Câu hỏi của Pha Le Chy

Question

$\frac{1+\frac{\sqrt{3}}{2}}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{1-\frac{\sqrt{3}}{2}}{1-\sqrt{1-\frac{\sqrt{3}}{2}}}$Hay cho minh thay canh tay cua may ban gioi toan naoooooTinh gia tri bieu thuc

Phạm Thị Thùy Linh · Accepted Answer

$\frac{1+\frac{\sqrt{3}}{2}}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{1-\frac{\sqrt{3}}{2}}{1-\sqrt{1-\frac{\sqrt{3}}{2}}}.$$=\frac{\frac{2+\sqrt{3}}{2}}{1+\sqrt{\frac{2+\sqrt{3}}{2}}}$$+\frac{\frac{2-\sqrt{3}}{2}}{1-\sqrt{\frac{2-\sqrt{3}}{2}}}$$=\frac{\frac{4+2\sqrt{3}}{4}}{1+\sqrt{\frac{4+\sqrt{3}}{4}}}$$+\frac{\frac{4-2\sqrt{3}}{4}}{1-\sqrt{\frac{4-2\sqrt{3}}{4}}}$$=\frac{\frac{3+2\sqrt{3}+1}{4}}{1+\sqrt{\frac{3+2\sqrt{3}+1}{4}}}$$+\frac{\frac{3-2\sqrt{3}+1}{4}}{1-\sqrt{\frac{3-2\sqrt{3}+1}{4}}}$$=\frac{\frac{\left(\sqrt{3}+1\right)^2}{4}}{1+\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{2}}$$+\frac{\frac{\left(\sqrt{3}-1\right)^2}{4}}{1+\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{2}}$$=\frac{\frac{\left(\sqrt{3}+1\right)^2}{4}}{1+\frac{\sqrt{3}+1}{2}}$$+\frac{\frac{\left(\sqrt{3}-1\right)^2}{4}}{1-\frac{\sqrt{3}-1}{2}}$$=\frac{\frac{\left(\sqrt{3}+1\right)^2}{4}}{\frac{2+\sqrt{3}}{2}}$$+\frac{\frac{\left(\sqrt{3}-1\right)^2}{4}}{\frac{2-\sqrt{3}}{2}}$$=\frac{\frac{\left(\sqrt{3}+1\right)^2}{4}}{\frac{\left(\sqrt{3}+1\right)^2}{4}}$$+\frac{\frac{\left(\sqrt{3}-1\right)^2}{4}}{\frac{\left(\sqrt{3}-1\right)^2}{4}}$$=1+1=2$Câu trả lời: $\frac{1+\frac{\sqrt{3}}{2}}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{1-\frac{\sqrt{3}}{2}}{1-\sqrt{1-\frac{\sqrt{3}}{2}}}.$$=\frac{\frac{2+\sqrt{3}}{2}}{1+\sqrt{\frac{2+\sqrt{3}}{2}}}$$+\frac{\frac{2-\sqrt{3}}{2}}{1-\sqrt{\frac{2-\sqrt{3}}{2}}}$$=\frac{\frac{4+2\sqrt{3}}{4}}{1+\sqrt{\frac{4+\sqrt{3}}{4}}}$$+\frac{\frac{4-2\sqrt{3}}{4}}{1-\sqrt{\frac{4-2\sqrt{3}}{4}}}$$=\frac{\frac{3+2\sqrt{3}+1}{4}}{1+\sqrt{\frac{3+2\sqrt{3}+1}{4}}}$$+\frac{\frac{3-2\sqrt{3}+1}{4}}{1-\sqrt{\frac{3-2\sqrt{3}+1}{4}}}$$=\frac{\frac{\left(\sqrt{3}+1\right)^2}{4}}{1+\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{2}}$$+\frac{\frac{\left(\sqrt{3}-1\right)^2}{4}}{1+\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{2}}$$=\frac{\frac{\left(\sqrt{3}+1\right)^2}{4}}{1+\frac{\sqrt{3}+1}{2}}$$+\frac{\frac{\left(\sqrt{3}-1\right)^2}{4}}{1-\frac{\sqrt{3}-1}{2}}$$=\frac{\frac{\left(\sqrt{3}+1\right)^2}{4}}{\frac{2+\sqrt{3}}{2}}$$+\frac{\frac{\left(\sqrt{3}-1\right)^2}{4}}{\frac{2-\sqrt{3}}{2}}$$=\frac{\frac{\left(\sqrt{3}+1\right)^2}{4}}{\frac{\left(\sqrt{3}+1\right)^2}{4}}$$+\frac{\frac{\left(\sqrt{3}-1\right)^2}{4}}{\frac{\left(\sqrt{3}-1\right)^2}{4}}$$=1+1=2$

Bui Huyen · Answer

$A=\frac{1+\frac{\sqrt{3}}{2}}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{1-\frac{\sqrt{3}}{2}}{1-\sqrt{1-\frac{\sqrt{3}}{2}}}$$A=\frac{2\left(1+\frac{\sqrt{3}}{2}\right)}{2+\sqrt{4+2\sqrt{3}}}+\frac{2\left(1-\frac{\sqrt{3}}{2}\right)}{2-\sqrt{4-2\sqrt{3}}}$$A=\frac{2+\sqrt{3}}{2+\sqrt{3}+1}+\frac{2-\sqrt{3}}{2-\sqrt{3}+1}$$A=\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}$$A=\frac{\left(3-\sqrt{3}\right)\left(2+\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{6}$$A=\frac{3+\sqrt{3}+3-\sqrt{3}}{6}$$A=\frac{6}{6}=1$Câu trả lời: $A=\frac{1+\frac{\sqrt{3}}{2}}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{1-\frac{\sqrt{3}}{2}}{1-\sqrt{1-\frac{\sqrt{3}}{2}}}$$A=\frac{2\left(1+\frac{\sqrt{3}}{2}\right)}{2+\sqrt{4+2\sqrt{3}}}+\frac{2\left(1-\frac{\sqrt{3}}{2}\right)}{2-\sqrt{4-2\sqrt{3}}}$$A=\frac{2+\sqrt{3}}{2+\sqrt{3}+1}+\frac{2-\sqrt{3}}{2-\sqrt{3}+1}$$A=\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}$$A=\frac{\left(3-\sqrt{3}\right)\left(2+\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{6}$$A=\frac{3+\sqrt{3}+3-\sqrt{3}}{6}$$A=\frac{6}{6}=1$

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