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Tugas Yosua

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by

Yonathan Happy Setiawan

on 11 August 2014

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Transcript of Tugas Yosua

√(

5

+

2√6

)
THANK YOU :D
Yosua Lucky Chandra
X-Nehemiah / 28
LANGKAH TERAKHIR
a = √2
b = √3
√(5+2√6)
Yosua Lucky Chandra
= √(

(2+3)
+
2√(2x3)

)
= √(
2
+
2√(2 𝑥3)
++

3

)
3
)
√(
2
+
2√(2 𝑥3)
++

3

)
3
)
a^2
+
2ab
+
b^2
= ( a + b )^2
by:
Yosua O'Luckyz Chandra
@lucky_ambigu
LANGKAH AWAL
SOAL
LANGKAH SELANJUTNYA
ingat persamaan ini...?
maka:
√(
2
+
2√(2 𝑥3)
++

3

)
3
)
= √( √2 + √3 ) ^2
√2 + √3
Full transcript