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có lộn đề ko bn ơi
c). 1081.46
d). 1000
e. P=(x−y)(x2+xy+y2)−2y3=x3−y3−2y3=x3−3y3=(12)3−3.(23)3=−5572
\(101^2=\left(100+1\right)^2=10000+200+1=10201\)
\(1001^2=\left(1000+1\right)^2=1000000+2000+1=1002001\)
\(102^2=\left(100+2\right)^2=10000+400+4=10404\)
\(99^2=\left(100-1\right)^2=10000-200+1=9801\)
\(19^2=\left(10+9\right)^2=100+180+81=361\)
\(999^2=\left(1000-1\right)^2=1000000-2000+1=998001\)
a) 1012 = (100 + 1)2 = 1002 + 2.100 + 1 = 10000 + 200 + 1 = 10201
b) 1992 = (200 – 1)2 = 2002 – 2.200 + 1 = 40000 – 400 + 1 = 39601
c) 47.53 = (50 – 3)(50 + 3) = 502 – 32 = 2500 – 9 = 2491.
\(a,101^2=101.\left(100+1\right)=10100+101=10201.\\ b,75^2-50.75+25^2\\ =75.\left(75-50\right)+25^2\\ =75.25+25^2\\ =25.\left(75+25\right)\\ =25.100\\ =2500.\)
\(c,103.97\\ =\left(100+3\right).97\\ =9700+291\\ =9991\)
Ta có :
\(A=\frac{2012.14+1997+2010.2011}{2011.5+2011.1008+1012.2011}\)
\(\Rightarrow A=\frac{\left(2011+1\right).14+1997+2010.2011}{2011.\left(5+1008+1012\right)}\)
\(\Rightarrow A=\frac{2011.14+14+1997+2010.2011}{2011.2025}\)
\(\Rightarrow A=\frac{2011.14+2011+2010.2011}{2011.2025}\)
\(\Rightarrow A=\frac{2011.\left(14+1+2011\right)}{2011.2025}\)
\(\Rightarrow A=\frac{2011.25}{2011.25}\)
\(\Rightarrow A=1\) ( tử số = mẫu số )
Vậy \(A=1\)
~ Ủng hộ nhé
Ta có: \(B=\frac12+\frac13-\frac14+\frac15-\frac16+\cdots-\frac{1}{2022}+\frac{1}{2023}\)
=>\(B=\frac12+\frac13+\frac14+\frac15+\frac16+\cdots+\frac{1}{2022}+\frac{1}{2023}-2\left(\frac14+\frac16+\cdots+\frac{1}{2022}\right)\)
\(=\frac12+\frac13+\frac14+\frac15+\cdots+\frac{1}{2022}+\frac{1}{2023}-\frac12-\frac13-\cdots-\frac{1}{1011}\)
\(=\frac{1}{1012}+\frac{1}{1013}+\cdots+\frac{1}{2022}+\frac{1}{2023}\)
=C
=>B-C=0
a) = 117 x 100 - 17 x 100
= ( 117 - 17 ) x 100
= 100 x 100
= 10000
b) ( 532 x 7 - 266 x 14 ) x ( 532 x 7 + 266 )
= 532 x 7 - 266 x ( 2 x 7 ) x 3990
= 532 x 7 - 266 x 2 x 7 x 3990
= (532 x 7 - 532 x 7) x 3990
= 0 x 3990
= 0
c) 506 x 5 + 1012 + 253 x 6
= 506 x 5 + (506 x 2) + 253 x ( 2 x 3 )
= 506 x 5 + 506 x 2 + 253 x 2 x 3
= 506 x 5 + 506 x 2 + 506 x 3
= 506 x ( 5 + 2 + 3 )
= 506 x 10
= 5060
Ta có: \(S=1-\frac12+\frac13-\frac14+\cdots+\frac{1}{2021}-\frac{1}{2022}\)
\(=1+\frac12+\frac13+\frac14+\cdots+\frac{1}{2021}+\frac{1}{2022}-2\left(\frac12+\frac14+\cdots+\frac{1}{2022}\right)\)
\(=1+\frac12+\cdots+\frac{1}{2022}-1-\frac12-\cdots-\frac{1}{1011}\)
\(=\frac{1}{1012}+\frac{1}{1013}+\cdots+\frac{1}{2022}=P\)
=>S-P=0
=>\(\left(S-P\right)^{2022}=0\)
