Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) 97.13 + 130.0,3
= 97.13 + 13.3
= 13. ( 97 + 3 )
= 13.100
=1300
b) 86.153 - 530.8,6
= 86.153 - 53.86
= 86. ( 153 - 53 )
= 86. 100
= 8600
( Toán lớp 8 đây hả ???)
~HT~
a) \(97\cdot13+130\cdot0,3=97\cdot13+13\cdot10\cdot0,3=13\cdot\left(97+10\cdot0,3\right)=13\cdot100=1300\)
b)\(86\cdot153-530\cdot8,6=8,6\cdot10\cdot153-53\cdot10\cdot8,6=10\cdot8,6\cdot\left(153-53\right)=10\cdot8,6\cdot100=8600\)
a) \(97.13+130.0,3\\ =97.13+13.3\\ =13.\left(97+3\right)\\ =13.100=1300\)
b) \(86.153-530.8,6\\ =8,6.1530-530.8,6\\ =8,6.\left(1530-530\right)\\ =8,6.1000\\ =8600\)
đây có lẽ là toán lớp 5
a)97*13+130*0,3
=97*13+13*10*0,3
=97*13+13*3
=13*(97+3)
=13*100
=1300
b)86*153-530*8,6
=86*153-53*10*8,6
=86*153-53*86
=86*(153-53)
=86*100
=8600
a, \(97.13+130.13\)
\(=13\left(97+130\right)\)
\(=13.227\)
\(=2951\)
b, \(86.153-530.8,6\)
\(=86.153-53.10.8,6\)
\(=86.153-53.86\)
\(=86\left(153-53\right)\)
\(=86.100\)
\(=8600\)
trả lời:
a) 97.13 + 130.13
= 13.( 97 + 130 )
= 13. 227
= 2951
b) 86.153 - 530.8,6
= 86.153 - 53.10.8,6
=86. 153 - 53.86
= 86. ( 153 - 53 )
= 86.100
= 8600
học tốt!
97.13 + 130.0,3 = 97.13 + 13.10.0,3
= 97.13 + 13.3 = 13.(97 + 3) = 13.100=1300
86.153 − 530.8,6 = 86.153 – 53.10.8,6
= 86.153 − 53.86 = 86.(153 − 53) = 86.100=8600
a)999x1001=(1000-1)(1000+1)=10002-12=1000000-1=999999
b)bạn viết đúng đề câu b k thế?
\(P=\left(x+\dfrac{1}{x}\right)^2+\left(y+\dfrac{1}{y}\right)^2+\left(z+\dfrac{1}{z}\right)^2-\left(x+\dfrac{1}{x}\right)\left(y+\dfrac{1}{y}\right)\left(z+\dfrac{1}{z}\right)\)
Ta có: \(xyz=1\Rightarrow x=\dfrac{1}{yz}\)
\(P=\left(\dfrac{1}{yz}+yz\right)^2+\left(y+\dfrac{1}{y}\right)^2+\left(z+\dfrac{1}{z}\right)^2-\left(yz+\dfrac{1}{yz}\right)\left(y+\dfrac{1}{y}\right)\left(z+\dfrac{1}{z}\right)\)
\(P=\dfrac{1}{y^2z^2}+2+1y^2z^2+y^2+2+\dfrac{1}{y^2}+z^2+2+\dfrac{1}{z^2}-\left(y^2z+z+\dfrac{1}{z}+\dfrac{1}{y^2z}\right)\left(z+\dfrac{1}{z}\right)\)
\(P=\dfrac{1}{y^2z^2}+y^2z^2+y^2+\dfrac{1}{y^2}+z^2+\dfrac{1}{z^2}+6-y^2z^2-y^2-z^2-1-1-\dfrac{1}{z^2}-\dfrac{1}{y^2}-\dfrac{1}{y^2z^2}\)\(P=\left(\dfrac{1}{y^2z^2}-\dfrac{1}{y^2z^2}\right)+\left(y^2z^2-y^2z^2\right)+\left(y^2-y^2\right)+\left(z^2-z^2\right)+\left(\dfrac{1}{y^2}-\dfrac{1}{y^2}\right)+\left(\dfrac{1}{z^2}-\dfrac{1}{z^2}\right)+4\)
\(P=4\)
Vậy: ...
a)=1300
b)=8600