Olympiad Maths - 27 වැඩ්ඩන්ට I Combined Maths I Ajantha Dissanayake

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[@sandeeshsewanagala8829]

5 scds
999999^2 + 999999 + (999999+1)
Let, 999999=x
x^2 + 2x + 1=(x+1)^2
sq root of (x+1)^2 is (x+1)
The answer is 1 000 000

[@iam_vinu_sandamith]

❤

[@JayanthaWanniarachchi-w5g]

I just took X to 999999 as a substitution. However the answer I took is 10 to the power 6...😊

[@ShashmikaRathnayaka-rn5jt]

Oke mehema karapiya
(999999*2 + 1999999)*1/2
(999999*2 + 1000000+ 999999)*/12
(999999*2 + 999999 + 1 + 999999)*1/2)
999999 - X Lesa ganin
(X*2 + 2x +1)*1/2
(X + 1)
999999+1
1000000//

[@YohanLakshan-fk6uz]

[(10^6-1)^2 + 2×10^6 -1]
[10^12 + 1 - 2×10^6 +2×10^6 -1] = [10^12]
[10^12]^(1÷2) = 10^6

[@wasanthathushari3081]

999999, කියන්නෙ 1000000-1, 1999999 කියන්නෙ 2×1000000 -1, ලක්ශ දහයට x ආදෙශ කරල වර්ගය ප්‍රසාරනය කරානම් ගානට x වර්ගයක් ඇවිත් උත්තරෙ එනව

[@thisaranimekhala8681]

Math hduwe mehem

[@PradeepaPalihakoon-ug9vn]

Niyamai sir

[@amilafernando4930]

ruclips.net/video/D2kOgpeWBh8/видео.htmlsi=xAqKeYXvGwOD1YuX

[@jayalathkamalsiri7339]

10^6

[@ushanhasaranga99]

999*999 + 999+999+1
999( 999+1) + (999+1)
1000*1000
1000 වර්ගය හැදෙනව නම් වර්ග මූලෙට වර්ගෙ කැපෙනව.
මං බින්දු අඩුවෙන් හැදුවෙ කම්මැලි නිසා
එතකොට හරි බින්දු ගාණ දැම්මම
1000000.

[@maheshsushantha4783]

🔥

[@lahiruabeywardana-fi9qb]

1000000

[@Mr.pavan0]

❤❤❤❤❤❤❤❤❤❤

[@amilafernando4930]

ruclips.net/video/D2kOgpeWBh8/видео.htmlsi=xAqKeYXvGwOD1YuX

[@nalakasandaruwan2762]

=999,999^2+1,999,999
=999,999^2+999,999+1,000,000
=999,999(999,999+1)+10^6
=999,999(10^6)+10^6
=10^6(999,999+1)
=10^6(10^6)
=10^6

[@madhurasenanayaka6632]

Sir.....achintha wijeesinha කියලා අයියා කෙනෙක් දන්නවා ද?...සර්ගේ class ආපු කෙනෙක්...

[@henaka_raala]

999,999^2 + 1,999,999
(1,000,000 - 1)^2 + (1,000,000 + 999,999)
10^{12} - 2,000,000 + 1 + 1,000,000 + 999,999
(10^6)^2 - 2,000,000 + 2,000,000
(10^6)^2 [වර්ගමූලය දැමූ විට වර්ගය ඉවත් වේ]
10^6 = 1,000,000

[@PHOENIXX-p8y]

=[(1000000-1)^2 + 1999999]^1%2
=1000000^2- 2000000+1 [phone]
=1000000

[@නිඛිල]

10^6