bunhiaxcopki

$((b-c)^2)/b^2$=$(2[1-cos(B-C)])/(1-cos(2B))$

Nhận dạng tam giác

a $\sqrt{x^{2}+x-1}+\sqrt{-x^{2}+x+1}=x^{2}- x-2$
b $\sqrt{1+\sqrt{1-x^{2}}} = x(1+2\sqrt{1-x^{2}})$
c $\sqrt{x-2}+\sqrt{4-x}=2x^{2}-5x-1$
d $\sqrt{x^{2}-x-6}+3\sqrt{x}=\sqrt{2(x^{2}+5x-3)}$
e $x^{3}-3x^{2}-3x+2\sqrt{(x+1)^{3}}=0$
f $\sqrt{3x^{2}-5x+1}-\sqrt{x^{2}-2}=\sqrt{3(x^{2}-x-1)}-\sqrt{x^{2}-3x+4}

a $\sqrt{x^{2}+x-1}+\sqrt{-x^{2}+x+1}=x^{2}- x-2$
b $\sqrt{1+\sqrt{1-x^{2}}} = x(1+2\sqrt{1-x^{2}})$
c $\sqrt{x-2}+\sqrt{4-x}=2x^{2}-5x-1$
d $\sqrt{x^{2}-x-6}+3\sqrt{x}=\sqrt{2(x^{2}+5x-3)}$
e $x^{3}-3x^{2}-3x+2\sqrt{(x+1)^{3}}=0$
f $\sqrt{3x^{2}-5x+1}-\sqrt{x^{2}-2}=\sqrt{3(x^{2}-x-1)}-\sqrt{x^{2}-3x+4}$