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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

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: \(89\cdot25+89\cdot74+89\)
\(=89\cdot\left(25+74+1\right)\)
\(=89\cdot100=8900\)

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

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)
= \(\frac12\)