\(\dfrac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\dfrac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)

\(=\dfrac{\sqrt{\left(\sqrt{2}\right)^2-2.\sqrt{2}.1+1^2}}{\sqrt{3^2-2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}}-\dfrac{\sqrt{\left(\sqrt{2}\right)^2+2.\sqrt{2}.1+1^2}}{\sqrt{3^2+2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}}\)

\(=\dfrac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(3-2\sqrt{2}\right)^2}}-\dfrac{\sqrt{\left(\sqrt{2}+1\right)^2}}{\sqrt{\left(3+2\sqrt{2}\right)^2}}=\dfrac{\sqrt{2}-1}{3-2\sqrt{2}}-\dfrac{\sqrt{2}+1}{3+2\sqrt{2}}\)

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

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

\(\sqrt{17+12\sqrt{2}}-\sqrt{17-12\sqrt{2}}\)

\(=\sqrt{3^2+2\cdot3\cdot2\sqrt{2}+\left(2\sqrt{2}\right)^2}-\sqrt{3^2-2\cdot3\cdot2\sqrt{2}+\left(2\sqrt{2}\right)^2}\)

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

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

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

\(=4\sqrt{2}\)

\(\sqrt{17+12\sqrt{2}}-\sqrt{17-12\sqrt{2}}\)

\(=\sqrt{3^2+2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}-\sqrt{3^2-2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}\)

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

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

\(=3+2\sqrt{2}-3+2\sqrt{2}=4\sqrt[]{2}\)

bdfbzdtvbeay           q4etwrtc3t5wtư

êrtechcgrgdcgtgư

nhân cả tử và mẫu với \(\sqrt{2}\)nha

\(\frac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)

\(=\frac{\sqrt{2-2.\sqrt{2}.1+1}}{\sqrt{17-3.2.2.\sqrt{2}}}-\)\(\frac{\sqrt{2+2.\sqrt{2}.1+1}}{\sqrt{17+3.2.2.\sqrt{2}}}\)

\(=\frac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{17-3.2.\sqrt{4}.\sqrt{2}}}\)\(-\frac{\sqrt{\left(\sqrt{2}+1\right)^2}}{\sqrt{17+3.2.\sqrt{4}.\sqrt{2}}}\)

\(=\frac{\sqrt{2}-1}{\sqrt{8-2.\sqrt{8}.3+9}}\)\(-\frac{\sqrt{2}+1}{\sqrt{8+2.\sqrt{8}.3+9}}\)

\(=\frac{\sqrt{2}-1}{\sqrt{\left(\sqrt{8}-3\right)^2}}\)\(-\frac{\sqrt{2}+1}{\sqrt{\left(\sqrt{8}+3\right)^2}}\)

\(=\frac{\sqrt{2}-1}{\sqrt{8}-3}\)\(-\frac{\sqrt{2}+1}{\sqrt{8}+3}\)

\(=\frac{\left(\sqrt{2}-1\right)\left(\sqrt{8}+3\right)-\left(\sqrt{2}+1\right)\left(\sqrt{8}-3\right)}{\left(\sqrt{8}-3\right)\left(\sqrt{8}+3\right)}\)

\(=\frac{\sqrt{16}+3\sqrt{2}-\sqrt{8}-3-\sqrt{16}+3\sqrt{2}-\sqrt{8}+3}{\left(\sqrt{8}-3\right)\left(\sqrt{8}+3\right)}\)

\(=\frac{6\sqrt{2}-2\sqrt{8}}{\left(\sqrt{8}-3\right)\left(\sqrt{8}+3\right)}\)

\(=\frac{6\sqrt{2}-2.2.\sqrt{2}}{\left(2\sqrt{2}-3\right)\left(2\sqrt{2}+3\right)}\)

\(=\frac{2\sqrt{2}}{\left(8-9\right)}=\frac{2\sqrt{2}}{-1}=-2\sqrt{2}\)

= - 0,3288755607 nha  Hà Phạm Như Ý ! ! !

K VÀ KB NHA ! ! !

bạn trúc tính sao ra kết quả vậy

Bạn tính sao vậy, nếu bạn đưa kết quả ko thì mình cũng tính dc

\(\dfrac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\dfrac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\) = \(\dfrac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(3-2\sqrt{2}\right)^2}}-\dfrac{\sqrt{\left(\sqrt{2}+1\right)^2}}{\sqrt{\left(3+2\sqrt{2}\right)^2}}\)

= \(\dfrac{\sqrt{2}-1}{3-2\sqrt{2}}-\dfrac{\sqrt{2}+1}{3+2\sqrt{2}}\) = \(\dfrac{\left(\sqrt{2}-1\right)\left(3+2\sqrt{2}\right)-\left(\sqrt{2}+1\right)\left(3-2\sqrt{2}\right)}{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}\)

= \(\dfrac{3\sqrt{2}+4-3-2\sqrt{2}-\left(3\sqrt{2}-4+3-2\sqrt{2}\right)}{9-8}\)

= \(\dfrac{3\sqrt{2}+4-3-2\sqrt{2}-3\sqrt{2}+4-3+2\sqrt{2}}{1}\)

= \(2\)

\(\sqrt{17-12\sqrt{2}}+\sqrt{17+12\sqrt{2}}\)

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

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

\(=6\)

Lời giải:

a. \(=|\sqrt{7}-5|+|2-\sqrt{7}|=5-\sqrt{7}+(\sqrt{7}-2)=3\)

b. \(=\sqrt{(3+\sqrt{2})^2}-\sqrt{(3-\sqrt{2})^2}=|3+\sqrt{2}|-|3-\sqrt{2}|\)

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

c.

\(=\sqrt{(3+2\sqrt{2})^2}+\sqrt{(3-2\sqrt{2})^2}=|3+2\sqrt{2}|+|3-2\sqrt{2}|\)

$=(3+2\sqrt{2})+(3-2\sqrt{2})=6$

d.

$=\sqrt{(\sqrt{5}+1)^2}-\sqrt{(\sqrt{5}-1)^2}$

$=|\sqrt{5}+1|-|\sqrt{5}-1|=\sqrt{5}+1-(\sqrt{5}-1)=2$

\(\frac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}=\frac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(3-2\sqrt{2}\right)^2}}-\frac{\sqrt{\left(\sqrt{2}+1\right)^2}}{\sqrt{\left(3+2\sqrt{2}\right)^2}}=\frac{\sqrt{2}-1}{3-2\sqrt{2}}-\frac{\sqrt{2}+1}{3+2\sqrt{2}}=\frac{\left(3+2\sqrt{2}\right)\left(\sqrt{2}-1\right)-\left(3-2\sqrt{2}\right)\left(\sqrt{2}+1\right)}{\left(3-2\sqrt{2}\right)\left(3+2\sqrt{2}\right)}=\frac{\sqrt{2}+1-\left(\sqrt{2}-1\right)}{9-8}=2\)