Mathematics
Junior High
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4
(1) (√2+3)(√5 +4)
をふくむ式の計算 次の計算をしなさい。
(2)
72
(3) (√5+3)²
(√3-2)(2√3-1)
(7) (√2-3)(√2-4)
(4) (√7-√6) ²
(5) (√10+√3)(√10-√3) (6) (√6-2)(√6 +2)
(8) (√5 +2)(√5-6)
141
(1) √√√2+3)/(√√5+4) = √/0 +4√2+ 3√5+ (2₁
(2) (13-2)215-1)=6-13-413+2=612-13-403=8-513m
(13)(√√5+3)=(√5) + 2 x (√5) + 3 + 3³² = 5 + 6 √5 + 9 = 14 +6 √5 11
(4) (√7-√√6)² = (1√7) ² 2X (UT) X (√6) + (√6)² = 7 -2√42 + 6 = "13-2042₁
(5) (√/0+√3) (V/0-√√3)= 10-√√30+√30-3 = 17 f
16/16-216+2)=0-4:24
(7) (V2-3)(1-4)=2-412312+12=14-7151
(8) (15+2)(15-6=5-6V5+2V5-12=5-12-6V5+205
=-7-415,
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