a, \(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{2}\sqrt{20}+\sqrt{5}\)
\(=\sqrt{5}+\dfrac{1}{2}.2\sqrt{5}+\sqrt{5}\)
\(=3\sqrt{5}\)
b, \(\sqrt{\dfrac{1}{2}}+\sqrt{4,5}+\sqrt{12,5}\)
\(=\sqrt{0,5}+3\sqrt{0,5}+5\sqrt{0,5}=9\sqrt{0,5}\)
c, \(\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}\)
\(=2\sqrt{5}-3\sqrt{5}+3\sqrt{18}+2\sqrt{18}\)
\(=-\sqrt{5}+5\sqrt{18}\)
d, \(0,1.\sqrt{200}+2\sqrt{0,08}+0,4\sqrt{50}\)
\(=\sqrt{0,01.200}+0,2.\sqrt{2}+0,4.5\sqrt{2}\)
\(=\sqrt{2}+0,2\sqrt{2}+2\sqrt{2}=3,2\sqrt{2}\)
Chúc bạn học tốt!!!
\(\frac{5}{\sqrt{10}}=\frac{5\sqrt{10}}{10}=\frac{\sqrt{10}}{2}\)
\(\frac{5}{2\sqrt{5}}=\frac{10\sqrt{5}}{20}=\frac{\sqrt{5}}{2}\)
\(\frac{1}{3\sqrt{20}}=\frac{3\sqrt{20}}{180}=\frac{\sqrt{20}}{60}=\frac{2\sqrt{5}}{60}=\frac{\sqrt{5}}{30}\)
\(\frac{2\sqrt{2}+2}{5\sqrt{2}}=\frac{10\sqrt{2}\left(\sqrt{2}+1\right)}{50}=\frac{20+10\sqrt{2}}{50}=\frac{10\left(2+\sqrt{2}\right)}{50}=\frac{2+\sqrt{2}}{5}\)
\(\frac{y+b\sqrt{y}}{b\sqrt{y}}=\frac{y\left(\sqrt{y}+b\right)}{by}=\frac{\sqrt{y}+b}{b}\)
+ Ta có:
5√10=5.√10√10.√10=5√10(√10)2=5√1010510=5.1010.10=510(10)2=51010
=5.√105.2=5.105.2=√102=102.
+ Ta có:
52√5=5.√52√5.√5=
LG a
12√48−2√75−√33√11+5√1131248−275−3311+5113;
Phương pháp giải:
+ Cách đổi hỗn số ra phân số: abc=a.c+bcabc=a.c+bc.
+ Sử dụng quy tắc đưa thừa số ra ngoài dấu căn:
√A2.B=A√BA2.B=AB, nếu A≥0, B≥0A≥0, B≥0.
√A2.B=−A√BA2.B=−AB, nếu A<0, B≥0A<0, B≥0.
+ √ab=√a√bab=ab, với a≥0, b>0a≥0, b>0.
+ √a.√b=√aba.b=ab, với a, b≥0a, b≥0.
+ A√B=A√BBAB=ABB, với B>0B>0.
Lời giải chi tiết:
Ta có:
12√48−2√75−√33√11+5√1131248−275−3311+5113
=12√16.3−2√25.3−√3.11√11+5√1.3+13=1216.3−225.3−3.1111+51.3+13
=12√42.3−2√52.3−√3.√11√11+5√43=1242.3−252.3−3.1111+543
=12.4√3−2.5√3−√3+5√4√3=12.43−2.53−3+543
=42√3−10√3−√3+5√4.√3√3.√3=423−103−3+54.33.3
=2√3−10√3−√3+52√33=23−103−3+5233
=2√3−10√3−√3+10√33=23−103−3+1033
=(2−10−1+103)√3=(2−10−1+103)3
=−173√3=−1733.
LG b
√150+√1,6.√60+4,5.√223−√6;150+1,6.60+4,5.223−6;
Phương pháp giải:
+ Cách đổi hỗn số ra phân số: abc=a.c+bcabc=a.c+bc.
+ Sử dụng quy tắc đưa thừa số ra ngoài dấu căn:
√A2.B=A√BA2.B=AB, nếu A≥0, B≥0A≥0, B≥0.
√A2.B=−A√BA2.B=−AB, nếu A<0, B≥0A<0, B≥0.
+ √ab=√a√bab=ab, với a≥0, b>0a≥0, b>0.
+ √a.√b=√aba.b=ab, với a, b≥0a, b≥0.
+ A√B=A√BBAB=ABB, với B>0B>0.
Lời giải chi tiết:
Ta có:
√150+√1,6.√60+4,5.√223−√6150+1,6.60+4,5.223−6
=√25.6+√1,6.60+4,5.√2.3+23−√6=25.6+1,6.60+4,5.2.3+23−6
=√52.6+√1,6.(6.10)+4,5√83−√6=52.6+1,6.(6.10)+4,583−6
=5√6+√(1,6.10).6+4,5√8√3−√6=56+(1,6.10).6+4,583−6
=5√6+√16.6+4,5√8.√33−√6=56+16.6+4,58.33−6
=5√6+√42.6+4,5√8.33−√6=56+42.6+4,58.33−6
=5√6+4√6+4,5.√4.2.33−√6=56+46+4,5.4.2.33−6
=5√6+4√6+4,5.√22.63−√6=56+46+4,5.22.63−6
=5√6+4√6+4,5.2√63−√6=56+46+4,5.263−6
=5√6+4√6+9√63−√6=56+46+963−6
=5√6+4√6+3√6−√6=56+46+36−6
=(5+4+3−1)√6=11√6.=(5+4+3−1)6=116.
Cách 2: Ta biến đổi từng hạng tử rồi thay vào biểu thức ban đầu:
+ √150=√25.6=5√6150=25.6=56
+ √1,6.60=√1,6.(10.6)=√(1,6.10).6=√16.61,6.60=1,6.(10.6)=(1,6.10).6=16.6
=4√6=46
+ 4,5.√223=4,5.√2.3+23=4,5.√83=4,5√8.334,5.223=4,5.2.3+23=4,5.83=4,58.33
=4,5.√4.2.33=4,5.2.√63=9.√63=3√6.=4,5.4.2.33=4,5.2.63=9.63=36.
Do đó:
√150+√1,6.√60+4,5.√223−√6150+1,6.60+4,5.223−6
=5√6+4√6+3√6−√6=56+46+36−6
=(5+4+3−1)√6=11√6=(5+4+3−1)6=116
LG c
(√28−2√3+√7)√7+√84;(28−23+7)7+84;
Phương pháp giải:
+ Cách đổi hỗn số ra phân số: abc=a.c+bcabc=a.c+bc.
+ Hằng đẳng thức số 1: (a+b)2=a2+2ab+b2(a+b)2=a2+2ab+b2.
+ Sử dụng quy tắc đưa thừa số ra ngoài dấu căn:
√A2.B=A√BA2.B=AB, nếu A≥0, B≥0A≥0, B≥0.
√A2.B=−A√BA2.B=−AB, nếu A<0, B≥0A<0, B≥0.
+ √ab=√a√bab=ab, với a≥0, b>0a≥0, b>0.
+ √a.√b=√aba.b=ab, với a, b≥0a, b≥0.
+ A√B=A√BBAB=ABB, với B>0B>0.
Lời giải chi tiết:
Ta có:
=(√28−2√3+√7)√7+√84=(28−23+7)7+84
=(√4.7−2√3+√7)√7+√4.21=(4.7−23+7)7+4.21
=(√22.7−2√3+√7)√7+√22.21=(22.7−23+7)7+22.21
=(2√7−2√3+√7)√7+2√21=(27−23+7)7+221
=2√7.√7−2√3.√7+√7.√7+2√21=27.7−23.7+7.7+221
=2.(√7)2−2√3.7+(√7)2+2√21=2.(7)2−23.7+(7)2+221
=2.7−2√21+7+2√21=2.7−221+7+221
=14−2√21+7+2√21=14−221+7+221
=14+7=21=14+7=21.
LG d
(√6+√5)2−√120.(6+5)2−120.
Phương pháp giải:
+ Cách đổi hỗn số ra phân số: abc=a.c+bcabc=a.c+bc.
+ Hằng đẳng thức số 1: (a+b)2=a2+2ab+b2(a+b)2=a2+2ab+b2.
+ Sử dụng quy tắc đưa thừa số ra ngoài dấu căn:
√A2.B=A√BA2.B=AB, nếu A≥0, B≥0A≥0, B≥0.
√A2.B=−A√BA2.B=−AB, nếu A<0, B≥0A<0, B≥0.
+ √a.√b=√aba.b=ab, với a, b≥0a, b≥0.
Lời giải chi tiết:
Ta có:
(√6+√5)2−√120(6+5)2−120
=(√6)2+2.√6.√5+(√5)2−√4.30=(6)2+2.6.5+(5)2−4.30
=6+2√6.5+5−2√30=6+26.5+5−230
=6+2√30+5−2√30=6+5=11.=6+230+5−230=6+5=11.
a: \(=10\sqrt{2}-4\sqrt{2}+6\sqrt{2}=12\sqrt{2}\)
b: \(=5\sqrt{7}-4\sqrt{7}+3\sqrt{7}=4\sqrt{7}\)
c: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{6}=\dfrac{1}{6}\sqrt{6}\)
d: \(=8\sqrt{5}-15\sqrt{5}+15\sqrt{5}-3\sqrt{5}=5\sqrt{5}\)
e: \(=\sqrt{5}+\dfrac{2}{5}\sqrt{5}+\sqrt{5}=2.4\sqrt{5}\)
f: \(=\dfrac{1}{5}\sqrt{5}+\dfrac{3}{2}\sqrt{2}+\dfrac{5}{2}\sqrt{2}=\dfrac{1}{5}\sqrt{5}+4\sqrt{2}\)
a) (√8 - 3√2 + √10)√2 - √5
= (√22.2 - 3√2 + √5.2)√2 - √5
= (2√2 - 3√2 + √5.√2)√2 - √5
= (2 - 3 + √5)√2.√2 - √5
= (-1 + √5).2 - √5
= -2 + 2√5 - √5
= -2 + √5
b) 0,2√((-10)2.3) + 2√(√3 - √52)
= 0,2.10√3 + 2|√3 - √5|
= 2√3 + 2(√5 - √3)
= 0,2.10.√3 + 2|√3 - √5|
= 2√3 + 2(√5 - √3)
= 2√3 + 2√5 - 2√3
= 2√5
\(\dfrac{2+\sqrt{2}}{1+\sqrt{2}}=\dfrac{\left(2+\sqrt{2}\right)\left(\sqrt{2}-1\right)}{2-1}=2\sqrt{2}-2+2-\sqrt{2}=\sqrt{2}\)
\(\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}=-\sqrt{5}\)
\(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}=\dfrac{\sqrt{6}}{2}\)
\(\dfrac{a-\sqrt{a}}{1-\sqrt{a}}=\dfrac{\left(a-\sqrt{a}\right)\left(1+\sqrt{a}\right)}{1-a}=\dfrac{a+a\sqrt{a}-\sqrt{a}-a}{1-a}=\dfrac{\sqrt{a}\left(a-1\right)}{1-a}=-\sqrt{a}\)
\(\dfrac{p-2\sqrt{p}}{\sqrt{p}-2}=\dfrac{\sqrt{p}\left(\sqrt{p}-2\right)}{\sqrt{p}-2}=\sqrt{p}\)
\(a,\left(\sqrt{8}-3.\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}\)
\(=\sqrt{8}.\sqrt{2}-3\sqrt{2}.\sqrt{2}+\sqrt{10}.\sqrt{2}-\sqrt{5}\)
\(=\sqrt{16}-3.2+\sqrt{20}-\sqrt{5}\)
\(=\sqrt{4^2}-6+\sqrt{2^2.5}-\sqrt{5}\)
\(=2-6+2\sqrt{5}-\sqrt{5}\)
\(=-2+\sqrt{5}\)
\(b,\)
\(0,2\sqrt{\left(-10^2\right).3}+2\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}\)
\(=0,2.\left|-10\right|.\sqrt{3}+2\left|\sqrt{3}-\sqrt{5}\right|\)
\(=0,2.10.\sqrt{3}+2\left(\sqrt{5}-\sqrt{3}\right)\)
\(=2\sqrt{3}+2\sqrt{5}-2\sqrt{3}\)
\(=2\sqrt{5}\)
a) xy−y√x+√x−1xy−yx+x−1
=y⋅√x⋅√x−y√x+√x−1=y⋅x⋅x−yx+x−1
=y√x(√x−1)+(√x−1)=yx(x−1)+(x−1)
=(√x−1)(y√x+1)=(x−1)(yx+1).
b) √ax−√by+√bx−√ayax−by+bx−ay
=(√
TRẢ LỜI :
\(=\sqrt{5}+\sqrt{5}+\sqrt{5}=3\sqrt{5}\)
c) √20 - √45 + 3√18 + √72
= √4.5 - √9.5 + 3√9.2 + √36.2
= 2√5 - 3√5 + 9√2 + 6√2
= -√5 + 15√2
a) 3√5 b) 9√2 / 2
c) -√5 + 15√2 d)
3,4√2
a) 3 \sqrt{5}35.
b) \dfrac{9}{\sqrt{2}}29 hay \dfrac{9 \sqrt{2}}{2}292.
c) 15 \sqrt{2}-\sqrt{5}152−5.
d) 3,4 \cdot \sqrt{2}3,4⋅2.
a) \(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{2}\sqrt{20}+\sqrt{5}=\sqrt{5}+\sqrt{5}+\sqrt[]{5}=3\sqrt{5}\)\(\sqrt{\dfrac{1}{2}}+\sqrt{4,5}+\sqrt{12,5}=\sqrt{0,5}+\sqrt{4,5}+\sqrt{12,5}=\sqrt{5}\left(\sqrt{0,1}+\sqrt{0,9}+\sqrt{2,5}\right)=\dfrac{1}{\sqrt{2}}\left(1+3+5\right)=\dfrac{9}{\sqrt{2}}\)
b)
a) 3\(\sqrt{5}\) b) \(\dfrac{9}{2}\)\(\sqrt{2}\) c)\(\sqrt{5}\) + 15\(\sqrt{2}\) d)\(\dfrac{17.\sqrt{2}}{5}\)
a) 3\(\sqrt{5}\)
b)\(\dfrac{9}{\sqrt{2}}\) ; \(\dfrac{9\sqrt{2}}{2}\)
c) 15\(\sqrt{2}\) - \(\sqrt{5}\)
d) \(\dfrac{17\sqrt{2}}{5}\)
c) \(-\sqrt{5}+15\sqrt{2}\)
d)\(\dfrac{17\sqrt{2}}{5}\)
a) \(3\sqrt{5}\)
b)\(\dfrac{9}{2}\sqrt{2}\)
a) 3\(\sqrt{5}\)
b) 9 \(\sqrt{\dfrac{1}{2}}\)
c) -\(\sqrt{5}\) +3\(\sqrt{2}\) (3 + \(\sqrt{3}\))
d) \(\sqrt{2}\) + 2\(\sqrt{0,08}\) + 4
a) 3\(\sqrt{5}\)
b) \(\dfrac{9\sqrt{2}}{2}\)
c)15\(\sqrt{2}\)\(-\sqrt{5}\)
a)\(\sqrt{5^2.\dfrac{1}{5}}+\sqrt{\dfrac{1}{2^2}.20}+\sqrt{5}=\sqrt{5}+\sqrt{5}+\sqrt{5}=3\sqrt{5}\)\(\sqrt{\dfrac{1}{2}}+\sqrt{\dfrac{9}{2}}+\sqrt{\dfrac{25}{2}}=\sqrt{\dfrac{1}{2}}+\sqrt{9.\dfrac{1}{2}}+\sqrt{25.\dfrac{1}{2}}=\dfrac{1}{2}+3\sqrt{\dfrac{1}{2}}+5\sqrt{\dfrac{1}{2}}=9\sqrt{\dfrac{1}{2}}=9\dfrac{\sqrt{1}}{\sqrt{2}}=\dfrac{9}{2}\sqrt{2}\)
a)\(3\sqrt{5}\)
b)\(\dfrac{9}{\sqrt{2}};\dfrac{9\sqrt{2}}{2}\)
c)\(15\sqrt{2}-\sqrt{5}\)
d)\(3,4.\sqrt{2}\)
a) 3\(\sqrt{5}\)
b)\(\dfrac{9\sqrt{2}}{2}\)
c)\(-\sqrt{5}+15\sqrt{2}\)
d)\(\dfrac{17\sqrt{2}}{5}\)
a. 3\(\sqrt{5}\)
b. \(\dfrac{9}{\sqrt{2}}\)
c. \(15\sqrt{2}-\sqrt{5}\)
d. \(3,4.\sqrt{2}\)
a/ 3\(\sqrt{5}\)
b/ 9\(\sqrt{0,5}\)
c/15\(\sqrt{2}\) - \(\sqrt{5}\)
d/ 3,4\(\sqrt{2}\)
a) 5√15+12√20+√5515+1220+5 ; b) √12+√4,5+√12,512+4,5+12,5 ;
c) √20−√45+3√18+√7220−45+318+72 ; d) 0,1.√200+2⋅√0,08+0,4⋅√500,1.200+2⋅0,08+0,4⋅50.
Gợi ý trả lời
a) 3√535.
b) 9<...
a, 3\(\sqrt{ }\)5
b. \(\dfrac{9\sqrt{2}}{2}\)
c. \(-\sqrt{5}+15\sqrt{2}\)
d. \(\dfrac{17\sqrt{2}}{5}\)
a) 5\(\sqrt{\dfrac{1}{5}}\)+\(\dfrac{1}{2}\sqrt{20}\)+\(\sqrt{5}\)
=\(\sqrt{25}\)\(\sqrt{\dfrac{1}{5}}+\)\(\dfrac{1}{2}\)\(\sqrt{2^2\cdot5}+\)\(\sqrt{5}\)
=\(\sqrt{25\cdot\dfrac{1}{5}}\)+\(\dfrac{1}{2}\cdot2\cdot\sqrt{5}+\sqrt{5}\)
=\(\sqrt{5}+\sqrt{5}+\sqrt{5}\)
=\(3\sqrt{5}\)
b)\(\sqrt{\dfrac{1}{2}}+\sqrt{4.5}+\sqrt{12.5}\)
=\(\sqrt{\dfrac{1}{2}}+\sqrt{\dfrac{9}{2}}+\sqrt{\dfrac{25}{2}}\)
=\(\sqrt{\dfrac{1}{2}}+\sqrt{3^2\cdot\dfrac{1}{2}}+\sqrt{5^2\cdot\dfrac{1}{2}}\)
=\(\sqrt{\dfrac{1}{2}}+3\sqrt{\dfrac{1}{2}}+5\sqrt{\dfrac{1}{2}}\)
=9\(\sqrt{\dfrac{1}{2}}\)
c)\(\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}\)
=\(\sqrt{2^2\cdot5}-\sqrt{3^2\cdot5}+3\sqrt{3^2\cdot2}+\sqrt{6^2\cdot2}\)
=\(\left(2\sqrt{5}-3\sqrt{5}\right)+\left(9\sqrt{2}+6\sqrt{2}\right)\)
=\(-\sqrt{5}+15\sqrt{2}\)
d)\(0.1\sqrt{200}+2\sqrt{0.08}+0.4\sqrt{50}\)
=\(0.1\sqrt{10^2\cdot2}+2\sqrt{\dfrac{1}{5^2}\cdot2}+0.4\sqrt{5^2\cdot2}\)
=\(0.1\cdot10\sqrt{2}+2\cdot\dfrac{1}{5}\sqrt{2}+0.4\cdot5\sqrt{2}\)
=\(\sqrt{2}+\dfrac{2}{5}\sqrt{2}+2\sqrt{2}\)
=\(\dfrac{17}{5}\sqrt{2}\) =\(\dfrac{17\sqrt{2}}{5}\)
a,5\(\sqrt{\dfrac{1}{5}}\)+\(\dfrac{1}{2}\)\(\sqrt{20}\)+\(\sqrt{5}\)=\(\sqrt{5}+\sqrt{5}+\sqrt{5}\)=3\(\sqrt{5}\)
b,\(\sqrt{\dfrac{1}{2}}+\sqrt{4,5}+\sqrt{12,5}=\dfrac{\sqrt{2}}{2}+\dfrac{3\sqrt{2}}{2}+\dfrac{5\sqrt{2}}{2}=\dfrac{\sqrt{2}+3\sqrt{2}+5\sqrt{2}}{2}=\dfrac{9\sqrt{2}}{2}\)
c,\(\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}=\left(2\sqrt{5}-3\sqrt{5}\right)+\left(9\sqrt{2}+6\sqrt{2}\right)=-\sqrt{5}+15\sqrt{2}\)
d,\(0,1\cdot\sqrt{200}+2\cdot\sqrt{0,08}+0,4\cdot\sqrt{50}=0,1\cdot10\sqrt{2}+2\cdot\dfrac{\sqrt{2}}{5}+0,4\cdot5\sqrt{2}=\sqrt{2}+\dfrac{2\sqrt{2}}{5}+2\sqrt{2}=\dfrac{17\sqrt{2}}{5}\)
a) \(\sqrt{5}+\sqrt{5}+\sqrt{5}\) = 3\(\sqrt{5}\)
b) \(\sqrt{\dfrac{1}{2}}+3\sqrt{\dfrac{1}{2}}+5\sqrt{\dfrac{1}{2}}=9\sqrt{\dfrac{1}{2}}\)
c) \(2\sqrt{5}-3\sqrt{5}+9\sqrt{2}+6\sqrt{2}=-\sqrt{5}+15\sqrt{2}\)
d)\(\sqrt{2}+2\sqrt{\dfrac{8}{100}}+2\sqrt{2}=3\sqrt{2}+\dfrac{2\sqrt{2}}{5}=\dfrac{17\sqrt{2}}{5}\)
a)5\(\sqrt{\dfrac{1}{5}}\)+\(\dfrac{1}{2}\)\(\sqrt{20}\)+\(\sqrt{5}\)
=\(\sqrt{\dfrac{1}{5}.5^2}\)+\(\dfrac{1}{2}\).2.\(\sqrt{5}\)+\(\sqrt{5}\)
=\(\sqrt{5}\)+\(\sqrt{5}\)+\(\sqrt{5}\)
=3\(\sqrt{5}\)
b)\(\sqrt{\dfrac{1}{2}}\)+\(\sqrt{4,5}\)+\(\sqrt{12,5}\)
=\(\sqrt{\dfrac{1}{2}}\)+3\(\sqrt{\dfrac{1}{2}}\)+5\(\sqrt{\dfrac{1}{2}}\)
=9\(\sqrt{\dfrac{1}{2}}\)=\(\dfrac{9\sqrt{2}}{2}\)
c)\(\sqrt{20}\)-\(\sqrt{45}\)+3\(\sqrt{18}\)+\(\sqrt{72}\)
=2\(\sqrt{5}\)-3\(\sqrt{5}\)+3.3\(\sqrt{2}\)+6\(\sqrt{2}\)
=15\(\sqrt{2}\)-\(\sqrt{5}\)
d)0,1\(\sqrt{200}\)+2\(\sqrt{0,08}\)+0,4\(\sqrt{50}\)
=0,1.10\(\sqrt{2}\)+2.0,2\(\sqrt{2}\)+0,4.5\(\sqrt{2}\)
=(1+0,4+2)\(\sqrt{2}\)
=3,4\(\sqrt{2}\)
a) 5√\(\dfrac{1}{5}\) + \(\dfrac{1}{2}\)√20 + √5
=√5 + \(\dfrac{1}{2}\) . 4√5 + √5
=√5 + 2√5 + √5
= 3√5
a,3 căn 5
b,(9 căn 2) : 2
c,- căn 5 + 15 căn 2
d, ( 17 căn 2 ) :5
a,\(3\sqrt{5}\) b,\(-5+15\sqrt{2}\) c,\(\dfrac{9\sqrt{2}}{2}\) d,\(\dfrac{17\sqrt{2}}{5}\)