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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: \(42^7:6^7\cdot49=7^7\cdot49=7^7\cdot7^2=7^{7+2}=7^9\)
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
Bài 5:
a: \(37\cdot146+46\cdot2-46\cdot37\)
\(=37\left(146-46\right)+46\cdot2\)
\(=37\cdot100+92=3700+92=3792\)
b: \(2\cdot5\cdot71+5\cdot18\cdot2+10\cdot11\)
\(=10\cdot71+10\cdot18+10\cdot11\)
\(=10\left(71+18+11\right)=10\cdot100=1000\)
c: \(135+360+65+40\)
=135+65+360+40
=200+400
=600
d: \(27\cdot75+25\cdot27-450\)
\(=27\left(75+25\right)-450\)
=2700-450
=2250
Bài 4:
a: \(32\cdot163+32\cdot837\)
\(=32\cdot\left(163+837\right)\)
\(=32\cdot1000=32000\)
b: \(2\cdot3\cdot4\cdot5\cdot25=2\cdot5\cdot4\cdot25\cdot3=3\cdot10\cdot100=3000\)
c: \(25\cdot27\cdot4=27\cdot100=2700\)
Bài 3:
a: \(128\cdot19+128\cdot41+128\cdot40\)
\(=128\cdot\left(19+41+40\right)=128\cdot100=12800\)
b: \(375+693+625+307\)
=375+625+693+307
=1000+1000
=2000
c: \(37+42-37+22\)
=37-37+42+22
=0+64
=64
d: \(21\cdot32+21\cdot68\)
\(=21\cdot\left(32+68\right)=21\cdot100=2100\)
Bài 2:
a: \(17\cdot85+15\cdot17-120\)
\(=17\left(85+15\right)-120\)
=1700-120
=1580
b: \(189+73+211+127\)
=189+211+73+127
=400+200
=600
c: \(38\cdot73+27\cdot38\)
\(=38\left(73+27\right)=38\cdot100=3800\)
Bài 1:
a: \(28\cdot76+23\cdot28-28\cdot13\)
\(=28\left(76+23-13\right)=28\cdot86=2408\)
b: \(39\cdot50+25\cdot39+75\cdot61\)
\(=39\left(50+25\right)+75\cdot61\)
\(=39\cdot75+75\cdot61=75\left(39+61\right)=75\cdot100=7500\)
c: \(32\cdot163+837\cdot32\)
\(=32\left(163+837\right)=32\cdot1000=32000\)
d: \(63+118+37+82\)
=63+37+118+82
=100+200
=300








Bài 6:
a: \(\frac{5}{x}=\frac{-y}{7}\)
=>-xy=35
=>xy=-35
mà y>0
nên (x;y)∈{(-35;1);(-7;5);(-5;7);(-1;35)}
b: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\frac{x}{4}=\frac{y}{3}=\frac{x+y}{4+3}=\frac{14}{7}=2\)
=>x=8; y=6
Bài 3: \(A=\frac{n-1}{n+4}\)
ĐKXĐ: n<>-4
Để A là số nguyên thì n-1⋮n+4
=>n+4-5⋮n+4
=>-5⋮n+4
=>n+4∈{1;-1;5;-5}
=>n∈{-3;-5;1;-9}
Bài 5:
a: \(\frac{18}{x}=\frac{9}{-7}\)
=>\(x=\frac{-7\cdot18}{9}=-7\cdot2=-14\)
b: \(\frac{3}{x}=\frac{y}{7}\)
=>\(x\cdot y=3\cdot7=21\)
=>(x;y)∈{(1;21);(21;1);(-1;-21);(-21;-1);(3;7);(7;3);(-3;-7);(-7;-3)}
Nhiều bài tập, em làm bài nào được rồi nhỉ? Thầy cô giúp bài em chưa làm được thôi em nhé!