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a. \(=\sqrt{2}.\left(\sqrt{7}+\sqrt{8}\right)\sqrt{5-\sqrt{3}\sqrt{7}}\)
\(=\left(\sqrt{7}+\sqrt{8}\right)\sqrt{3-2\sqrt{3}.\sqrt{7}+7}\)
\(=\left(\sqrt{7}+\sqrt{8}\right)\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)
\(=\left(\sqrt{7}+\sqrt{8}\right)\left(\sqrt{7}-\sqrt{3}\right)\)
Rồi nhân ra. bạn làm tiếp nhé. Tuy nhiên minh nghĩ bạn bị nhầm đề. là \(\sqrt{6}\) chứ không phải căn 16
b. \(=\frac{5\left(\sqrt{21}+1\right)}{21-16}+\frac{\sqrt{3}.\sqrt{7}\left(\sqrt{3}-\sqrt{7}\right)}{-\left(\sqrt{3}-\sqrt{7}\right)}\)
\(=\sqrt{21}+4-\sqrt{21}=4\)
a: \(=\dfrac{2+\sqrt{3}}{2-\sqrt{3}}-\dfrac{2-\sqrt{3}}{2+\sqrt{3}}\)
\(=\dfrac{7+4\sqrt{3}-7+4\sqrt{3}}{1}=8\sqrt{3}\)
b: \(=\sqrt{2}-1-\sqrt{2}=-1\)
\(A=\sqrt{3+\sqrt{5}}+\sqrt{7-3.\sqrt{5}}-\sqrt{2}\)
\(\sqrt{2}.A=\sqrt{5+2\sqrt{5}.1+1}+\sqrt{9-2.3.\sqrt{5}+5}-2\)
\(\sqrt{2}.A=\sqrt{5}+1+3-\sqrt{5}-2=2\)
\(\Rightarrow A=\sqrt{2}\)
ĐKXĐ: \(\hept{\begin{cases}2x-4\ge0\\x+2.\sqrt{2x-4}\ge0\\x-2\sqrt{2x-4}\end{cases}}\Leftrightarrow x\ge2\)
\(\sqrt{x+2.\sqrt{2x-4}}+\sqrt{x-2.\sqrt{2x-4}}\)
\(=\sqrt{x-2+2.\sqrt{x-2}.\sqrt{2}+2}+\sqrt{x-2-2.\sqrt{x-2}.\sqrt{2}+2}\)
\(=\sqrt{x-2}+\sqrt{2}+\left|\sqrt{x-2}-\sqrt{2}\right|\)
Tự phá trị tuyệt đối
a) \(A=\sqrt{4-\sqrt{15}}-\sqrt{2+\sqrt{3}}\)
\(\Rightarrow\)\(\sqrt{2}A=\sqrt{8-2\sqrt{15}}-\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\sqrt{5}-\sqrt{3}-\left(\sqrt{3}+1\right)=\sqrt{5}-1\)
\(\Rightarrow\)\(A=\frac{\sqrt{5}-1}{\sqrt{2}}\)
b) tương tự câu a
c) \(\sqrt{6+2\sqrt{5-\sqrt{13+4\sqrt{3}}}}-\sqrt{6-2\sqrt{5+\sqrt{13-4\sqrt{3}}}}\)
\(=\sqrt{6+2\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}-\sqrt{6-2\sqrt{5+\sqrt{\left(\sqrt{12}-1\right)^2}}}\)
\(=\sqrt{6+2\sqrt{5-\left(\sqrt{12}+1\right)}}-\sqrt{6-2\sqrt{5+\left(\sqrt{12}-1\right)}}\)
\(=\sqrt{6+2\sqrt{4-2\sqrt{3}}}-\sqrt{6-2\sqrt{4+2\sqrt{3}}}\)
\(=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}-\sqrt{6-2\sqrt{\left(\sqrt{3}+1\right)^2}}\)
\(=\sqrt{6+2\left(\sqrt{3}-1\right)}-\sqrt{6-2\left(\sqrt{3}+1\right)}\)
\(=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\left(\sqrt{3}+1\right)-\left(\sqrt{3}-1\right)=2\)
1 a/ Trục căn thức ở mẫu
\(VT=\frac{-\sqrt{1}+\sqrt{2}}{2-1}+\frac{-\sqrt{2}+\sqrt{3}}{3-2}+...+\frac{-\sqrt{47}+\sqrt{48}}{48-47}\)\(=-\sqrt{1}+\sqrt{2}-\sqrt{2}+\sqrt{3}-....-\sqrt{47}+\sqrt{48}=\sqrt{48}-1>3=VP\)
b/
\(2\left(10+3\sqrt{11}\right)=11+2.\sqrt{11}.3+9=\left(\sqrt{11}+3\right)^2\)
\(VT=\left(\sqrt{11}-3\right)\sqrt{2}\sqrt{10+3\sqrt{11}}=\left(\sqrt{11}-3\right)\left(\sqrt{11}+3\right)=11-9=2=VP\)
2/
\(B=\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\sqrt{2\left(5+\sqrt{3}.\sqrt{7}\right)}\)
\(2\left(5+\sqrt{21}\right)=7+2\sqrt{7}.\sqrt{3}+3=\left(\sqrt{7}+\sqrt{3}\right)^2\)
\(B=\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{7}+\sqrt{3}\right)=\left(5+\sqrt{21}\right).4\)
\(=20+4\sqrt{21}\)
A chắc không rút gọn được.
\(A=\dfrac{\sqrt{10+2\sqrt{21}}}{\sqrt{2}}+\dfrac{\sqrt{10-2\sqrt{21}}}{\sqrt{2}}-\dfrac{2}{\sqrt{2}}\sqrt{8-2\sqrt{7}}\)
\(A=\dfrac{\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}}{\sqrt{2}}+\dfrac{\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}}{\sqrt{2}}-\dfrac{2}{\sqrt{2}}\sqrt{\left(\sqrt{7}-1\right)^2}\)
\(A=\dfrac{1}{\sqrt{2}}\left(\sqrt{7}+\sqrt{3}+\sqrt{7}-\sqrt{3}-2\sqrt{7}+2\right)=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
\(B=\dfrac{\sqrt[3]{2}\left(\sqrt[3]{2}+1+\sqrt[3]{2^2}\right)}{\sqrt[3]{4}+\sqrt[3]{2}+1}=\dfrac{\sqrt[3]{2}\left(\sqrt[3]{4}+\sqrt[3]{2}+1\right)}{\sqrt[3]{4}+\sqrt[3]{2}+1}=\sqrt[3]{2}\)