\(45-4x=\frac{375}{15}=25\)

\(4x=45-25=20\)

\(x=20:4=5\)

bài 15:

giá tiền 125 chiếc điện thoại là:

125 x 2350000=293750000 (đồng)

giá tiền 250 chiếc máy tính bảng là:

250 x 4950000 = 1237500000 (đồng)

tổng số tiền mà cửa hàng phải trả cho số điện thoại và máy tính trên là:

293750000 + 1237500000 = 1531250000 (đồng)

đáp số: 1531250000 đồng

\(1,2^3-5^3:5^2+12\cdot2^2\)

\(=8-5+12\cdot4=3+48=51\)

\(2)5\cdot\left\lbrack\left(85-35:7\right):8+90\right\rbrack-50\)

\(=5\cdot\left\lbrack\left(85-5\right):8+90\right\rbrack-50\)

\(=5\cdot\left(80:8+90\right)-50\)

\(=5\cdot\left(10+90\right)-50=5\cdot100-50\)

\(=500-50=450\)

\(3)2\cdot\left\lbrack\left(7-3^3:3^2\right):2^2+99\right\rbrack-100\)

\(=2\cdot\left\lbrack\left(7-3\right):4+99\right\rbrack-100\)

\(=2\cdot\left(1+99\right)-100=2\cdot100-100\)

\(=200-100=100\)

\(4)2^7:2^2+5^4:5^3\cdot2^4-3\cdot2^5\)

\(=2^5+5^2\cdot16-3\cdot32=32+25\cdot16-96\)

\(=32+400-96=336\)

\(5)5\cdot2^2\cdot2^3-4\cdot\left(5^8:5^6\right)=5\cdot2^5-4\cdot5^2\)

\(=5\cdot32-4\cdot25=160-100=60\)

\(6)\left(3^5\cdot3^7\right):3^{10}+5\cdot2^4-7^3:7\)

\(=3^{12}:3^{10}+5\cdot16-7^2=9+80-49=40\)

\(7)15:\left(3^5:3^4\right)-2^9:2^7\)

\(=15:3-4=5-4=1\)

\(8)5\cdot3^5:\left(3^8:3^5\right)-2^3\cdot5\)

\(=5\cdot3^2-40=5\cdot9-40=45-40=5\)

\(9)4\left\lbrack\left(3+3^7:3^4\right):10+97\right\rbrack-300\)

\(=4\left\lbrack\left(3+3^3\right):10+97\right\rbrack-300\)

\(=4\left\lbrack\left(3+27\right):10+97\right\rbrack-300\)

\(=4\cdot\left(3+97\right)-300=400-300=100\)

\(10)5\left\lbrack\left(92+2^5:2^2\right):5^2+2^4\right\rbrack-7^2\)

\(=5\left\lbrack\left(92+8\right):25+16\right\rbrack-49\)

\(=5\cdot\left(100:25+16\right)-49\)

\(=5\cdot\left(4+16\right)-49=100-49=51\)

\(11)3^2\cdot\left\lbrack5^2-3\right):11]-2^4+2\cdot10^3\)

\(=9\cdot\left(22:11\right)-16+2000=9\cdot2-16+2000\)

\(=18-16+2000=2002\)

\(12)2^2\cdot5\left\lbrack\left(5^2+2^3\right):11-2\right\rbrack-3^2\cdot2\)

\(=4\cdot5\left\lbrack\left(25+8\right):11-2\right\rbrack-18\)

\(=20\cdot\left(3-2\right)-18=20-18=2\)

\(13)\left(6^{2007}-6^{2006}\right):6^{2006}\)

\(=\left\lbrack6^{2006}\cdot\left(6-1\right)\right\rbrack:6^{2006}=5\)

14) \(\left(5^{2001}-5^{2000}\right):5^{2000}\)

\(=\left\lbrack5^{2000}\cdot\left(5-1\right)\right\rbrack:5^{2000}=4\)

\(15)\left(7^{2005}+7^{2004}\right):7^{2004}\)

\(=\left\lbrack7^{2004}\cdot\left(7+1\right)\right\rbrack:7^{2004}=8\)

\(16)\left(11^{2023}+11_{}^{2022}\right):11^{2022}\)

\(=\left\lbrack11^{2022}\cdot\left(11+1\right)\right\rbrack:11^{2022}=12\)