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a: 11+13+15+17+19
=(11+19)+(13+17)+15
=30+30+15
=75
b: \(122+2116+278+84\)
=(122+278)+(2116+84)
=400+2200
=2600
c: \(12\times125\times54\)
=1500x54
=81000
d: \(27\times36+27\times64=27\times\left(36+64\right)=27\times100=2700\)
e: \(25\text{x}37+25\text{x}63-150\)
=25x(37+63)-150
=25x100-150
=2500-150
=2350
f: \(425\text{x}74-27\text{x}425-1\)
=425x(74-27-1)
=425x46=19550
g: \(8\text{x}9\text{x}14+6\text{x}17\text{x}12+19\text{x}4\text{x}18\)
=14x72+17x72+19x72
=72x(14+17+19)
=50x72=3600
a) $11+13+15+17+19$
$=(11+19)+(13+17)+15$
$=30+30+15$
$=60+15=75$
b) $122+2116+278+84$
$=(122+278)+(2116+84)$
$=400+2200=2600$
c) $12\times125\times54$
$=3\times4\times125\times2\times27$
$=(4\times2)\times125\times(3\times27)$
$=8\times125\times81$
$=1000\times81=81000$
d) $27\times36+27\times64$
$=27\times(36+64)$
$=27\times100=2700$
e) $25\times37+25\times63-150$
$=25\times(37+63)-150$
$=25\times100-150$
$=2500-150=2350$
f) $425\times74-27\times425-425$
$=425\times(74-27-1)$
$=425\times(47-1)$
$=425\times46=19550$
g) $8\times9\times14+6\times17\times12+19\times4\times18$
$=(8\times9)\times14+(6\times12)\times17+(4\times18)\times19$
$=72\times14+72\times17+72\times19$
$=72\times(14+17+19)$
$=72\times(31+19)$
$=72\times50=3600$
$\mathtt{Toru}$
Ta có :
\(\dfrac{1300}{1500}=\dfrac{13}{15}=1-\dfrac{2}{15}\)
\(\dfrac{1333}{1555}=1-\dfrac{222}{1555}\)
Vì \(\dfrac{222}{1555}>\dfrac{2}{15}\)
\(\Rightarrow1-\dfrac{222}{1555}< 1-\dfrac{2}{15}\)
\(\dfrac{\Rightarrow1333}{1555}< \dfrac{1300}{1500}\)
\(\dfrac{29}{2}=\dfrac{28+1}{2}=14+\dfrac{1}{2}=14\dfrac{1}{2}\)
\(\dfrac{15}{4}=\dfrac{12+3}{4}=3+\dfrac{3}{4}=3\dfrac{3}{4}\)
\(\dfrac{31}{2}=\dfrac{30+1}{2}=15\dfrac{1}{2}\)
\(\dfrac{29}{3}=\dfrac{27+2}{3}=9\dfrac{2}{3}\)
\(\dfrac{125}{8}=\dfrac{120+5}{8}=15+\dfrac{5}{8}=15\dfrac{5}{8}\)
\(\dfrac{36}{27}=\dfrac{27+9}{27}=1+\dfrac{9}{27}=1\dfrac{9}{27}\)
\(\dfrac{124}{15}=\dfrac{120+4}{15}=8+\dfrac{4}{15}=8\dfrac{4}{15}\)
\(\dfrac{96}{3}=\dfrac{93+3}{3}=31\dfrac{3}{3}\)
\(\dfrac{129}{24}=\dfrac{120+9}{24}=5+\dfrac{9}{24}=5\dfrac{9}{24}\)
\(\dfrac{78}{13}=\dfrac{65+13}{13}=5+\dfrac{13}{13}=5\dfrac{13}{13}\)
\(\dfrac{91}{4}=\dfrac{88+3}{4}=22+\dfrac{3}{4}=22\dfrac{3}{4}\)
\(\dfrac{115}{8}=\dfrac{112+3}{8}=14+\dfrac{3}{8}=14\dfrac{3}{8}\)
a: \(B=2021\times2025=\left(2023-2\right)\times\left(2023+2\right)=2023\times2023-2\times2\)
=>\(B=A-4\)
=>A lớn hơn B 4 đơn vị
b: \(C=35\times53-18=35\times35+35\times18-18\)
\(=35\times35+18\times\left(35-1\right)\)
\(=35\times35+18\times34\)
\(D=35+53\times34\)
\(=35+\left(35-1\right)\times\left(35+18\right)\)
\(=35+35\times35+35\times18-35\times1-18\)
\(=35\times35+35\times17+17=35\times35+36\times17\)
\(=35\times35+18\times34\)
=C
=>C=D
Gọi số cần tìm có dạng là \(X=\overline{ab}\)
Vì viết thêm số 7 vào bên trái số đó thì sẽ được số mới gấp 36 lần số cần tìm nên ta có: \(\overline{7ab}=36\times\overline{ab}\)
=>\(700+\overline{ab}=36\times\overline{ab}\)
=>\(35\times X=700\)
=>X=20
Vậy: Số cần tìm là 20
a: x+(x+1)+(x+2)+...+(x+30)=496
=>(x+x+...+x)+(1+2+3+...+30)=496
=>\(31x+30\times\dfrac{31}{2}=496\)
=>\(31x+465=496\)
=>31x=31
=>x=1
b: \(x+\left(x-1\right)+\left(x-2\right)+...+\left(x-50\right)=1530\)
=>\(51x-\left(1+2+3+...+50\right)=1530\)
=>\(51x-\dfrac{50\times51}{2}=1530\)
=>\(51x-1275=1530\)
=>51x=1275+1530=2805
=>x=2805:51=55
a, \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+30\right)=496\)
\(\Leftrightarrow31x+1+2+...+30=496\Leftrightarrow31x+\dfrac{\left(30+1\right).30}{2}=496\)
\(\Leftrightarrow31x+465=496\Leftrightarrow31x=31\Leftrightarrow x=1\)
b, \(x+\left(x-1\right)+\left(x-2\right)+...+\left(x-50\right)=1530\)
\(\Leftrightarrow51x+\dfrac{\left(-1-50\right).50}{2}=1530\Leftrightarrow51x-1275=1530\Leftrightarrow51x=2805\Leftrightarrow x=55\)
b, \(275-5\left(2x-1\right)=200\Leftrightarrow5\left(2x-1\right)=75\Leftrightarrow2x-1=15\Leftrightarrow x=8\)
c, \(3x-38:2=206\Leftrightarrow3x-19=206\Leftrightarrow3x=225\Leftrightarrow x=75\)
d, \(2x+x+5x=400\Leftrightarrow8x=400\Leftrightarrow x=50\)