So sánh các phân số sau bằng cách hợp lý nhất
1300/1500 và 1333/1555
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\(\left(\dfrac{1}{2}-1\right).\left(\dfrac{1}{3}-1\right).\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{2008}-1\right).\left(\dfrac{1}{2009}-1\right)\)
\(=-\dfrac{1}{2}.-\dfrac{2}{3}.-\dfrac{3}{4}...-\dfrac{2007}{2008}.-\dfrac{2008}{2009}\)
\(=-\left(\dfrac{1.2.3...2007.2008}{2.3.4...2008.2009}\right)\)
\(=-\dfrac{1}{2009}\)
4(x-2)-3(x+1)=5
=>\(4x-8-3x-3=5\)
=>\(x-11=5\)
=>x=11+5=16
\(4\left(x-2\right)-3\left(x+1\right)=5\)
\(\Leftrightarrow4x-8-3x-3=5\)
\(\Leftrightarrow\left(4x-3x\right)=5+8+3\)
\(\Leftrightarrow x=16\)
Vậy \(x=16\)
Ta có:
\(\dfrac{1}{2}=\dfrac{1\times2}{2\times2}=\dfrac{2}{4};\dfrac{1}{4}=\dfrac{1}{4}\)
Vì \(\dfrac{2}{4}>\dfrac{1}{4}\) nên \(\dfrac{1}{2}>\dfrac{1}{4}\)
\(\dfrac{29}{2}=\dfrac{28+1}{2}=14+\dfrac{1}{2}=14\dfrac{1}{2}\)
\(\dfrac{15}{4}=\dfrac{12+3}{4}=3+\dfrac{3}{4}=3\dfrac{3}{4}\)
\(\dfrac{31}{2}=\dfrac{30+1}{2}=15\dfrac{1}{2}\)
\(\dfrac{29}{3}=\dfrac{27+2}{3}=9\dfrac{2}{3}\)
\(\dfrac{125}{8}=\dfrac{120+5}{8}=15+\dfrac{5}{8}=15\dfrac{5}{8}\)
\(\dfrac{36}{27}=\dfrac{27+9}{27}=1+\dfrac{9}{27}=1\dfrac{9}{27}\)
\(\dfrac{124}{15}=\dfrac{120+4}{15}=8+\dfrac{4}{15}=8\dfrac{4}{15}\)
\(\dfrac{96}{3}=\dfrac{93+3}{3}=31\dfrac{3}{3}\)
\(\dfrac{129}{24}=\dfrac{120+9}{24}=5+\dfrac{9}{24}=5\dfrac{9}{24}\)
\(\dfrac{78}{13}=\dfrac{65+13}{13}=5+\dfrac{13}{13}=5\dfrac{13}{13}\)
\(\dfrac{91}{4}=\dfrac{88+3}{4}=22+\dfrac{3}{4}=22\dfrac{3}{4}\)
\(\dfrac{115}{8}=\dfrac{112+3}{8}=14+\dfrac{3}{8}=14\dfrac{3}{8}\)
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
\(A=\dfrac{1}{2\cdot6}+\dfrac{1}{3\cdot8}+...+\dfrac{1}{2023\cdot4048}\)
\(=\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{4046\cdot4048}\)
\(=\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{4046}-\dfrac{1}{4048}\)
\(=\dfrac{1}{4}-\dfrac{1}{4048}=\dfrac{1012-1}{4048}=\dfrac{1011}{4048}\)
\(A=\dfrac{1}{2\cdot6}+\dfrac{1}{3\cdot8}+\dfrac{1}{4\cdot10}+...+\dfrac{1}{2023\cdot4048}\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{2023\cdot2024}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2023}-\dfrac{1}{2024}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2024}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{1012-1}{2024}\)
\(=\dfrac{1011}{4048}\)
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, \(x+\left(x-1\right)+\left(x-2\right)+...+\left(x-50\right)=1530\)
\(\Leftrightarrow51x+\dfrac{\left(-1-50\right).50}{2}=1530\Leftrightarrow51x-1275=1530\Leftrightarrow51x=2805\Leftrightarrow x=55\)
Ta có :
\(\dfrac{1300}{1500}=\dfrac{13}{15}=1-\dfrac{2}{15}\)
\(\dfrac{1333}{1555}=1-\dfrac{222}{1555}\)
Vì \(\dfrac{222}{1555}>\dfrac{2}{15}\)
\(\Rightarrow1-\dfrac{222}{1555}< 1-\dfrac{2}{15}\)
\(\dfrac{\Rightarrow1333}{1555}< \dfrac{1300}{1500}\)
hi