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NL
Nguyễn Lê Phước Thịnh
CTVHS
22 tháng 9 2025
Bài 1:
a: 2x+3=15
=>2x=15-3=12
=>\(x=\frac{12}{2}=6\)
b: 28-3x=13
=>\(3x=28-13=15\)
=>\(x=\frac{15}{3}=5\)
c: \(\left(x-1954\right)\cdot5=50\)
=>x-1954=50:5=10
=>x=10+1954=1964
d: 30(60-x)=30
=>60-x=1
=>x=60-1=59
Bài 2:
a: x+99:3=55
=>x+33=55
=>x=55-33=22
b: (x-25):15=20
=>\(x-25=20\cdot15=300\)
=>x=300+25=325
c: \(7\left(3x-15\right)=42\)
=>\(3x-15=\frac{42}{7}=6\)
=>3x=15+6=21
=>\(x=\frac{21}{3}=7\)
d: \(x\left(x+1\right)=2+4+6+\cdots+2500\)
=>\(x\left(x+1\right)=2\left(1+2+3+\cdots+1250\right)\)
=>\(x\left(x+1\right)=2\cdot1250\cdot\frac{1251}{2}=1250\cdot1251\)
=>\(x^2+x-1250\cdot1251=0\)
=>(x+1251)(x-1250)=0
=>\(\left[\begin{array}{l}x+1251=0\\ x-1250=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-1251\\ x=1250\end{array}\right.\)
Bài 1:
a: \(46\cdot99=46\cdot\left(100-1\right)=46\cdot100-46\)
=4600-46
=4554
b: \(47\cdot98=47\cdot\left(100-2\right)=47\cdot100-47\cdot2\)
=4700-94
=4604
c: \(18\cdot19=18\left(20-1\right)=18\cdot20-18\)
=360-18
=342
d: \(24\cdot198=24\left(200-2\right)\)
\(=24\cdot200-24\cdot2\)
=4800-48
=4752
Bài 2:
a: \(1800\cdot5=18\cdot100\cdot5=90\cdot100=9000\)
b: \(36\cdot25=25\cdot4\cdot9=100\cdot9=900\)
c: 36600:50=3660:5=732
d: 220000:5000=220:5=44
NL
Nguyễn Lê Phước Thịnh
CTVHS
20 tháng 8 2025
c: \(\left(x-1\right)^3=\left(-9\right)^3\)
=>x-1=-9
=>x=-9+1=-8
f: \(3x-2^3=7+\left(-9\right)\)
=>3x-8=7-9=-2
=>3x=-2+8=6
=>x=2
NL
Nguyễn Lê Phước Thịnh
CTVHS
21 tháng 9 2025
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
NL
Nguyễn Lê Phước Thịnh
CTVHS
21 tháng 9 2025
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
NL
Nguyễn Lê Phước Thịnh
CTVHS
21 tháng 9 2025
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
NL
Nguyễn Lê Phước Thịnh
CTVHS
22 tháng 9 2025
d: \(48\cdot26+24\cdot148\)
\(=48\cdot26+48\cdot74\)
\(=48\cdot\left(26+74\right)=48\cdot100=4800\)
e: \(23\cdot48+92\cdot88\)
\(=23\cdot4\cdot12+92\cdot88\)
\(=92\cdot12+92\cdot88=92\cdot100=9200\)
b: \(89\cdot25+89\cdot74+89\)
\(=89\cdot\left(25+74+1\right)\)
\(=89\cdot100=8900\)
NL
Nguyễn Lê Phước Thịnh
CTVHS
24 tháng 8 2025
Sửa đề: \(3^{n+2}-2^{n+2}+3^{n}-2^{n}\)
Ta có: \(3^{n+2}+3^{n}-2^{n+2}-2^{n}\)
\(=3^{n}\cdot3^2+3^{n}-2^{n}\cdot4-2^{n}\)
\(=3^{n}\left(3^2+1\right)-2^{n}\cdot\left(4+1\right)\)
\(=3^{n}\cdot10-2^{n}\cdot5=3^{n}\cdot10-2^{n-1}\cdot10=10\left(3^{n}-2^{n-1}\right)\) ⋮10
TP
Thảo Phương
VIP
24 tháng 8 2025
Sửa đề : 3^n+2 - 2^n+2 + 3^n - 2^n
Ta có : 3^n+2 + 3^n - 2^n+2 - 2^n
= 3^n . 3^2 + 3^n - 2^n . 4 - 2^n
= 3^n . ( 3^2 + 1 ) - 2^n . ( 4 + 1 )
= 3^n . 10 - 2^n . 5 = 3^n . 10 - 2^n-1 . 10 = 10 . ( 3^n - 2^n-1 ) chia hết cho 10
a: \(A=1\dfrac{1}{2}\cdot1\dfrac{1}{3}\cdot...\cdot1\dfrac{1}{15}\)
\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{16}{15}\)
\(=\dfrac{16}{2}=8\)
b: \(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{1000}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{999}{1000}\)
\(=\dfrac{1}{1000}\)
c: \(C=\dfrac{1}{3}:\left(-\dfrac{4}{3}\right):\dfrac{5}{4}:\left(-\dfrac{6}{5}\right):\dfrac{7}{6}:\left(-\dfrac{8}{7}\right):...:\dfrac{99}{98}:\left(-\dfrac{100}{99}\right)\)
\(=-\dfrac{1}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\cdot...\cdot\dfrac{98}{99}\cdot\dfrac{99}{100}\)
\(=-\dfrac{1}{100}\)
d: \(\dfrac{\dfrac{2}{5}+\dfrac{2}{27}-\dfrac{2}{9}-\dfrac{2}{11}}{\dfrac{3}{5}+\dfrac{3}{27}-\dfrac{3}{9}-\dfrac{3}{11}}=\dfrac{2\left(\dfrac{1}{5}+\dfrac{1}{27}-\dfrac{1}{9}-\dfrac{1}{11}\right)}{3\left(\dfrac{1}{5}+\dfrac{1}{27}-\dfrac{1}{9}-\dfrac{1}{11}\right)}=\dfrac{2}{3}\)
e: \(D=\dfrac{1}{2}\cdot\dfrac{1}{3}+\dfrac{1}{3}\cdot\dfrac{1}{4}+...+\dfrac{1}{2022}\cdot\dfrac{1}{2023}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\)
\(=\dfrac{1}{2}-\dfrac{1}{2023}=\dfrac{2021}{4046}\)