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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.
\(1.A=\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}=\dfrac{1}{2}\sqrt{\dfrac{1}{3}.144}-2\sqrt{\dfrac{1}{3}.225}-\sqrt{\dfrac{1}{3}.9}+5\sqrt{\dfrac{4}{3}}=6\sqrt{\dfrac{1}{3}}-30\sqrt{\dfrac{1}{3}}-3\sqrt{\dfrac{1}{3}}+10\sqrt{\dfrac{1}{3}}=-17\sqrt{\dfrac{1}{3}}\) \(2.B=\left(2\sqrt{27}-3\sqrt{48}+3\sqrt{75}-\sqrt{192}\right)\left(1-\sqrt{3}\right)=\left(6\sqrt{3}-12\sqrt{3}+15\sqrt{3}-8\sqrt{3}\right)\left(1-\sqrt{3}\right)=\sqrt{3}\left(1-\sqrt{3}\right)=\sqrt{3}-3\) \(3.C=\left(2\sqrt{7}-2\sqrt{6}\right).\sqrt{6}-\sqrt{168}=2\sqrt{42}-12-2\sqrt{42}=-12\) \(4.D=\left(\sqrt{28}-2\sqrt{8}+\sqrt{7}\right).\sqrt{7}+4\sqrt{14}=\left(3\sqrt{7}-4\sqrt{2}\right)\sqrt{7}=21-4\sqrt{14}+4\sqrt{14}=21\)
i: =-12*căn 3/2căn 3=-6
h: =72căn 2/12căn 2=6
g: =25căn 12/5căn 6=5căn 2
f: =(15:5)*căn 6:3=3căn 2
d: =-1/2*6*căn 10=-3căn 10
a) \(\dfrac{2\sqrt{125}-3\sqrt{5}-\sqrt{180}}{-\sqrt{5}}+\sqrt{8}=\dfrac{2\sqrt{25.5}-3\sqrt{5}-\sqrt{36.5}}{-\sqrt{5}}+\sqrt{8}\)
\(=\dfrac{10\sqrt{5}-3\sqrt{5}-6\sqrt{5}}{-\sqrt{5}}+2\sqrt{2}=\dfrac{\sqrt{5}}{-\sqrt{5}}+2\sqrt{2}=2\sqrt{2}-1\)
b) \(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}+\sqrt{18}=\left|\sqrt{2}-\sqrt{3}\right|+\sqrt{9.2}\)
\(=\sqrt{3}-\sqrt{2}+3\sqrt{2}=2\sqrt{2}+\sqrt{3}\)
c) \(\sqrt{48}-6\sqrt{\dfrac{1}{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}=\sqrt{16.3}-2\sqrt{9.\dfrac{1}{3}}+\dfrac{\sqrt{3}\left(1-\sqrt{3}\right)}{\sqrt{3}}\)
\(=4\sqrt{3}-2\sqrt{3}+1-\sqrt{3}=1+\sqrt{3}\)
d) \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right):\dfrac{1}{\sqrt{5}-\sqrt{2}}=\left(\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}-\sqrt{5}\right).\left(\sqrt{5}-\sqrt{2}\right)\)
\(=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)=-\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)=-3\)
a) \(\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}\)
= \(2\sqrt{5}-3\sqrt{5}+9\sqrt{2}+6\sqrt{2}\)
= \(-\sqrt{5}+15\sqrt{2}\)
b) \(\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}\)
= \(\left(2\sqrt{7}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\)
= \(2.7-2\sqrt{21}+7+2\sqrt{21}=14+7=21\)
c) \(\left(\sqrt{6}+\sqrt{5}\right)^2-\sqrt{120}\)
= \(6+2\sqrt{6}.\sqrt{5}+5-2\sqrt{30}\)
= \(11+2\sqrt{30}-2\sqrt{30}=11\)
d) \(\left(\dfrac{1}{2}-\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}\sqrt{2}+\dfrac{4}{5}\sqrt{200}\right):\dfrac{1}{8}\)
= \(\left(\dfrac{1}{2}-\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}\sqrt{2}+8\sqrt{2}\right).8\)
= \(4-4\sqrt{2}-12\sqrt{2}+64\sqrt{2}=4+48\sqrt{2}\)
Bài này dễ ẹc ( đâu có khó đâu :)) )
a) \(\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}\)
\(=\sqrt{2^2.5}-\sqrt{3^2.5}+3\sqrt{3^2.2}+\sqrt{6^2.2}\)
\(=2\sqrt{5}-3\sqrt{5}+9\sqrt{2}+6\sqrt{2}\)
\(=\left(2-3\right)\sqrt{5}+\left(9+6\right)\sqrt{2}\)
\(=15\sqrt{2}-\sqrt{5}\)
b) \(\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}\)
\(=\sqrt{2^2.7}.\sqrt{7}-2\sqrt{3}.\sqrt{7}+\sqrt{7}.\sqrt{7}+\sqrt{2^2.21}\)
\(=2.7-2\sqrt{21}+7+2\sqrt{21}\)
\(=14+7+\left(2-2\right)\sqrt{21}=21\)
c) \(\left(\sqrt{6}+\sqrt{5}\right)^2-\sqrt{120}\)
\(=6+2\sqrt{30}+5-\sqrt{2^2.30}\)
\(=6+5+2\sqrt{30}-2\sqrt{30}=11\)
d) \(\left(\dfrac{1}{2}\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}\sqrt{2}+\dfrac{4}{5}\sqrt{200}\right):\dfrac{1}{8}\)
\(=\left(\dfrac{1}{2}\sqrt{\dfrac{2}{2^2}}-\dfrac{3}{2}\sqrt{2}+\dfrac{4}{5}\sqrt{10^2.2}\right):\dfrac{1}{8}\)
\(=\left(\dfrac{1}{4}\sqrt{2}-\dfrac{3}{2}\sqrt{2}+8\sqrt{2}\right).8\)
\(=2\sqrt{2}-12\sqrt{2}+64\sqrt{2}=54\sqrt{2}\)
Hok tốt
Bài 1:
a) Ta có: \(\left(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{2}\sqrt{20}-\dfrac{5}{4}\sqrt{\dfrac{4}{5}}+\sqrt{5}\right)\)
\(=\left(\sqrt{5}+\sqrt{5}-\dfrac{5}{4}\cdot\dfrac{2}{\sqrt{5}}+\sqrt{5}\right)\)
\(=3\sqrt{5}-\dfrac{1}{2}\sqrt{5}\)
\(=\dfrac{5}{2}\sqrt{5}\)
c) Ta có: \(\dfrac{5\sqrt{7}-7\sqrt{5}+2\sqrt{70}}{\sqrt{35}}\)
\(=\dfrac{\sqrt{35}\left(\sqrt{5}-\sqrt{7}+2\sqrt{2}\right)}{\sqrt{35}}\)
\(=2\sqrt{2}+\sqrt{5}-\sqrt{7}\)
Bài 2:
e) ĐKXĐ: \(\dfrac{4}{3}\le x\le6\)
Ta có: \(\sqrt{6-x}=3x-4\)
\(\Leftrightarrow6-x=\left(3x-4\right)^2\)
\(\Leftrightarrow9x^2-24x+16+6-x=0\)
\(\Leftrightarrow9x^2-25x+22=0\)
\(\Delta=\left(-25\right)^2-4\cdot9\cdot22=625-792< 0\)
Vậy: Phương trình vô nghiệm
a)A=\(2\sqrt{3}-8\sqrt{3}+7\sqrt{3}=\sqrt{3}\)
b)B\(=\sqrt{\left(3-\sqrt{5}\right)^2}+\sqrt{\left(2-\sqrt{5}\right)^2}=3-\sqrt{5}+\sqrt{5}-2=1\)
d)\(=\dfrac{\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)}{1}+1-\sqrt{5}-\dfrac{11\left(2\sqrt{5}-3\right)}{11}=5\sqrt{5}+5-10-2\sqrt{5}+1-\sqrt{5}-2\sqrt{5}+3=-1\)
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, \(\left(\sqrt{6}+\sqrt{5}\right)^2-\sqrt{120}\)
= \(6+2\sqrt{30}+5-2\sqrt{30}=11\)