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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, \(2x+x+5x=400\Leftrightarrow8x=400\Leftrightarrow x=50\)
(x-15):12=79 dư 8
=>\(x-15=79\times12+8=956\)
=>x=956+15=971
(x - 15) : 12 = 79 dư 8
x - 15 = 79 x 12 + 8
x - 15 = 956
x = 956 + 15
x= 971
a) A = 199 × 201
= (200 - 1) × 201
= 200 × 201 - 201
= 200 × (200 + 1) - 201
= 200 × 200 + 200 - 201
= 200 × 200 - 1 < 200 × 200
Vậy A < B
b) Ta có:
2015 × 2025 = (2020 - 5) × (2020 + 5)
= 2020 × 2020 + 2020 × 5 - 5 × 2020 - 25
= 2020 × 2020 - 25 < 2020 × 2020
Vậy A < B
dạo này nhìu btvn quá , nhờ mn 1 tí nhá
