問題文全文(内容文)：
$(x+y)^2=$
$(x-y)^2=$
$(x+y) (x-y)=$
$(x+a) (X+b)=$
⑤$(\sqrt{5}-\sqrt{3})^2=$
⑥$(\sqrt{7}+\sqrt{2}) (\sqrt{7}-\sqrt{2}) =$
⑦$(\sqrt{2}+5) (\sqrt{2}+4)=$
⑧$\sqrt{2}(\sqrt{12 }-\sqrt{3}) =$
⑨$(2\sqrt{2}+3) (2\sqrt{2}-3)=$
⑩$(\sqrt{2}+4\sqrt{2})^2=$
11$(4\sqrt{3}-1) (-2\sqrt{3}+3)=$
12$(\sqrt{3}-4) (\sqrt{3}+1) -\sqrt{3}(2-5\sqrt{3}) =$
$(x+y)^2=$
$(x-y)^2=$
$(x+y) (x-y)=$
$(x+a) (X+b)=$
⑤$(\sqrt{5}-\sqrt{3})^2=$
⑥$(\sqrt{7}+\sqrt{2}) (\sqrt{7}-\sqrt{2}) =$
⑦$(\sqrt{2}+5) (\sqrt{2}+4)=$
⑧$\sqrt{2}(\sqrt{12 }-\sqrt{3}) =$
⑨$(2\sqrt{2}+3) (2\sqrt{2}-3)=$
⑩$(\sqrt{2}+4\sqrt{2})^2=$
11$(4\sqrt{3}-1) (-2\sqrt{3}+3)=$
12$(\sqrt{3}-4) (\sqrt{3}+1) -\sqrt{3}(2-5\sqrt{3}) =$
単元：
#数学(中学生)#中３数学#平方根
指導講師：
とある男が授業をしてみた
問題文全文(内容文)：
$(x+y)^2=$
$(x-y)^2=$
$(x+y) (x-y)=$
$(x+a) (X+b)=$
⑤$(\sqrt{5}-\sqrt{3})^2=$
⑥$(\sqrt{7}+\sqrt{2}) (\sqrt{7}-\sqrt{2}) =$
⑦$(\sqrt{2}+5) (\sqrt{2}+4)=$
⑧$\sqrt{2}(\sqrt{12 }-\sqrt{3}) =$
⑨$(2\sqrt{2}+3) (2\sqrt{2}-3)=$
⑩$(\sqrt{2}+4\sqrt{2})^2=$
11$(4\sqrt{3}-1) (-2\sqrt{3}+3)=$
12$(\sqrt{3}-4) (\sqrt{3}+1) -\sqrt{3}(2-5\sqrt{3}) =$
$(x+y)^2=$
$(x-y)^2=$
$(x+y) (x-y)=$
$(x+a) (X+b)=$
⑤$(\sqrt{5}-\sqrt{3})^2=$
⑥$(\sqrt{7}+\sqrt{2}) (\sqrt{7}-\sqrt{2}) =$
⑦$(\sqrt{2}+5) (\sqrt{2}+4)=$
⑧$\sqrt{2}(\sqrt{12 }-\sqrt{3}) =$
⑨$(2\sqrt{2}+3) (2\sqrt{2}-3)=$
⑩$(\sqrt{2}+4\sqrt{2})^2=$
11$(4\sqrt{3}-1) (-2\sqrt{3}+3)=$
12$(\sqrt{3}-4) (\sqrt{3}+1) -\sqrt{3}(2-5\sqrt{3}) =$
投稿日：2013.05.30
