\(=\sqrt{7^2-2.7.3\sqrt{5}+\left(3\sqrt{5}\right)^2}+\sqrt{7^2+2.7.3\sqrt{5}+\left(3\sqrt{5}\right)^2}\)

\(=\sqrt{\left(7-3\sqrt{5}\right)^2}+\sqrt{\left(7+3\sqrt{5}\right)^2}\)

\(=7-3\sqrt{5}+7+3\sqrt{5}=14\)

chịu thôi

Đặt: \(P=\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}\) 

\(P^2=\left(\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}\right)^2\)

\(P^2=94-42\sqrt{5}-2\sqrt{94-42\sqrt{5}}.\sqrt{94+42\sqrt{5}}+94+42\sqrt{5}\) 

\(P^2=188-2\sqrt{\left(94-42\sqrt{5}\right)\left(94+42\sqrt{5}\right)}\) 

\(P^2=188-2\sqrt{94^2+3948\sqrt{5}-3948\sqrt{5}-8820}\) 

\(P^2=188-2\sqrt{8836-8820}\) 

\(P^2=188-2\sqrt{16}\) 

\(P^2=188-8\) 

\(P^2=180\) 

\(P=\orbr{\begin{cases}6\sqrt{5}\\-6\sqrt{5}\end{cases}}\)  .

Mà theo bài ra: \(\sqrt{94-42\sqrt{5}}< \sqrt{94+42\sqrt{5}}\)

\(\Rightarrow\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}< 0\)

\(\Rightarrow P=-6\sqrt{5}\)
 

Làm gì phức tạp thế

94 - 42\(\sqrt{5}\)= 49 - 2×7×3×\(\sqrt{5}\)+ 45 = (7 - \(3\sqrt{5}\))²

Tương tự 94 + 42\(\sqrt{5}\) = (7 + \(3\sqrt{5}\))²

Từ đó suy ra đáp số là 6\(\sqrt{5}\)

Đáp số là -6\(\sqrt{5}\)

\(\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}=\sqrt{\left(7-3\sqrt{5}\right)^2}-\sqrt{\left(7+3\sqrt{5}\right)^2}\)

\(=7-3\sqrt{5}-7-3\sqrt{5}=-6\sqrt{5}\)

=0 nha bạn

\(a,\sqrt{33+20\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)

\(=\sqrt{8+2.2\sqrt{2}.5+25}-\sqrt{2-2.\sqrt{2}.3+9}\)

\(=\sqrt{\left[2\sqrt{2}+5\right]^2}-\sqrt{\left[\sqrt{2}-3\right]^2}\)

\(=2\sqrt{2}+5-\left(3-\sqrt{2}\right)\)

\(=2+\sqrt{2}\)

chúc bn học tốt

a) \(\sqrt{\left(2\sqrt{2}+5\right)^2}\) \(-\) \(\sqrt{\left(3-\sqrt{2}\right)^2}\)= \(|2\sqrt{2}+5|\)\(-\)\(|3-\sqrt{2}|\)

\(=\)\(2\sqrt{2}+5-3+\sqrt{2}=2+3\sqrt{2}\)

b)\(\sqrt{\left(7-3\sqrt{5}\right)^2}-\sqrt{\left(7+3\sqrt{5}\right)^2}=7-3\sqrt{5}-7-3\sqrt{5}=-6\sqrt{5}\)

a/ \(=\sqrt{\left(5+2\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)

\(=5+2\sqrt{2}-3+\sqrt{2}=2+3\sqrt{2}\)

b/ \(=\sqrt{\left(7-3\sqrt{5}\right)^2}-\sqrt{\left(7+3\sqrt{5}\right)}=7-3\sqrt{5}-7-3\sqrt{5}=-6\sqrt{5}\)

a)

\(\sqrt{33+20\sqrt{2}}-\sqrt{11-6\sqrt{2}}\\ =\sqrt{33+2\sqrt{200}}-\sqrt{11-2\sqrt{18}}\\ =\sqrt{\sqrt{8^2}+2\sqrt{8}\sqrt{25}+5^2}-\sqrt{\sqrt{2^2}-2\sqrt{2}\sqrt{9}+3^2}\\ =\sqrt{\left(\sqrt{8}+5\right)^2}-\sqrt{\left(\sqrt{2}-3\right)^2}\\ =\left|\sqrt{8}+5\right|-\left|\sqrt{2}-3\right|\\ =\sqrt{8}+5-3+\sqrt{2}\\ =3\sqrt{2}+2\)

b)

\(\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}\\ =\sqrt{\left(7-\sqrt{45}\right)^2}-\sqrt{\left(7+\sqrt{45}\right)^2}\\ =\left|7-\sqrt{45}\right|-\left|7+\sqrt{45}\right|\\ =7-\sqrt{45}-7-\sqrt{45}\\ =-2\sqrt{45}\)

a)  \(\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(\sqrt{5}+2\right)^2}\)

\(=\left(\sqrt{5}-2\right)-\left(\sqrt{5}+2\right)=-4\)

b)   \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}=\frac{1}{\sqrt{2}}.\left(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\right)\)

\(=\frac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\right)\)

\(=\frac{1}{\sqrt{2}}\left(\sqrt{7}-1-\sqrt{7}-1\right)=-\sqrt{2}\)

c)  \(\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}=\sqrt{\left(7-3\sqrt{5}\right)^2}-\sqrt{\left(7+3\sqrt{5}\right)^2}\)

\(=7-3\sqrt{5}-\left(7+3\sqrt{5}\right)=-6\sqrt{5}\)

\(a.\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}=\sqrt{49-2.7.3\sqrt{5}+45}-\sqrt{49+2.7.3\sqrt{5}+45}=7-3\sqrt{5}-7-3\sqrt{5}=-6\sqrt{5}\) \(b.\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=\dfrac{\sqrt{7+2\sqrt{7}+1}-\sqrt{7-2\sqrt{7}+1}}{\sqrt{2}}=\dfrac{\sqrt{7}+1-\sqrt{7}+1}{\sqrt{2}}=\sqrt{2}\) \(c.\sqrt{5-\sqrt{13+\sqrt{48}}}=\sqrt{5-\sqrt{12+2.2\sqrt{3}+1}}=\sqrt{3-2\sqrt{3}+1}=\sqrt{3}-1\)

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