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a. \(\sqrt{13^2-12^2}\)
=\(\sqrt{\left(13+12\right).\left(13-12\right)}\)
=\(\sqrt{25.1}\)
=\(\sqrt{25}.\sqrt{1}\)
=5.1
=5
b. \(\sqrt{17^2-8^2}\)
=\(\sqrt{\left(17+8\right).\left(17-8\right)}\)
=\(\sqrt{25.9}\)
=\(\sqrt{25}.\sqrt{9}\)
=5.3
=15
c. \(\sqrt{117^2-108^2}\)
=\(\sqrt{\left(117+108\right).\left(117-108\right)}\)
=\(\sqrt{225.9}\)
=\(\sqrt{225}.\sqrt{9}\)
=15.3
=45
d. \(\sqrt{313^2-312^2}\)
=\(\sqrt{\left(313+312\right).\left(313-312\right)}\)
=\(\sqrt{625.1}\)
=\(\sqrt{625}.\sqrt{1}\)
=25.1
=25
c.\(\sqrt{117^2-108^2}\)
a) \(\sqrt{13^2-12^2}\)=\(\sqrt{\left(13-12\right)\left(13+12\right)}\)=\(\sqrt{1x25}\)=5
Câu a: Ta có:
√132−122=√(13+12)(13−12)132−122=(13+12)(13−12)
=√25.1=√25=25.1=25
=√52=|5|=5=52=|5|=5.
Câu b: Ta có:
√172−82=√(17+8)(17−8)172−82=(17+8)(17−8)
=√25.9=√25.√9=25.9=25.9
=√52.√32=|5|.|3|=52.32=|5|.|3|.
=5.3=15=5.3=15.
Câu c: Ta có:
√1172−1082=√(117−108)(117+108)1172−1082=(117−108)(117+108)
=√9.225=9.225 =√9.√225=9.225
=√32.√152=|3|.|15|=32.152=|3|.|15|
=3.15=45=3.15=45.
Câu d: Ta có:
√3132−3122=√(313−312)(313+312)3132−3122=(313−312)(313+312)
=√1.625=√625=1.625=625
=√252=|25|=25=252=|25|=25.
a) √54 = √9.6 = 3√6
b) √108 = √36.3 = 6√3
c) 0,1√20000 = 0,1√10000.2= 0,1.100√2 = 10√2
d) -0,05.√28800 = -0,05.√14400.2 = -0,05.120√2 = -6√2
e)√7.63.a2 = √7.7.9.a2 = 7.3|a| = 21|a|
a: \(=\sqrt{9\cdot6}=3\sqrt{6}\)
b: \(=\sqrt{36\cdot3}=6\sqrt{3}\)
c: \(=\dfrac{1}{10}\cdot\sqrt{10000\cdot2}=\dfrac{1}{10}\cdot100\cdot\sqrt{2}=10\sqrt{2}\)
d: \(=-\dfrac{1}{20}\cdot\sqrt{14400\cdot2}=-\dfrac{1}{20}\cdot120\cdot\sqrt{2}=-6\sqrt{2}\)
e: \(=\sqrt{7\cdot7\cdot9\cdot a^2}=21\left|a\right|\)
\(13-4\sqrt{3}=\left(2\sqrt{3}\right)^2-2.2\sqrt{2}.1+1^2=\left(2\sqrt{3}-1\right)^2\)
a) \(\left(\sqrt{5}+\sqrt{3}\right)\sqrt{8-2\sqrt{15}}=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=5-3=2\)
câu này \(\sqrt{15}\)đúng hơn \(\sqrt{5}\)
b) \(\sqrt{3-\sqrt{5}}-\sqrt{3+\sqrt{5}}=\frac{\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}}{\sqrt{2}}=\frac{\sqrt{5}-1-\sqrt{5}-1}{\sqrt{2}}=\frac{-2}{\sqrt{2}}=-\sqrt{2}\)c) \(\sqrt{5-2\sqrt{6}}-\sqrt{5+2\sqrt{6}}=\sqrt{3}-\sqrt{2}-\sqrt{3}-\sqrt{2}=-2\sqrt{2}\)
b: \(5+2\sqrt{6}=\left(\sqrt{3}+\sqrt{2}\right)^2\)
c: \(13+\sqrt{48}=13+4\sqrt{3}=\left(2\sqrt{3}+1\right)^2\)
d: \(4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\)
a) Ta có:
5√15+12√20+√5515+1220+5
=√52.15+√(12)2.20+√5=√25.15+√14.20+√5=√255+√204+√5=√5+√5+√5=(1+1+1)√5=3√5=52.15+(12)2.20+5=25.15+14.20+5=255+204+5=5+5+5=(1+1+1)5=35
b) Ta có:
√12+√4,5+√12,512+4,5+12,5
=√12+√92+√252=√12+√9.12+√25.12=√12+√32.12+√52.12=√12+3√12+5√12=(1+3+5).√12=9√12=91√2=9.√22=9√22=12+92+252=12+9.12+25.12=12+32.12+52.12=12+312+512=(1+3+5).12=912=912=9.22=922
c) Ta có:
√20−√45+3√18+√72=√4.5−√9.5+3√9.2+√36.2=√22.5−√32.5+3√32.2+√62.2=2√5−3√5+3.3√2+6√2=2√5−3√5+9√2+6√2=(2√5−3√5)+(9√2+6√2)=(2−3)√5+(9+6)√2=−√5+15√2=15√2−√520−45+318+72=4.5−9.5+39.2+36.2=22.5−32.5+332.2+62.2=25−35+3.32+62=25−35+92+62=(25−35)+(92+62)=(2−3)5+(9+6)2=−5+152=152−5
d) Ta có:
0,1√200+2√0,08+0,4.√50=0,1√100.2+2√0,04.2+0,4√25.2=0,1√102.2+2√0,22.2+0,4√52.2=0,1.10√2+2.0,2√2+0,4.5√2=1√2+0,4√2+2√2=(1+0,4+2)√2=3,4√2
a) \(\sqrt{117^2-108^2}=\sqrt{\left(117-108\right)\left(117+108\right)}=\sqrt{9\cdot225}=\sqrt{3^2\cdot15^2}=\left|3\cdot15\right|=45\)
b) \(\sqrt{9-4\sqrt{5}}+2=\sqrt{5-4\sqrt{5}+4}+2=\sqrt{\left(\sqrt{5}-2\right)^2}+2=\left|\sqrt{5}-2\right|+2=\sqrt{5}\)
\(a,\sqrt{117^2-108^2}\\ =\sqrt{\left(117-108\right)\left(117+108\right)}\\ =\sqrt{9.225}\\ =\sqrt{3^2}.\sqrt{15^2}\\ =3.15\\ =45\)
\(b,\sqrt{9-4\sqrt{5}}+2=\sqrt{5}\)
\(VT=\sqrt{9-4\sqrt{5}}+2\\ =\sqrt{\sqrt{5^2}-2.2\sqrt{5}+2^2}+2\\ =\sqrt{\left(\sqrt{5}-2\right)^2}+2\\ =\left|\sqrt{5}-2\right|+2\\ =\sqrt{5}-2+2\\ =\sqrt{5}=VP\left(dpcm\right)\)