BT1: Cho A = \(\sqrt{4+\sqrt{10+2\sqrt{5}}}\)
B = \(\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
Tính A.B và A-B
BT2: Tính A = \(\sqrt{4+\sqrt{7}}\)- \(\sqrt{4-\sqrt{7}}\) - \(\sqrt{2}\)
BT3: Rút gọn:
C = \(\frac{x+2+\sqrt{x^2-4}}{x+2-\sqrt{x^2-4}}\) + \(\frac{x+2-\sqrt{x^2-4}}{x+2+\sqrt{x^2-4}}\)
GIÚP MÌNH VỚI Ạ MÌNH CẦN GẤP LẮM
Chờ từ trưa không idol nào đụng thì thôi em xin vậy :))
BT1:
Ta có: \(A\cdot B=\sqrt{4+\sqrt{10+2\sqrt{5}}}\cdot\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(=\sqrt{16-10-2\sqrt{5}}\)
\(=\sqrt{6-2\sqrt{5}}=\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-1\)
Từ đó thay vào: \(\left(A-B\right)^2\)
\(=A^2-2AB+B^2\)
\(=4+\sqrt{10+2\sqrt{5}}-2\left(\sqrt{5}-1\right)+4-\sqrt{10+2\sqrt{5}}\)
\(=10-2\sqrt{5}\)
\(\Rightarrow A-B=\sqrt{10-2\sqrt{5}}\)
BT2:
Đặt \(B=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
\(\Leftrightarrow B^2=4+\sqrt{7}-2\sqrt{\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}+4-\sqrt{7}\)
\(=8-2\sqrt{16-7}=8-2\cdot3=2\)
\(\Rightarrow B=\sqrt{2}\)
\(\Rightarrow A=B-\sqrt{2}=\sqrt{2}-\sqrt{2}=0\)
BT3:
đk: \(\orbr{\begin{cases}x\ge2\\x< -2\end{cases}}\)
\(C=\frac{x+2+\sqrt{x^2-4}}{x+2-\sqrt{x^2-4}}+\frac{x+2-\sqrt{x^2-4}}{x+2+\sqrt{x^2-4}}\)
\(C=\frac{\left(x+2+\sqrt{x^2-4}\right)^2}{\left(x+2\right)^2-\left(x^2-4\right)}+\frac{\left(x+2-\sqrt{x^2-4}\right)^2}{\left(x+2\right)^2-\left(x^2-4\right)}\)
\(C=\frac{\left(x+2\right)^2+2\left(x+2\right)\sqrt{x^2-4}+x^2-4+\left(x+2\right)^2-2\left(x+2\right)\sqrt{x^2-4}+x^2-4}{x^2+4x+4-x^2+4}\)
\(C=\frac{2x^2+8x+8+2x^2-8}{4x+8}\)
\(C=\frac{4x^2+8x}{4x+8}=x\)
Vậy C = x
