b) 13.65-120+35.13
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a: \(13\cdot65+35\cdot12\)
\(=13\cdot65+35\cdot13-35\)
=1300-35
=1265
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a: Sửa đề: \(A=\dfrac{35^3+13^3}{48}-35\cdot13\)
\(=\dfrac{\left(35+13\right)\left(35^2-35\cdot13+13^2\right)}{48}-35\cdot13\)
\(=35^2-2\cdot35\cdot13+13^2=22^2\)
b: \(B=\dfrac{\left(68-52\right)\left(68^2+68\cdot52+52^2\right)}{16}+68\cdot52\)
\(=68^2+2\cdot68\cdot52+52^2\)
\(=110^2\)
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a: Sửa đề: \(A=\dfrac{35^3+13^3}{48}-35\cdot13\)
\(=\dfrac{\left(35+13\right)\left(35^2-35\cdot13+13^2\right)}{48}-35\cdot13\)
\(=35^2-2\cdot35\cdot13+13^2=22^2\)
b: \(B=\dfrac{\left(68-52\right)\left(68^2+68\cdot52+52^2\right)}{16}+68\cdot52\)
\(=68^2+2\cdot68\cdot52+52^2\)
\(=110^2\)
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a, 125+70+375+230
= (125+375)+(70+230)
= 500+300 = 800
b, 7 2 . 33 + 7 2 . 67
= 49.33+49.67 = 49.(33+67) = 49.100 = 4900
c, 35.11+65.18+35.13+65.6
= 35.(11+13)+65.(18+6) = 35.24+65.24 = 24.(35+65) = 24.100 = 2400
d, 490 – {[(128+22):3. 2 2 ] – 7}
= 490 – {[150:3.4] – 7}
= 490 – {[50.4] – 7}
= 490 – {200 – 7}
= 490 – 193 = 297
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a) Ta có: A= \(\frac{35^3+13^3}{48}-35\cdot13\)=\(\frac{\left(35+13\right)\left(35^2-35\cdot13+13^2\right)}{48}-35\cdot13\)
=\(35^2-35\cdot13+13^2+35\cdot13\)=\(35^2+13^2=1394\)
b) Ta có: B=\(\frac{68^3-52^3}{16}+68\cdot52\)=\(\frac{\left(68-52\right)\left(68^2+68\cdot52+52^2\right)}{16}+68\cdot52\)
=\(68^2+2\cdot68\cdot52+52^2\)= \(\left(68+52\right)^2=120^2=14400\)
cho mk hỏi tại sao ở dòng thứ nhất là trừ 35.13 xong dưới lại là cộng
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Bài 1 a)75+256=331 b)845-120+455=1180 Bài 2 a)40+144=184 b)222-150+675=747 Bài 3 a)18.27-27.12=486-324=126 b)855-160+1045=1740 4 a)378-180=190 b)1188-290+1512=2410 5 a)147+48=195 b)1168-370+432=1230 6 a)153+75=228 b)2015-210+1085=2890
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\(A=\dfrac{\left(35+13\right)\left(35^2-35\cdot13+13^2\right)}{48}-35\cdot13\)
\(=35^2-2\cdot35\cdot13+13^2\)
\(=22^2=484\)