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

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

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

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

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

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) đặt \(A=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)

nhân cả hai vế với \(\sqrt{2}\), ta được:

\(\sqrt{2}A=\sqrt{2}\sqrt{4-\sqrt{7}}-\sqrt{2}\sqrt{4+\sqrt{7}}\)

\(=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)

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

\(=\left|1-\sqrt{7}\right|-\left|1+\sqrt{7}\right|\)

\(=\sqrt{7}-1-\sqrt{7}-1\)

\(=-2\)

\(\Rightarrow A=-\frac{2}{\sqrt{2}}=-\sqrt{2}\)

a) \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)

\(=\frac{\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}}{\sqrt{2}}\)

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

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

\(=\frac{-2}{\sqrt{2}}=-\sqrt{2}\)

wwreftr

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

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

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)

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

1) ĐKXĐ: \(x\notin\left\{0;1\right\}\)

2) Ta có: \(A=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\left(1-\dfrac{3-\sqrt{x}}{\sqrt{x}+1}\right)\)

\(=\dfrac{x+\sqrt{x}+1-\left(x-\sqrt{x}+1\right)}{\sqrt{x}}:\dfrac{\sqrt{x}+1-3+\sqrt{x}}{\sqrt{x}+1}\)

\(=2\cdot\dfrac{\sqrt{x}+1}{2\sqrt{x}-2}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

Bài 1: 

Ta có: \(D=\sqrt{16x^4}-2x^2+1\)

\(=4x^2-2x^2+1\)

\(=2x^2+1\)