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Bài 5:
a: \(B=2009\cdot2011\)
\(=\left(2010-1\right)\left(2010+1\right)\)
\(=2010\cdot2010-1=A-1\)
=>B<A
b: \(B=2019\cdot2021\)
\(=\left(2020-1\right)\left(2020+1\right)\)
\(=2020\cdot2020-1\)
=A-1
=>B<A
c: \(A=234234\cdot233=234\cdot233\cdot1001\)
\(B=233233\cdot234=233\cdot234\cdot1001\)
Do đó: A=B
d: \(A=123\cdot137137=123\cdot137\cdot1001\)
\(B=137\cdot123123=137\cdot123\cdot1001\)
Do đó: A=B
Bài 4:
a: \(391-125<\overline{26x}<184+84\)
=>\(266<\overline{26x}<268\)
=>\(\overline{26x}=267\)
=>x=7
b: \(935+167<\overline{110x}<1240-135\)
=>\(1102<\overline{110x}<1105\)
=>x∈{3;4}
c: \(135\cdot12<\overline{162x}<4869:3\)
=>\(1620<\overline{162x}<1623\)
=>x∈{1;2}
d: \(11268:3<\overline{375x}<235\cdot16\)
=>\(3756<\overline{375x}<3760\)
=>x∈{7;8;9}
Bài 3:
l: x+125=492
=>x=492-125=367
m: 327-x=129
=>x=327-129=198
n: 124+(118-x)=217
=>118-x=217-124=93
=>x=118-93=25
o: 89-(73-x)=20
=>73-x=89-20=69
=>x=73-69=4
p: 198-(x+4)=120
=>x+4=198-120=78
=>x=78-4=74
q: (x+7)-25=23
=>x+7=25+23=48
=>x=48-7=41
r: 140:(x-8)=7
=>x-8=140:7=20
=>x=20+8=28
s: 4(x+41)=400
=>x+41=400:4=100
=>x=100-41=59
t: 4(3x-4)-2=18
=>4(3x-4)=2+18=20
=>3x-4=5
=>3x=9
=>x=3
u: 123-5(x+4)=38
=>5(x+4)=123-38=85
=>x+4=85:5=17
=>x=17-4=13
v: 231-(x-6)=1339:13
=>231-(x-6)=103
=>x-6=231-103=128
=>x=128+6=134
w: (x-36):18+12=14
=>(x-36):18=2
=>\(x-36=2\cdot18=36\)
=>x=36+36=72

Bài 5:
a: \(B=2009\cdot2011\)
\(=\left(2010-1\right)\left(2010+1\right)\)
\(=2010\cdot2010-1=A-1\)
=>B<A
b: \(B=2019\cdot2021\)
\(=\left(2020-1\right)\left(2020+1\right)\)
\(=2020\cdot2020-1\)
=A-1
=>B<A
c: \(A=234234\cdot233=234\cdot233\cdot1001\)
\(B=233233\cdot234=233\cdot234\cdot1001\)
Do đó: A=B
d: \(A=123\cdot137137=123\cdot137\cdot1001\)
\(B=137\cdot123123=137\cdot123\cdot1001\)
Do đó: A=B
Bài 4:
a: \(391-125<\overline{26x}<184+84\)
=>\(266<\overline{26x}<268\)
=>\(\overline{26x}=267\)
=>x=7
b: \(935+167<\overline{110x}<1240-135\)
=>\(1102<\overline{110x}<1105\)
=>x∈{3;4}
c: \(135\cdot12<\overline{162x}<4869:3\)
=>\(1620<\overline{162x}<1623\)
=>x∈{1;2}
d: \(11268:3<\overline{375x}<235\cdot16\)
=>\(3756<\overline{375x}<3760\)
=>x∈{7;8;9}
Bài 3:
l: x+125=492
=>x=492-125=367
m: 327-x=129
=>x=327-129=198
n: 124+(118-x)=217
=>118-x=217-124=93
=>x=118-93=25
o: 89-(73-x)=20
=>73-x=89-20=69
=>x=73-69=4
p: 198-(x+4)=120
=>x+4=198-120=78
=>x=78-4=74
q: (x+7)-25=23
=>x+7=25+23=48
=>x=48-7=41
r: 140:(x-8)=7
=>x-8=140:7=20
=>x=20+8=28
s: 4(x+41)=400
=>x+41=400:4=100
=>x=100-41=59
t: 4(3x-4)-2=18
=>4(3x-4)=2+18=20
=>3x-4=5
=>3x=9
=>x=3
u: 123-5(x+4)=38
=>5(x+4)=123-38=85
=>x+4=85:5=17
=>x=17-4=13
v: 231-(x-6)=1339:13
=>231-(x-6)=103
=>x-6=231-103=128
=>x=128+6=134
w: (x-36):18+12=14
=>(x-36):18=2
=>\(x-36=2\cdot18=36\)
=>x=36+36=72

Bài 5:
a: \(B=2009\cdot2011\)
\(=\left(2010-1\right)\left(2010+1\right)\)
\(=2010\cdot2010-1=A-1\)
=>B<A
b: \(B=2019\cdot2021\)
\(=\left(2020-1\right)\left(2020+1\right)\)
\(=2020\cdot2020-1\)
=A-1
=>B<A
c: \(A=234234\cdot233=234\cdot233\cdot1001\)
\(B=233233\cdot234=233\cdot234\cdot1001\)
Do đó: A=B
d: \(A=123\cdot137137=123\cdot137\cdot1001\)
\(B=137\cdot123123=137\cdot123\cdot1001\)
Do đó: A=B
Bài 4:
a: \(391-125<\overline{26x}<184+84\)
=>\(266<\overline{26x}<268\)
=>\(\overline{26x}=267\)
=>x=7
b: \(935+167<\overline{110x}<1240-135\)
=>\(1102<\overline{110x}<1105\)
=>x∈{3;4}
c: \(135\cdot12<\overline{162x}<4869:3\)
=>\(1620<\overline{162x}<1623\)
=>x∈{1;2}
d: \(11268:3<\overline{375x}<235\cdot16\)
=>\(3756<\overline{375x}<3760\)
=>x∈{7;8;9}
Bài 3:
l: x+125=492
=>x=492-125=367
m: 327-x=129
=>x=327-129=198
n: 124+(118-x)=217
=>118-x=217-124=93
=>x=118-93=25
o: 89-(73-x)=20
=>73-x=89-20=69
=>x=73-69=4
p: 198-(x+4)=120
=>x+4=198-120=78
=>x=78-4=74
q: (x+7)-25=23
=>x+7=25+23=48
=>x=48-7=41
r: 140:(x-8)=7
=>x-8=140:7=20
=>x=20+8=28
s: 4(x+41)=400
=>x+41=400:4=100
=>x=100-41=59
t: 4(3x-4)-2=18
=>4(3x-4)=2+18=20
=>3x-4=5
=>3x=9
=>x=3
u: 123-5(x+4)=38
=>5(x+4)=123-38=85
=>x+4=85:5=17
=>x=17-4=13
v: 231-(x-6)=1339:13
=>231-(x-6)=103
=>x-6=231-103=128
=>x=128+6=134
w: (x-36):18+12=14
=>(x-36):18=2
=>\(x-36=2\cdot18=36\)
=>x=36+36=72

20.
a.
\(4^{n}=256\)
\(4^{n}=4^4\)
\(n=4\)
b.
\(9^{5n-8}=81\)
\(9^{5n-8}=9^2\)
5n-8=2
5n=10
n=2
c.
\(3^{n+2}:27=3\)
\(3^{n+2}=27.3\)
\(3^{n+2}=81\)
\(3^{n+2}=3^4\)
n+2=4
n=2
d.
\(8^{n+2}.2^3=8^5\)
\(8^{n+2}=8^5:2^3\)
\(8^{n+2}=8^4\)
n+2=4
n=2
21.
a.
\(30-2x^2=12\)
\(2x^2=30-12\)
\(2x^2=18\)
\(x^2=18:2=9\)
\(x^2=3^2\)
\(x=\pm3\)
b.
\(\left(9-2x\right)^3=125\)
\(\left(9-2x\right)^3=5^3\)
\(9-2x=5\)
2x=9-5=4
x=2
c.
\(\left(2x-2\right)^4=0\)
2x-2=0
2x=2
x=1
d.
\(\left(x+5\right)^3=\left(2x\right)^3\)
x+5=2x
2x-x=5
x=5

20.
4^n=256
4^n=4^4
n=4
9^5n-8=81
9^5n-8=9^2
5n-8=2
5n=10
n=2
3^n+2:27=3
3^n+2:3^3=3
3^n+2-3=3
n+2-3=1
n=2
8^n+2.2^3=8^5
8^n+2.8=8^5
8^n+2+1=8^5
n+2+1=5
n=2
21.
30-2x^2=12
2x^2=30-12
2x^2=18
x^2=9
x^2=3^2
x=3
(9-2x)^3=125
(9-2x)^3=5^3
(9-2x)=5
2x=4
x=2
(2x-2)^4=0
(2x-2)=0
2x=2
x=1
(x+5)^3=(2x)^3
x+5=2x
x+5-2x=0
(x-2x)=-5
-x=-5
x=5
20:
a: \(4^{n}=256\)
=>\(4^{n}=4^4\)
=>n=4
b: \(9^{5n-8}=81\)
=>\(9^{5n-8}=9^2\)
=>5n-8=2
=>5n=10
=>n=2
c: \(3^{n+2}:27=3\)
=>\(3^{n+2}=27\cdot3=81=3^4\)
=>n+2=4
=>n=2
d: \(8^{n+2}\cdot2^3=8^5\)
=>\(8^{n+2}=8^5:8=8^4\)
=>n+2=4
=>n=2
Bài 21:
a: \(30-2x^2=12\)
=>\(2x^2=30-12=18\)
=>\(x^2=9\)
mà x>=0(do x là số tự nhiên)
nên x=3
b: \(\left(9-2x\right)^3=125\)
=>9-2x=5
=>2x=4
=>x=2
c: \(\left(2x-2\right)^4=0\)
=>2x-2=0
=>2x=2
=>x=1
d: \(\left(x+5\right)^3=\left(2x\right)^3\)
=>2x=x+5
=>2x-x=5
=>x=5

a) \(M=1+2+2^2+2^3+\cdots+2^{100}\)
\(2M=2+2^2+2^3+2^4+\cdots+2^{101}\)
\(2M-M=\left(2+2^2+2^3+2^4+\cdots+2^{101}\right)-\left(1+2+2^2+2^3+\cdots+2^{100}\right)\)
\(\Rightarrow M=2^{101}-1\)
Vậy \(M=2^{101}-1\)
b) \(N=1+3^2+3^4+3^6+\cdots+3^{100}\)
\(3N=3+3^2+3^4+3^6+3^8+\cdots+3^{102}\)
\(3N-N=\left(3+3^2+3^4+3^6+3^8+\cdots+3^{102}\right)-\left(1+3+3^2+3^4+3^6+\cdots+3^{100}\right)\)
\(\Rightarrow2N=3^{102}-1\)
\(\Rightarrow N=\frac{3^{102}-1}{2}\)
Vậy \(N=\frac{3^{102}-1}{2}\)
c) \(P=1+5^3+5^6+5^9+\cdots+5^{99}\)
\(5^3\cdot P=5^3+5^6+5^9+5^{12}\cdots+5^{102}\)
\(125P-P=\left(5^3+5^6+5^9+5^{12}\cdots+5^{102}\right)-\left(1+5^3+5^6+5^9+\cdots+5^{99}\right)\)
\(\Rightarrow124P=5^{102}-1\)
\(\Rightarrow P=\frac{5^{102}-1}{124}\)
Vậy \(P=\frac{5^{102}-1}{124}\)
a: \(M=1+2+2^2+\cdots+2^{100}\)
=>\(2M=2+2^2+2^3+\cdots+2^{101}\)
=>\(2M-M=2+2^2+2^3+\cdots+2^{101}-1-2-\cdots-2^{100}\)
=>\(M=2^{101}-1\)
b: \(N=1+3^2+3^4+\cdots+3^{100}\)
=>\(9N=3^2+3^4+3^6+\cdots+3^{102}\)
=>\(9N-N=3^2+3^4+\cdots+3^{102}-1-3^2-\cdots-3^{100}\)
=>\(8N=3^{102}-1\)
=>\(N=\frac{3^{102}-1}{8}\)
c: \(P=1+5^3+5^6+\cdots+5^{99}\)
=>\(125P=5^3+5^6+5^9+\cdots+5^{102}\)
=>\(125P-P=5^3+5^6+\cdots+5^{102}-1-5^3-\cdots-5^{99}\)
=>\(124P=5^{102}-1\)
=>\(P=\frac{5^{102}-1}{124}\)
Học lớp 4 thì sao
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