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NL
Nguyễn Lê Phước Thịnh
CTVHS
28 tháng 9 2025
Bài 7: Ta có: \(a\cdot b=ƯCLN\left(a;b\right)\cdot BCN\mathbb{N}\left(a;b\right)\)
=>\(a\cdot b=10\cdot900=9000\)
ƯCLN(a;b)=10
=>a⋮10; b⋮10
ab=9000
mà a⋮10 và b⋮10 và a<b
nên (a;b)∈{(10;900);(20;450);(30;300);(50;180);(60;150);(90;100)}
mà ƯCLN(a;b)=10
nên (a;b)∈{(10;900);(20;450);(50;180);(90;100)}
Bài 5:
ƯCLN(a;b)=6
=>a⋮6; b⋮6
ab=216
mà a⋮6; b⋮6
nên (a;b)∈{(6;36);(12;18);(18;12);(36;6)}
Bài 4: Gọi hai số tự nhiên cần tìm là a,b
ƯCLN(a;b)=5
=>a⋮5; b⋮5
Ta có: ab=300
mà a⋮5; b⋮5
nên (a;b)∈{(5;60);(60;5);(10;30);(30;10);(15;20);(20;15)}
Bài 1:
a: ƯCLN(a;b)=6
=>a⋮6 và b⋮6
a+b=96
mà a⋮6 và b⋮6
nên (a;b)∈{(6;90);(90;6);(12;84);(84;12);(18;78);(78;18);(24;72);(72;24);(30;66);(66;30);(36;60);(60;36);(42;54);(54;42);(48;48)}
mà ƯCLN(a;b)=6
nên (a;b)∈{(6;90);(90;6);(18;78);(78;18);(30;66);(66;30);(42;54);(54;42)}
b: ƯCLN(a;b)=4
=>a⋮4 và b⋮4
a+b=16
mà a⋮4; b⋮4 và a>b
nên (a;b)∈{(12;4);(8;8)}
mà ƯCLN(a;b)=4
nên (a;b)=(12;4)
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
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
16 tháng 8 2025
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
c: \(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{200}\left(1+2+...+200\right)\)
\(=1+\dfrac{1}{2}\cdot\dfrac{2\cdot3}{2}+\dfrac{1}{3}\cdot\dfrac{3\cdot4}{2}+...+\dfrac{1}{200}\cdot\dfrac{200\cdot201}{2}\)
\(=1+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{201}{2}\)
\(=\dfrac{2+3+...+201}{2}=\dfrac{\left(201-2+1\right)\cdot\dfrac{\left(201+2\right)}{2}}{2}\)
\(=\dfrac{200\cdot203}{4}=50\cdot203=10150\)
d: \(\dfrac{2^7\cdot3+2^{10}}{13\cdot2^7-14\cdot2^6}=\dfrac{2^7\left(3+2^3\right)}{2^6\left(13\cdot2-14\right)}=2\cdot\dfrac{3+8}{26-14}\)
\(=2\cdot\dfrac{11}{12}=\dfrac{11}{6}\)
Bài 2:
1:
a: \(\left(x-\dfrac{1}{2}\right):\dfrac{2}{3}+0,75=-3\dfrac{3}{4}\)
=>\(\left(x-\dfrac{1}{2}\right):\dfrac{2}{3}=-3-\dfrac{3}{4}-\dfrac{3}{4}=-3-\dfrac{3}{2}=-\dfrac{9}{2}\)
=>\(x-\dfrac{1}{2}=-\dfrac{9}{2}\cdot\dfrac{2}{3}=-\dfrac{9}{3}=-3\)
=>\(x=-3+\dfrac{1}{2}=-\dfrac{5}{2}\)
b: \(\left(2x-15\right)^3=\left(2^2\cdot3^3-2^3\cdot3^2\right):\left(-36\right)\)
=>\(\left(2x-15\right)^3=\left(4\cdot27-8\cdot9\right):\left(-36\right)\)
=>\(\left(2x-15\right)^3=\dfrac{\left(108-72\right)}{-36}=\dfrac{36}{-36}=-1\)
=>2x-15=-1
=>2x=14
=>x=7
c: \(\left(4x+1\right)^4=\left(4x+1\right)^6\)
=>\(\left(4x+1\right)^6-\left(4x+1\right)^4=0\)
=>\(\left(4x+1\right)^4\cdot\left[\left(4x+1\right)^2-1\right]=0\)
=>\(\left(4x+1\right)^4\left(4x+1-1\right)\left(4x+1+1\right)=0\)
=>\(4x\left(4x+2\right)\left(4x+1\right)^4=0\)
=>\(\left[{}\begin{matrix}x=0\\4x+2=0\\4x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\\x=-\dfrac{1}{4}\end{matrix}\right.\)
d: \(\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)\cdot x^2=99\)
=>\(\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\cdot x^2=99\)
=>\(\left(1-\dfrac{1}{100}\right)\cdot x^2=99\)
=>\(x^2=99:\dfrac{99}{100}=100\)
=>\(x=\pm10\)
2: 2xy-x-y=2
=>\(x\left(2y-1\right)-y+\dfrac{1}{2}=2+\dfrac{1}{2}\)
=>\(2x\left(y-\dfrac{1}{2}\right)-\left(y-\dfrac{1}{2}\right)=\dfrac{5}{2}\)
=>\(\left(2x-1\right)\left(y-\dfrac{1}{2}\right)=\dfrac{5}{2}\)
=>(2x-1)(2y-1)=5
=>\(\left(2x-1;2y-1\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)
=>\(\left(x;y\right)\in\left\{\left(1;3\right);\left(3;1\right);\left(0;-2\right);\left(-2;0\right)\right\}\)
Bài 1:
a; 56.712 - 7.50.80 - 7.12.8
= 56.712 - (7.8).(50.10) - (7.8).12
= 56.712 - 56. 500 - 56.12
= 56.(712 - 500 - 12)
= 56.(212 - 12)
= 56.200
= 11200
b; 5\(^0\): \(\frac{3}{\left(-1\right)^{2024}}\) + (6\(\frac16\) - 2\(\frac49\) .2\(\frac{5}{11}\))
= 1:\(\frac31\) + (\(6+\frac16\) - \(\frac{22}{9}\).\(\frac{27}{11}\))
= \(\frac13\) + (\(6+\frac16\) - 6)
=(\(\frac26\) + \(\frac16\)) + ( 6 - 6)
= \(\frac12\)
Bài 1ý 3:
2023 - (2022 - 2021)\(^{2020}\) + (2022+1)\(^0\)
= 2023 - 1\(^{2020}\) + 1
= 2023 - 1 + 1
= 2023 - (1 - 1)
= 2023 - 0
= 2023
: 1 + 1 2 ( 1 + 2 ) + 1 3 ( 1 + 2 + 3 ) + . . . + 1 200 ( 1 + 2 + . . . + 200 ) 1+ 2 1 (1+2)+ 3 1 (1+2+3)+...+ 200 1 (1+2+...+200) = 1 + 1 2 ⋅ 2 ⋅ 3 2 + 1 3 ⋅ 3 ⋅ 4 2 + . . . + 1 200 ⋅ 200 ⋅ 201 2 =1+ 2 1 ⋅ 2 2⋅3 + 3 1 ⋅ 2 3⋅4 +...+ 200 1 ⋅ 2 200⋅201 = 1 + 3 2 + 4 2 + . . . + 201 2 =1+ 2 3 + 2 4 +...+ 2 201 = 2 + 3 + . . . + 201 2 = ( 201 − 2 + 1 ) ⋅ ( 201 + 2 ) 2 2 = 2 2+3+...+201 = 2 (201−2+1)⋅ 2 (201+2) = 200 ⋅ 203 4 = 50 ⋅ 203 = 10150 = 4 200⋅203 =50⋅203=10150 d: 2 7 ⋅ 3 + 2 10 13 ⋅ 2 7 − 14 ⋅ 2 6 = 2 7 ( 3 + 2 3 ) 2 6 ( 13 ⋅ 2 − 14 ) = 2 ⋅ 3 + 8 26 − 14 13⋅2 7 −14⋅2 6 2 7 ⋅3+2 10 = 2 6 (13⋅2−14) 2 7 (3+2 3 ) =2⋅ 26−14 3+8 = 2 ⋅ 11 12 = 11 6 =2⋅ 12 11 = 6 11 Bài 2: 1: a: ( x − 1 2 ) : 2 3 + 0 , 75 = − 3 3 4 (x− 2 1 ): 3 2 +0,75=−3 4 3 => ( x − 1 2 ) : 2 3 = − 3 − 3 4 − 3 4 = − 3 − 3 2 = − 9 2 (x− 2 1 ): 3 2 =−3− 4 3 − 4 3 =−3− 2 3 =− 2 9 => x − 1 2 = − 9 2 ⋅ 2 3 = − 9 3 = − 3 x− 2 1 =− 2 9 ⋅ 3 2 =− 3 9 =−3 => x = − 3 + 1 2 = − 5 2 x=−3+ 2 1 =− 2 5 b: ( 2 x − 15 ) 3 = ( 2 2 ⋅ 3 3 − 2 3 ⋅ 3 2 ) : ( − 36 ) (2x−15) 3 =(2 2 ⋅3 3 −2 3 ⋅3 2 ):(−36) => ( 2 x − 15 ) 3 = ( 4 ⋅ 27 − 8 ⋅ 9 ) : ( − 36 ) (2x−15) 3 =(4⋅27−8⋅9):(−36) => ( 2 x − 15 ) 3 = ( 108 − 72 ) − 36 = 36 − 36 = − 1 (2x−15) 3 = −36 (108−72) = −36 36 =−1 =>2x-15=-1 =>2x=14 =>x=7 c: ( 4 x + 1 ) 4 = ( 4 x + 1 ) 6 (4x+1) 4 =(4x+1) 6 => ( 4 x + 1 ) 6 − ( 4 x + 1 ) 4 = 0 (4x+1) 6 −(4x+1) 4 =0 => ( 4 x + 1 ) 4 ⋅ [ ( 4 x + 1 ) 2 − 1 ] = 0 (4x+1) 4 ⋅[(4x+1) 2 −1]=0 => ( 4 x + 1 ) 4 ( 4 x + 1 − 1 ) ( 4 x + 1 + 1 ) = 0 (4x+1) 4 (4x+1−1)(4x+1+1)=0 => 4 x ( 4 x + 2 ) ( 4 x + 1 ) 4 = 0 4x(4x+2)(4x+1) 4 =0 => [ x = 0 4 x + 2 = 0 4 x + 1 = 0 ⇔ [ x = 0 x = − 1 2 x = − 1 4 x=0 4x+2=0 4x+1=0 ⇔ x=0 x=− 2 1 x=− 4 1 d: ( 1 1 ⋅ 2 + 1 2 ⋅ 3 + . . . + 1 99 ⋅ 100 ) ⋅ x 2 = 99 ( 1⋅2 1 + 2⋅3 1 +...+ 99⋅100 1 )⋅x 2 =99 => ( 1 − 1 2 + 1 2 − 1 3 + . . . + 1 99 − 1 100 ) ⋅ x 2 = 99 (1− 2 1 + 2 1 − 3 1 +...+ 99 1 − 100 1 )⋅x 2 =99 => ( 1 − 1 100 ) ⋅ x 2 = 99 (1− 100 1 )⋅x 2 =99 => x 2 = 99 : 99 100 = 100 x 2 =99: 100 99 =100 => x = ± 10 x=±10 2: 2xy-x-y=2 => x ( 2 y − 1 ) − y + 1 2 = 2 + 1 2 x(2y−1)−y+ 2 1 =2+ 2 1 => 2 x ( y − 1 2 ) − ( y − 1 2 ) = 5 2 2x(y− 2 1 )−(y− 2 1 )= 2 5 => ( 2 x − 1 ) ( y − 1 2 ) = 5 2 (2x−1)(y− 2 1 )= 2 5 =>(2x-1)(2y-1)=5 => ( 2 x − 1 ; 2 y − 1 ) ∈ { ( 1 ; 5 ) ; ( 5 ; 1 ) ; ( − 1 ; − 5 ) ; ( − 5 ; − 1 ) } (2x−1;2y−1)∈{(1;5);(5;1);(−1;−5);(−5;−1)} => ( x ; y ) ∈ { ( 1 ; 3 ) ; ( 3 ; 1 ) ; ( 0 ; − 2 ) ; ( − 2 ; 0 ) } (x;y)∈{(1;3);(3;1);(0;−2);(−2;0)}
=> ( 2 x − 1 ) ( y − 1 2 ) = 5 2 (2x−1)(y− 2 1 )= 2 5 =>(2x-1)(2y-1)=5 => ( 2 x − 1 ; 2 y − 1 ) ∈ { ( 1 ; 5 ) ; ( 5 ; 1 ) ; ( − 1 ; − 5 ) ; ( − 5 ; − 1 ) } (2x−1;2y−1)∈{(1;5);(5;1);(−1;−5);(−5;−1)} => ( x ; y ) ∈ { ( 1 ; 3 ) ; ( 3 ; 1 ) ; ( 0 ; − 2 ) ; ( − 2 ; 0 ) } (x;y)∈{(1;3);(3;1);(0;−2);(−2;0)}