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Ta có : \(94-42\sqrt{5}=45-2.7.3\sqrt{5}+49=\left(3\sqrt{5}\right)^2-2.7.3\sqrt{5}+7^2=\left(7-3\sqrt{5}\right)^2\)
\(94+42\sqrt{5}=\left(7+3\sqrt{5}\right)^2\)
\(\Rightarrow\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}\)
\(=\sqrt{\left(7-3\sqrt{5}\right)^2}-\sqrt{\left(7+3\sqrt{5}\right)^2}=7-3\sqrt{5}-7-3\sqrt{5}=-6\sqrt{5}\)
\(=\sqrt{7^2-2.7.3\sqrt{5}+\left(3\sqrt{5}\right)^2}+\sqrt{7^2+2.7.3\sqrt{5}+\left(3\sqrt{5}\right)^2}\)
\(=\sqrt{\left(7-3\sqrt{5}\right)^2}+\sqrt{\left(7+3\sqrt{5}\right)^2}\)
\(=7-3\sqrt{5}+7+3\sqrt{5}=14\)
\(a,\sqrt{33+20\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{8+2.2\sqrt{2}.5+25}-\sqrt{2-2.\sqrt{2}.3+9}\)
\(=\sqrt{\left[2\sqrt{2}+5\right]^2}-\sqrt{\left[\sqrt{2}-3\right]^2}\)
\(=2\sqrt{2}+5-\left(3-\sqrt{2}\right)\)
\(=2+\sqrt{2}\)
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)
\(\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}\\ =\sqrt{\left(7-\sqrt{45}\right)^2}-\sqrt{\left(7+\sqrt{45}\right)^2}\\ =\left|7-\sqrt{45}\right|-\left|7+\sqrt{45}\right|\\ =7-\sqrt{45}-7-\sqrt{45}\\ =-2\sqrt{45}\)
\(\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}=\sqrt{49-2.7.3\sqrt{5}+45}-\sqrt{49+2.7.3\sqrt{5}+45}=7-3\sqrt{5}-7-3\sqrt{5}=-6\sqrt{5}\)
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\(\sqrt{5}\)
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) \(\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}=\sqrt{\left(7-3\sqrt{5}\right)^2}-\sqrt{\left(7+3\sqrt{5}\right)^2}\)
\(=7-3\sqrt{5}-\left(7+3\sqrt{5}\right)=-6\sqrt{5}\)
Dễ thấy \(A>0\)
\(A^2=94-42\sqrt{5}+94+42\sqrt{5}+2\sqrt{\left(94^2-42^2.5\right)}\)
\(A^2=188+2\sqrt{16}=196\)
\(\Rightarrow A=14\)