Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, \(\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}\)
= \(2\sqrt{3}-10\sqrt{3}-\dfrac{\sqrt{3}\cdot\sqrt{11}}{\sqrt{11}}+5\sqrt{\dfrac{4}{3}}\)
= \(2\sqrt{3}-10\sqrt{3}-\sqrt{3}+5\sqrt{\dfrac{12}{3^2}}\)
= \(2\sqrt{3}-10\sqrt{3}-\sqrt{3}+5\dfrac{2\sqrt{3}}{3}\)
= \(2\sqrt{3}-10\sqrt{3}-\sqrt{3}+\dfrac{10\sqrt{3}}{3}\)
= \(-9\sqrt{3}+\dfrac{10\sqrt{3}}{3}=\dfrac{-27\sqrt{3}}{3}+\dfrac{10\sqrt{3}}{3}=\dfrac{-17\sqrt{3}}{3}\)
b, \(\sqrt{150}+\sqrt{1,6}\cdot\sqrt{60}+4.5\sqrt{2\dfrac{2}{3}}-\sqrt{6}\)
= \(5\sqrt{6}+\dfrac{2\sqrt{10}}{5}\cdot2\sqrt{15}+4,5\sqrt{\dfrac{8}{3}}-\sqrt{6}\)
= \(5\sqrt{6}+4\sqrt{6}+4,5\sqrt{\dfrac{24}{3^2}}-\sqrt{6}\)
= \(5\sqrt{6}+4\sqrt{6}+4,5\cdot\dfrac{2\sqrt{6}}{3}-\sqrt{6}\)
= \(5\sqrt{6}+4\sqrt{6}+3\sqrt{6}-\sqrt{6}=11\sqrt{6}\)
c, \(\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\cdot\sqrt{7}+\sqrt{84}\)
= \(\left(2\sqrt{7}-2\sqrt{3}+\sqrt{7}\right)\cdot\sqrt{7}+2\sqrt{21}\)
= \(\left(3\sqrt{7}-2\sqrt{3}\right)\cdot\sqrt{7}+2\sqrt{21}\)
= \(21-2\sqrt{21}+2\sqrt{21}=21\)
d, \(\left(\sqrt{6}+\sqrt{5}\right)^2-\sqrt{120}\)
= \(6+2\sqrt{30}+5-2\sqrt{30}=11\)
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
\(\frac{5}{\sqrt{10}}=\frac{5\sqrt{10}}{10}=\frac{\sqrt{10}}{2}\)
\(\frac{5}{2\sqrt{5}}=\frac{10\sqrt{5}}{20}=\frac{\sqrt{5}}{2}\)
\(\frac{1}{3\sqrt{20}}=\frac{3\sqrt{20}}{180}=\frac{\sqrt{20}}{60}=\frac{2\sqrt{5}}{60}=\frac{\sqrt{5}}{30}\)
\(\frac{2\sqrt{2}+2}{5\sqrt{2}}=\frac{10\sqrt{2}\left(\sqrt{2}+1\right)}{50}=\frac{20+10\sqrt{2}}{50}=\frac{10\left(2+\sqrt{2}\right)}{50}=\frac{2+\sqrt{2}}{5}\)
\(\frac{y+b\sqrt{y}}{b\sqrt{y}}=\frac{y\left(\sqrt{y}+b\right)}{by}=\frac{\sqrt{y}+b}{b}\)
+ 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=
\(\dfrac{2+\sqrt{2}}{1+\sqrt{2}}=\dfrac{\left(2+\sqrt{2}\right)\left(\sqrt{2}-1\right)}{2-1}=2\sqrt{2}-2+2-\sqrt{2}=\sqrt{2}\)
\(\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}=-\sqrt{5}\)
\(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}=\dfrac{\sqrt{6}}{2}\)
\(\dfrac{a-\sqrt{a}}{1-\sqrt{a}}=\dfrac{\left(a-\sqrt{a}\right)\left(1+\sqrt{a}\right)}{1-a}=\dfrac{a+a\sqrt{a}-\sqrt{a}-a}{1-a}=\dfrac{\sqrt{a}\left(a-1\right)}{1-a}=-\sqrt{a}\)
\(\dfrac{p-2\sqrt{p}}{\sqrt{p}-2}=\dfrac{\sqrt{p}\left(\sqrt{p}-2\right)}{\sqrt{p}-2}=\sqrt{p}\)
a) a) Biến đổi vế trái thành 32√6+23√6−42√6326+236−426 và làm tiếp.
b) Biến đổi vế trái thành (√6x+13√6x+√6x):√6x(6x+136x+6x):6x và làm tiếp
+ Ta có:
2√6−√5=2(√6+√5)(√6−√5)(√6+√5)26−5=2(6+5)(6−5)(6+5)
=2(√6+√5)(√6)2−(√5)2=2(√6+√5)6−5=2(6+5)(6)2−(5)2=2(6+5)6−5
=2(√6+√5)1=2(√6+√5)=2(6+5)1=2(6+5).
+ Ta có:
3√10+√7=3(√10−√7)(√10+√7)(√10−√7)310+7=3(10−7)(10+7)(10−7)
=3(√10−√7)(√10)2−(√7)2=3(10−7)(10)2−(7)2=3(√10−√7)10−7=3(10−7)10−7
=3(√10−√7)3=√10−√7=3(10−7)3=10−7.
+ Ta có:
1√x−√y=1.(√x+√y)(√x−√y)(√x+√y)1x−y=1.(x+y)(x−y)(x+y)
=√x+√y(√x)2−(√y)2=√x+√yx−y=x+y(x)2−(y)2=x+yx−y
+ Ta có:
2ab√a−√b=2ab(√a+√b)(√a−√b)(√a+√b)2aba−b=2ab(a+b)(a−b)(a+b)
=2ab(√a+√b)(√a)2−(√b)2=2ab(√a+√b)a−b=2ab(a+b)(a)2−(b)2=2ab(a+b)a−b.
\(\frac{2}{\sqrt{6}-\sqrt{5}}=\frac{2\left(\sqrt{6}+\sqrt{5}\right)}{\left(\sqrt{6}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{5}\right)}=\frac{2\left(\sqrt{6}+\sqrt{5}\right)}{6-5}=2\left(\sqrt{6}+\sqrt{5}\right)\)
\(\frac{3}{\sqrt{10}+\sqrt{7}}=\frac{3\left(\sqrt{10}-\sqrt{7}\right)}{\left(\sqrt{10}-\sqrt{7}\right)\left(\sqrt{10}+\sqrt{7}\right)}=\frac{3\left(\sqrt{10}-\sqrt{7}\right)}{10-7}=\sqrt{10}-\sqrt{7}\)
\(\frac{1}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{x}+\sqrt{y}}{x-y}\)
\(\frac{2ab}{\sqrt{a}-\sqrt{b}}=\frac{2ab\left(\sqrt{a}+\sqrt{b}\right)}{a-b}\)
\(1.A=\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}=\dfrac{1}{2}\sqrt{\dfrac{1}{3}.144}-2\sqrt{\dfrac{1}{3}.225}-\sqrt{\dfrac{1}{3}.9}+5\sqrt{\dfrac{4}{3}}=6\sqrt{\dfrac{1}{3}}-30\sqrt{\dfrac{1}{3}}-3\sqrt{\dfrac{1}{3}}+10\sqrt{\dfrac{1}{3}}=-17\sqrt{\dfrac{1}{3}}\) \(2.B=\left(2\sqrt{27}-3\sqrt{48}+3\sqrt{75}-\sqrt{192}\right)\left(1-\sqrt{3}\right)=\left(6\sqrt{3}-12\sqrt{3}+15\sqrt{3}-8\sqrt{3}\right)\left(1-\sqrt{3}\right)=\sqrt{3}\left(1-\sqrt{3}\right)=\sqrt{3}-3\) \(3.C=\left(2\sqrt{7}-2\sqrt{6}\right).\sqrt{6}-\sqrt{168}=2\sqrt{42}-12-2\sqrt{42}=-12\) \(4.D=\left(\sqrt{28}-2\sqrt{8}+\sqrt{7}\right).\sqrt{7}+4\sqrt{14}=\left(3\sqrt{7}-4\sqrt{2}\right)\sqrt{7}=21-4\sqrt{14}+4\sqrt{14}=21\)
1) Ta có: \(\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}\)
\(=\dfrac{1}{2}\cdot4\sqrt{3}-2\cdot5\sqrt{3}-\sqrt{3}+5\cdot\sqrt{\dfrac{4}{3}}\)
\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}+\dfrac{10}{\sqrt{3}}\)
\(=\dfrac{-27+10}{\sqrt{3}}\)
\(=\dfrac{-17\sqrt{3}}{3}\)
b) Ta có: \(\dfrac{\sqrt{2}-1}{\sqrt{2}+2}-\dfrac{1}{\sqrt{2}+1}+\dfrac{\sqrt{2}+1}{\sqrt{2}}\)
\(=\dfrac{\sqrt{2}-1-\sqrt{2}+3+2\sqrt{2}}{\sqrt{2}\left(\sqrt{2}+1\right)}\)
\(=\dfrac{2+2\sqrt{2}}{2+2\sqrt{2}}=1\)
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.
-17√3/3 b) 11√6
c) 21 d) 11 C4:
a)
) 12⋅4√3−2.5√3−√3+5⋅23⋅√3=2√3−10√3−√3+103√312⋅43−2.53−3+5⋅23⋅3=23−103−3+1033
=−17√33=−1733.
b) 11√6116.
c) 2121.
d) 1111.
a) \(\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}=2\sqrt{3}-10\sqrt{3}-\sqrt{3}+\dfrac{10}{3}\sqrt{3}=\sqrt{3}\left(2-10-1+\dfrac{10}{3}\right)=\)\(\dfrac{-17}{\sqrt{3}}\)
b) \(\sqrt{150}+\sqrt{1,6}.\sqrt{60}+4,5.\sqrt{2\dfrac{2}{3}}-\sqrt{6}=5\sqrt{6}+4\sqrt{6}+3\sqrt{6}-\sqrt{6}=11\sqrt{6}\)
c) \(\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}=14-2\sqrt{21}+7+2\sqrt{21}=21\)
d) \(\left(\sqrt{6}+\sqrt{5}\right)^2-\sqrt{120}=6+2\sqrt{30}+5-2\sqrt{30}=11\)
c)
a)\(\dfrac{-17\sqrt{3}}{3}\)
b)\(11\sqrt{6}\)
c)21
d)11
a) \(-\dfrac{17\sqrt{3}}{3}\)
b) 11\(\sqrt{6}\)
c) 21
d) 11
a) \(\dfrac{17}{3}\)\(\sqrt{3}\)
b) 11\(\sqrt{6}\)
c) 21
d) 11
a)\(\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}\)
=\(\dfrac{1}{2}\sqrt{4^2\cdot3}-2\sqrt{5^2\cdot3}-\sqrt{\dfrac{33}{11}}+5\sqrt{\dfrac{4}{3}}\)
=\(\dfrac{1}{2}\cdot4\sqrt{3}-2\cdot5\sqrt{3}-\sqrt{3}+5\dfrac{2\sqrt{3}}{3}\)
=\(2\sqrt{3}-10\sqrt{3}-\sqrt{3}+\dfrac{10}{3}\sqrt{3}\)
=\(\left(2-10-1+\dfrac{10}{3}\right)\sqrt{3}\)
=\(\dfrac{-17\sqrt{3}}{3}\)
b)\(\sqrt{150}+\sqrt{1.6}\cdot\sqrt{60}+4.5\cdot\sqrt{2\dfrac{2}{3}}-\sqrt{6}\)
=\(\sqrt{5^2\cdot6}+\sqrt{\left(\dfrac{2}{5}\right)^2\cdot10}\cdot\sqrt{2^2\cdot15}+\dfrac{9}{2}\sqrt{\dfrac{8}{3}}-\sqrt{6}\)
a, (-17 căn 3):3
b,11 căn 6
c, 21
d, 11
a) \(\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}\)
=\(\dfrac{1}{2}4\sqrt{3}-2.5\sqrt{3}-\sqrt{\dfrac{33}{11}}+5\sqrt{\dfrac{4}{3}}\)
=2\(\sqrt{3}-10\sqrt{3}-\sqrt{3}+\dfrac{10}{3}\sqrt{3}\)
=-\(\dfrac{17}{3}\sqrt{3}\)
b)\(\sqrt{150}+\sqrt{1,6}.\sqrt{60}+4,5.\sqrt{2\dfrac{2}{3}}-\sqrt{6}\)
=5\(\sqrt{6}+4\sqrt{6}+4,5.2.\dfrac{\sqrt{6}}{3}-\sqrt{6}\)
=11\(\sqrt{6}\)
c) (\(\sqrt{28}-2\sqrt{3}+\sqrt{7})\sqrt{7}+\sqrt{84}\)
=(2\(\sqrt{7}-2\sqrt{3}+\sqrt{7})\sqrt{7}+2\sqrt{21}\)
=2.7+2\(\sqrt{21}+7+2\sqrt{21}\)
=21
d)\((\sqrt{6}+\sqrt{5})^2-\sqrt{120}\)
=6+2\(\sqrt{30}+5-2\sqrt{30}\)
=11
a. = \(\dfrac{1}{2}\sqrt{16\times3}\) - 2\(\times\)5\(\sqrt{3}\) - \(\sqrt{3}\) - \(\dfrac{10}{3}\sqrt{3}\)
= \(9\sqrt{3}\) + \(\dfrac{10}{3}\sqrt{3}\)
= \(\dfrac{-17}{3}\sqrt{3}\)
a) \dfrac{1}{2} \cdot 4 \sqrt{3}-2.5 \sqrt{3}-\sqrt{3}+5 \cdot \dfrac{2}{3} \cdot \sqrt{3}=2 \sqrt{3}-10 \sqrt{3}-\sqrt{3}+\dfrac{10}{3} \sqrt{3}21⋅43−2.53−3+5⋅32⋅3=23−103−3+3103
=\dfrac{-17 \sqrt{3}}{3}=3−173.
b) 11 \sqrt{6}116.
c) 2121.
d) 1111.
bài 5
\(a)\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{\dfrac{4}{3}}\\ =2\sqrt{3}-10\sqrt{3}-\sqrt{3}+\dfrac{10\sqrt{3}}{3}\\ =\sqrt{3}\left(2-10-1+\dfrac{10}{3}\right)\\ =\left(-\dfrac{17}{3}\right)\sqrt{3}\) \(c)\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}\\ =\left(2\sqrt{7}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}\\ =\left(3\sqrt{7}-2\sqrt{3}\right)\sqrt{7}+\sqrt{84}\\ =3.7-2\sqrt{21}+2\sqrt{21}\\ =21\)
\(b)\\ \sqrt{150}+\sqrt{1,6}.\sqrt{60}+4,5\sqrt{2^2_3}-\sqrt{6}\\ =\sqrt{25.6}+\sqrt{1,6.60}+\dfrac{4}{5}\sqrt{\dfrac{8}{3}}-\sqrt{6}\\ =5\sqrt{6}+\sqrt{16.6}4,5.\dfrac{1}{3}\sqrt{3^2\dfrac{\left(4.2\right)}{3}}-\sqrt{6}\\ =9\sqrt{6}+3\sqrt{6}-\sqrt{6}\\ =11\sqrt{6}\) \(d)\left(\sqrt{6}+\sqrt{5}\right)^2-\sqrt{120}\\ =\left(\sqrt{6}\right)^2+2\sqrt{6}\sqrt{5}+\left(\sqrt{5}\right)^2-\sqrt{4.30}\\ =6+2\sqrt{30}+5-2\sqrt{30}\\ =11\)
= 9√6 + 3√6 - √6 = 11√6
c) (√28 - 2√3 + √7)√7 + √84
= (√4.7 - 2√3 + √7)√7 + √4.21
= (2√7 - 2√3 + √7)√7 + 2√21
= (3√7 - 2√3)√7 + 2√21
= 3.7 - 2√21 + 2√21 = 21
a)\(\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}\)=\(2\sqrt{3}-10\sqrt{3}-\sqrt{3}+\dfrac{10\sqrt{3}}{3}\)
=\(-\dfrac{17\sqrt{3}}{3}\)
b)\(\sqrt{150}+\sqrt{1,6}.\sqrt{60}+4,5.\sqrt{2\dfrac{2}{3}}-\sqrt{6}\) =\(5\sqrt{6}+4\sqrt{6}+3\sqrt{6}-\sqrt{6}\)
=\(11\sqrt{6}\)
c)\(\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}\) =\(14-2\sqrt{21}+7+\sqrt{84}\)
=21
d)\(\left(\sqrt{6}+\sqrt{5}\right)^2-\sqrt{120}\) =\(11+2\sqrt{30}-2\sqrt{30}=11\)
\(a)\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}\)
=\(\dfrac{1}{2}\sqrt{16.3}-2\sqrt{25.3}-\sqrt{3}-\dfrac{10}{3}\sqrt{3}\)
=\(2\sqrt{3}-10\sqrt{3}-\sqrt{3}-\dfrac{10}{3}\sqrt{3}\)
=\(-9\sqrt{3}+\dfrac{10}{3}\sqrt{3}=(-9+\dfrac{10}{3})\sqrt{3}=(\dfrac{-27}{3}+\dfrac{10}{3})\sqrt{3}=\dfrac{-17}{3}\sqrt{3}\)
\(b)\sqrt{150}+\sqrt{1,6}.\sqrt{60}+4,5.\sqrt{2\dfrac{2}{3}}-\sqrt{6}\)
=\(\sqrt{25.6}+\sqrt{1,6.60}+4,5\sqrt{\dfrac{8}{3}}-\sqrt{6}\)
=\(5\sqrt{6}+\sqrt{16.6}+4,5.\dfrac{1}{3}\sqrt{3^2.\dfrac{4.2}{3}}-\sqrt{6}\)
=\(5\sqrt{6}+4\sqrt{6}+4,5.\dfrac{1}{3}\sqrt{3^2.\dfrac{4.2}{3}}-\sqrt{6}\)
=\(9\sqrt{6}+3\sqrt{6}-\sqrt{6}=11\sqrt{6}\)
\(c)(\sqrt{28}-2\sqrt{3}+\sqrt{7})\sqrt{7}+\sqrt{84}=(\sqrt{4.7}-2\sqrt{3}+\sqrt{7})\sqrt{7}+\sqrt{4.21}\)
=\(\left(2\sqrt{7}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+2\sqrt{21}=\left(3\sqrt{7}-2\sqrt{3}\right)\sqrt{7}+2\sqrt{21}\)
=\(3.7-2\sqrt{21}+2\sqrt{21}=21\)
\(d)\left(\sqrt{6}+\sqrt{5}\right)^2-\sqrt{120}=6+2\sqrt{6}\sqrt{5}+5-\sqrt{4.30}\)
=\(6+2\sqrt{30}+5-2\sqrt{30}=11\)
a,\(\dfrac{-17\sqrt{3}}{3}\)
b,11\(\sqrt{6}\)
c,21
d,11
=\(11\sqrt{6}\)
c)21