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1: \(=\dfrac{\sqrt{8+2\sqrt{7}}+\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}+1+\sqrt{7}-1}{\sqrt{2}}=\dfrac{2\sqrt{7}}{\sqrt{2}}=\sqrt{14}\)
3: \(=\sqrt{6+2\sqrt{2\cdot\sqrt{3-\sqrt{3}-1}}}\)
\(=\sqrt{6+2\sqrt{2\cdot\sqrt{2-\sqrt{3}}}}\)
\(=\sqrt{6+2\sqrt{\sqrt{2}\left(\sqrt{3}-1\right)}}\)
\(=\sqrt{6+2\sqrt{\sqrt{6}-\sqrt{2}}}\)
1: \(=\dfrac{\sqrt{8+2\sqrt{7}}+\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}+1+\sqrt{7}-1}{\sqrt{2}}=\dfrac{2\sqrt{7}}{\sqrt{2}}=\sqrt{14}\)
3: \(=\sqrt{6+2\sqrt{2\cdot\sqrt{3-\sqrt{3}-1}}}\)
\(=\sqrt{6+2\sqrt{2\cdot\sqrt{2-\sqrt{3}}}}\)
\(=\sqrt{6+2\sqrt{\sqrt{2}\left(\sqrt{3}-1\right)}}\)
\(=\sqrt{6+2\sqrt{\sqrt{6}-\sqrt{2}}}\)
a.\(\left(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\right).\left(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\right)\)
\(=\left(\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}\right).\left(\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\right)\)
\(=\left(\sqrt{3}+1-\sqrt{3}+1\right)\left(\sqrt{3}-1+\sqrt{3}+1\right)\)
\(=2.2\sqrt{3}=4\sqrt{3}\)
b.\(\left(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\right)^2=\left[\frac{\sqrt{8+2\sqrt{7}}}{\sqrt{2}}-\frac{\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\right]^2\)
\(=\left(\frac{\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}-\frac{\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}\right)^2\)
\(=\left(\frac{\sqrt{7}+1-\sqrt{7}+1}{\sqrt{2}}\right)^2=\left(\sqrt{2}\right)^2=2\)
c.\(\sqrt{5-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{5-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{5-\sqrt{3-\left(2\sqrt{5}-3\right)}}=\sqrt{5-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{5-\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{5-\sqrt{5}+1}=\sqrt{6-\sqrt{5}}\)
\(a\sqrt{b}-b\sqrt{a}=\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\)
\(7\sqrt{7}+3\sqrt{3}=\left(\sqrt{7}+\sqrt{3}\right)\left(7-\sqrt{21}+3\right)=\left(\sqrt{7}+\sqrt{3}\right)\left(10-\sqrt{21}\right)\)
\(a\sqrt{a}-b\sqrt{b}=\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)\)
\(1-a\sqrt{a}=\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)\)
\(x^2-\sqrt{x}=\sqrt{x}\left(x\sqrt{x}-1\right)=\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\)
\(\left(\sqrt{2}+1\right)^2-4\sqrt{2}=\left(\sqrt{2}-1\right)^2\)
\(\left(\sqrt{5}+2\right)^2-8\sqrt{5}=\left(\sqrt{5}-2\right)^2\)
2 cái trên đều áp dụng HĐT \(\left(a+b\right)^2-4ab=\left(a-b\right)^2\)
\(5\sqrt{2}-2\sqrt{5}=\sqrt{10}\left(\sqrt{5}-\sqrt{2}\right)\)
\(\frac{A}{\sqrt{2}}=\frac{1+\sqrt{7}}{2+\sqrt{8+2\sqrt{7}}}+\frac{1-\sqrt{7}}{2-\sqrt{8-2\sqrt{7}}}\)
\(=\frac{1+\sqrt{7}}{2+1+\sqrt{7}}+\frac{1-\sqrt{7}}{2-\sqrt{7}+1}\)
\(=\frac{1+\sqrt{7}}{3+\sqrt{7}}+\frac{1-\sqrt{7}}{3-\sqrt{7}}\)
=\(\frac{\left(1+\sqrt{7}\right)\left(3-\sqrt{7}\right)+\left(1-\sqrt{7}\right)\left(3+\sqrt{7}\right)}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}\)
\(=\frac{-8}{2}=-4\)
\(\Rightarrow A=-4\sqrt{2}\)
\(B=\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}+2\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\sqrt{3}+2+2-\sqrt{3}\)
\(=4\)
Còn cách nữa là bình phương
Đag làm thì ấn nhầm trả lời .V
Cách bình phương đây
\(B=\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)
\(\Rightarrow B^2=7+4\sqrt{3}+2\sqrt{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}+7-4\sqrt{3}\)
\(=14+2\sqrt{49-48}\)
\(=14+2\)
\(=16\)
\(\Rightarrow B=\sqrt{16}=4\)
3.
\(•x=3+\sqrt{2}\\ x^2=\left(3+\sqrt{2}\right)^2\\ x^2=9+2.3.\sqrt{2}+2\\ x^2=11+6\sqrt{2}\\• y=\sqrt{11+6\sqrt{2}}\\ y^2=\left(\sqrt{11+6\sqrt{2}}\right)^2\\ y^2=11+6\sqrt{2}\)
\(\Rightarrow x^2=y^2=11+6\sqrt{2}\)
1. ta có : \(4\sqrt{7}=\sqrt{112}\)
\(3\sqrt{3}=\sqrt{27}\)
ta thấy : \(\sqrt{112}>\sqrt{27}\) hay \(4\sqrt{7}>3\sqrt{3}\)
2. \(\dfrac{1}{4}\sqrt{82}=\sqrt{\dfrac{41}{8}}\)
\(6\sqrt{\dfrac{1}{7}}=\sqrt{\dfrac{36}{7}}\)
ta thấy :\(\sqrt{\dfrac{41}{8}}< \sqrt{\dfrac{36}{7}}\) hay \(\dfrac{1}{4}\sqrt{82}< 6\sqrt{\dfrac{1}{7}}\)
3. \(x^2=\left(3+\sqrt{2}\right)^2\)
\(y^2=11+6\sqrt{2}\)=\(\left(3+\sqrt{2}\right)^2\)
ta thấy : \(x^2=y^2\Rightarrow x=y\)
\(S^3=\left(\sqrt[3]{7+4\sqrt{3}+}\sqrt[3]{7-4\sqrt{3}}\right)^3\)
= \(7+4\sqrt{3}+7-4\sqrt{3}+3.\sqrt{7+4\sqrt{3}}.\sqrt{7-4\sqrt{3}}.\left(a+b\right)\)
= 14+\(3.\sqrt{49-48}.S\)
= 14+3S
=> S3-3S=14+3S-3S=14
\(P=S^3-3S\)
\(P=\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)^3-3\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\)
\(P=7+4\sqrt{3}+3\left(\sqrt[3]{7+4\sqrt{3}}\right)^2.\sqrt[3]{7-4\sqrt{3}}+3.\sqrt[3]{7+4\sqrt{3}}\left(\sqrt[3]{7-4\sqrt{3}}\right)^2+7-4\sqrt{3}\text{}\text{}-3\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\)
\(P=14+3\sqrt[3]{7+4\sqrt{3}}.\sqrt[3]{7-4\sqrt{3}}\text{}\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\text{}\text{}-3\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\)
\(P=14+3\sqrt[3]{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\text{}\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\text{}\text{}-3\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\)
\(P=14+3\sqrt[3]{49-48}\text{}\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\text{}\text{}-3\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\)
\(P=14+3\text{}\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\text{}\text{}-3\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\)
\(P=14\)