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Đương làm thì lại nhấn hủy TvT
Bài 1.
a) \(\sqrt{\left(4-3\sqrt{2}\right)^2}\)
\(=\left|4-3\sqrt{2}\right|\)
\(=-\left(4-3\sqrt{2}\right)=3\sqrt{2}-4\)( vì \(3\sqrt{2}>4\))
b) \(\sqrt{\left(\sqrt{3-1}\right)^2}+\sqrt{\left(\sqrt{3-2}\right)^2}\)
\(=\sqrt{\left(\sqrt{2}\right)^2}+\sqrt{1^2}\)
\(=\left|\sqrt{2}\right|+\left|1\right|\)
\(=\sqrt{2}+1=1+\sqrt{2}\)
Bài 2.
Sửa VP = \(\left(\sqrt{5}+2\right)^2\)
VT = \(5+4\sqrt{5}+4=\left(\sqrt{5}\right)^2+2\cdot2\cdot\sqrt{5}+2^2=\left(\sqrt{5}+2\right)^2\)= VP ( đpcm )
Còn ý b) em chưa làm được :((
Không hiểu sao cứ gửi ảnh nó lại bị lộn xộn nên bạn cố nhìn nhé
( ͡°( ͡° ͜ʖ( ͡° ͜ʖ ͡°)ʖ ͡°) ͡°)
d/ \(x=\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)
\(\Leftrightarrow x^3=3+\sqrt{9+\frac{125}{27}}+3-\sqrt{9+\frac{125}{27}}-3\left(\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\right)\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}.\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)
\(\Leftrightarrow x^3=6-3x\sqrt[3]{9-9-\frac{125}{27}}\)
\(\Leftrightarrow x^3=6-5x\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+6\right)=0\)
\(\Leftrightarrow x=1\)
c/
\(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{12}+4}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{4+2\sqrt{3}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
\(=3-1=2\)
Câu 2:
a: \(=2+\sqrt{17-4\left(\sqrt{5}+2\right)}\)
\(=2+\sqrt{17-4\sqrt{5}-8}\)
\(=2+\sqrt{9-4\sqrt{5}}\)
\(=2+\sqrt{5}-2=\sqrt{5}\)
b: \(=\sqrt{2}+1+1-\sqrt{2}=2\)
c: \(=\sqrt{6-2\sqrt{5}}\cdot\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\)
\(=\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
\(=18+6\sqrt{5}-6\sqrt{5}-10=8\)
a) \(H=\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}=\sqrt{\left(9-\sqrt{17}\right)\left(9+\sqrt{17}\right)}\)
\(=\sqrt{81-17}=\sqrt{64}=8\)
b) \(K=\left(\sqrt{20}-3\sqrt{5}+\sqrt{80}\right).\sqrt{5}\)
\(=\sqrt{20}.\sqrt{5}-3\sqrt{5}.\sqrt{5}+\sqrt{80}.\sqrt{5}\)
\(=\sqrt{100}-3.5+\sqrt{400}=\sqrt{10^2}-15+\sqrt{20^2}\)
\(=10-15+20=15\)
\(H=\sqrt{9-\sqrt{17}}\cdot\sqrt{9+\sqrt{17}}\)
\(=\sqrt{\left(9-\sqrt{17}\right)\left(9+\sqrt{17}\right)}\)
\(=\sqrt{9^2-\left(\sqrt{17}\right)^2}\)
\(=\sqrt{81-17}\)
\(=\sqrt{64}=8\)
\(K=\left(\sqrt{20}-3\sqrt{5}+\sqrt{80}\right)\cdot\sqrt{5}\)
\(=\sqrt{20}\cdot\sqrt{5}-3\sqrt{5}\cdot\sqrt{5}+\sqrt{80}\cdot\sqrt{5}\)
\(=\sqrt{20\cdot5}-3\sqrt{5\cdot5}+\sqrt{80\cdot5}\)
\(=\sqrt{100}-3\sqrt{25}+\sqrt{400}\)
\(=10-3\cdot5+20\)
\(=15\)