Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, \(\sqrt{\dfrac{289}{225}}=\sqrt{\dfrac{17^2}{15^2}}=\dfrac{17}{15}\)
b, \(\sqrt{2\dfrac{14}{25}}=\sqrt{\dfrac{64}{25}}=\sqrt{\dfrac{8^2}{5^2}}=\dfrac{8}{5}\)
c, \(\sqrt{\dfrac{0,25}{9}}=\sqrt{\dfrac{0,5^2}{3^2}}=\dfrac{0,5}{3}\)
d, \(\sqrt{\dfrac{8,1}{1,6}}=\sqrt{\dfrac{0,1}{0,1}.\dfrac{81}{16}}=\sqrt{1.\dfrac{81}{16}}=\dfrac{9}{4}\)
Chúc bạn học tốt!!!
a) \(\sqrt{\dfrac{289}{225}}\)
\(=\dfrac{\sqrt{289}}{\sqrt{225}}\)
\(=\dfrac{\sqrt{17^2}}{\sqrt{15^2}}\)
\(=\dfrac{17}{15}\)
b) \(\sqrt{2\dfrac{14}{15}}\)
\(=\sqrt{\dfrac{44}{15}}\)
\(=\dfrac{\sqrt{44}}{\sqrt{15}}\)
\(=\dfrac{2\sqrt{11}}{\sqrt{15}}\)
\(=\dfrac{2\sqrt{165}}{15}\)
c) \(\sqrt{\dfrac{0,25}{9}}\)
\(=\sqrt{\dfrac{1}{\dfrac{4}{9}}}\)
\(=\dfrac{\dfrac{1}{2}}{3}\)
\(=\dfrac{1}{6}\)
d) \(\sqrt{\dfrac{8,1}{1,6}}\)
\(=\sqrt{5,0625}\)
\(=\sqrt{\dfrac{81}{16}}\)
\(=\dfrac{9}{4}\)
Bài 1 :
Câu a : \(\sqrt{\dfrac{1,44}{3,61}}=\sqrt{\dfrac{144}{361}}=\dfrac{\sqrt{144}}{\sqrt{361}}=\dfrac{12}{19}\)
Câu b : \(\sqrt{\dfrac{0,25}{9}}=\sqrt{\dfrac{25}{900}}=\dfrac{\sqrt{25}}{\sqrt{900}}=\dfrac{5}{30}=\dfrac{1}{6}\)
Câu c : \(\sqrt{1\dfrac{13}{36}}.\sqrt{3\dfrac{13}{36}}=\sqrt{\dfrac{49}{36}}.\sqrt{\dfrac{121}{46}}=\dfrac{\sqrt{49}}{\sqrt{36}}.\dfrac{\sqrt{121}}{36}=\dfrac{7}{6}.\dfrac{11}{6}=\dfrac{77}{36}\)
Câu d : \(\sqrt{\dfrac{1}{121}.3\dfrac{6}{25}}=\sqrt{\dfrac{1}{121}.\dfrac{81}{25}}=\dfrac{1}{\sqrt{121}}.\dfrac{\sqrt{81}}{\sqrt{25}}=\dfrac{1}{11}.\dfrac{9}{5}=\dfrac{9}{55}\)
Câu e : \(\sqrt{1\dfrac{13}{36}.2\dfrac{2}{49}.2\dfrac{7}{9}}=\sqrt{\dfrac{49}{36}.\dfrac{100}{49}.\dfrac{25}{9}}=\dfrac{\sqrt{49}}{\sqrt{36}}.\dfrac{\sqrt{100}}{\sqrt{49}}.\dfrac{\sqrt{25}}{\sqrt{9}}=\dfrac{7}{6}.\dfrac{10}{7}.\dfrac{5}{3}=\dfrac{25}{9}\)
Bài 2 :
Câu a : \(\dfrac{\sqrt{245}}{\sqrt{5}}=\sqrt{\dfrac{245}{5}}=\sqrt{49}=7\)
Câu b : \(\dfrac{\sqrt{3}}{\sqrt{75}}=\sqrt{\dfrac{3}{75}}=\sqrt{\dfrac{1}{25}}=\dfrac{1}{5}\)
Câu c : \(\dfrac{\sqrt{10,8}}{\sqrt{0,3}}=\sqrt{\dfrac{10,8}{0,3}}=\sqrt{\dfrac{108}{3}}=\sqrt{36}=6\)
Câu d : \(\dfrac{\sqrt{6,5}}{\sqrt{58,5}}=\sqrt{\dfrac{6,5}{58,5}}=\sqrt{\dfrac{65}{585}}=\sqrt{\dfrac{1}{9}}=\dfrac{1}{3}\)
a) \(\sqrt{\dfrac{25}{81}.\dfrac{16}{49}.\dfrac{196}{9}}=\sqrt{\dfrac{25}{81}}.\sqrt{\dfrac{16}{49}}.\sqrt{\dfrac{196}{9}}=\dfrac{5}{9}.\dfrac{4}{7}.\dfrac{14}{3}=\dfrac{40}{27}\)
b) \(\sqrt{3\dfrac{1}{16}.2\dfrac{14}{25}.2\dfrac{34}{81}}=\sqrt{\dfrac{49}{16}.\dfrac{64}{25}.\dfrac{196}{81}}=\sqrt{\dfrac{49}{16}}.\sqrt{\dfrac{64}{25}}.\sqrt{\dfrac{196}{81}}=\dfrac{7}{4}.\dfrac{8}{5}.\dfrac{14}{9}=\dfrac{196}{45}\)
c) \(\dfrac{\sqrt{640}.\sqrt{34,3}}{\sqrt{567}}=\sqrt{\dfrac{640.34,3}{567}}=\sqrt{\dfrac{64.49}{81}}=\dfrac{\sqrt{64}.\sqrt{49}}{\sqrt{81}}=\dfrac{8.7}{9}=\dfrac{56}{9}\)
d) \(\sqrt{21,6}.\sqrt{810}.\sqrt{11^2-5^2}=\sqrt{21,6.810.\left(11^2-5^2\right)}=\sqrt{216.81.\left(11+5\right)\left(11-5\right)}=\sqrt{36^2.9^2.4^2}=36.9.4=1296\)
a) \(\sqrt{\dfrac{59}{25}+\dfrac{6}{5}\sqrt{2}}=\sqrt{2+2.\dfrac{3}{5}\sqrt{2}+\dfrac{9}{25}}=\sqrt{\left(\sqrt{2}+\dfrac{3}{5}\right)^2}\)
= / \(\sqrt{2}+\dfrac{3}{5}\) / = \(\sqrt{2}+\dfrac{3}{5}\)
b) \(\sqrt{\dfrac{129}{16}+\sqrt{2}}=\sqrt{8+2.2\sqrt{2}.\dfrac{1}{4}+\dfrac{1}{16}}\)
= \(\sqrt{\left(2\sqrt{2}+\dfrac{1}{4}\right)^2}\) = / \(2\sqrt{2}+\dfrac{1}{4}\) / = \(2\sqrt{2}+\dfrac{1}{4}\)
c) Tương tự , mình bận rồi , nếu chưa biết tẹo mk làm cho.
c) \(\sqrt{\dfrac{289+4\sqrt{72}}{16}}=\sqrt{\dfrac{289}{16}+\dfrac{1}{4}\sqrt{72}}=\sqrt{\dfrac{289}{16}+\dfrac{1}{4}.6\sqrt{2}}=\sqrt{18+2.\dfrac{1}{4}.3\sqrt{2}+\dfrac{1}{16}}=\sqrt{\left(3\sqrt{2}+\dfrac{1}{4}\right)^2}\) = / \(3\sqrt{2}+\dfrac{1}{4}\) / = \(3\sqrt{2}+\dfrac{1}{4}\)
b: \(=\left(\sqrt{ab}+\dfrac{2\sqrt{ab}}{a}-\sqrt{\dfrac{a^2+1}{ab}}\right)\cdot\sqrt{ab}\)
\(=ab+\dfrac{2ab}{a}-\sqrt{a^2+1}=ab+2b-\sqrt{a^2+1}\)
c: \(=2\sqrt{6b}-6\sqrt{18}+10\sqrt{12}-\sqrt{48}\)
\(=2\sqrt{6b}-18\sqrt{2}+20\sqrt{3}-4\sqrt{3}\)
\(=2\sqrt{6n}-18\sqrt{2}+16\sqrt{3}\)
d: \(=\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\dfrac{\sqrt{21}}{7}\)
\(b>=\sqrt{2}\sqrt{4+\sqrt{15}}\)
\(\Leftrightarrow\sqrt{2\left(4+\sqrt{15}\right)}\)
\(\Leftrightarrow\sqrt{8+2\sqrt{15}}\)
\(\Leftrightarrow\sqrt{5}+\sqrt{3}\)
\(a,2\sqrt{\dfrac{27}{4}}-\sqrt{\dfrac{48}{9}}-\dfrac{2}{5}.\sqrt{\dfrac{75}{16}}\)
\(\Leftrightarrow2.\dfrac{\sqrt{27}}{2}-\sqrt{\dfrac{48}{3}}-\dfrac{2}{5}.\dfrac{\sqrt{75}}{4}\)
\(\Leftrightarrow\sqrt{27}-\dfrac{4\sqrt{3}}{3}-\dfrac{1}{5}.\dfrac{5\sqrt{3}}{2}\)
\(\Leftrightarrow3\sqrt{3}-\dfrac{4\sqrt{3}}{3}-\dfrac{\sqrt{3}}{2}\)
\(\Leftrightarrow\dfrac{7\sqrt{3}}{6}\)
\(b,\left(1+\dfrac{5-\sqrt{5}}{1-\sqrt{5}}\right).\left(\dfrac{5+\sqrt{5}}{1+\sqrt{5}}+1\right)\)
\(\Leftrightarrow\)\(\left[1+\dfrac{\left(5-\sqrt{5}\right)\left(1+\sqrt{5}\right)}{-4}\right].\left[\dfrac{\left(5+\sqrt{5}\right).\left(1-\sqrt{5}\right)}{-4}+1\right]\)
\(\Leftrightarrow\)\(\left(1+\dfrac{5+5\sqrt{5}-\sqrt{5}-5}{-4}\right).\left(\dfrac{5-5\sqrt{5}+\sqrt{5}-5}{-4}+1\right)\)
\(\Leftrightarrow\)\(\left(1+\dfrac{4\sqrt{5}}{-4}\right)\left(\dfrac{-4\sqrt{5}}{-4}+1\right)\)
\(\Leftrightarrow\left(1-\sqrt{5}\right)\left(\sqrt{5}+1\right)\)
\(\Leftrightarrow\left(1-\sqrt{5}\right).\left(1+\sqrt{5}\right)\)
<=> 1-5
=-4
\(a,\sqrt{4,9.360}=\sqrt{49.36}=\sqrt{49}.\sqrt{36}=7.6=42\)
b,\(\sqrt{2,25.0,04}=\sqrt{0.09}=0.3\)
c, \(\sqrt{3\dfrac{1}{16}.2\dfrac{4}{15}}=\sqrt{\dfrac{49}{16}.\dfrac{44}{15}}=\sqrt{\dfrac{49}{16}}.\sqrt{\dfrac{44}{15}}=\dfrac{7}{4}.1,7=2,99\approx3\)
e, \(\sqrt{\dfrac{144}{169}}=\dfrac{\sqrt{144}}{\sqrt{169}}=\dfrac{12}{13}\)
g,\(\dfrac{\sqrt{27}}{\sqrt{3}}=\sqrt{\dfrac{27}{3}}=\sqrt{9}=3\)
f,\(\sqrt{2,25}=\dfrac{3}{2}\)
n,\(\sqrt{\dfrac{25}{529}}=\dfrac{\sqrt{25}}{\sqrt{529}}=\dfrac{5}{23}\)
1)
a. \(\sqrt{\dfrac{25}{7}}.\sqrt{\dfrac{7}{9}}=\sqrt{\dfrac{25.7}{7.9}}=\sqrt{\dfrac{25}{9}}=\dfrac{5}{3}\)
b. \(\left(\sqrt{\dfrac{9}{2}}+\sqrt{\dfrac{1}{2}}-\sqrt{2}\right).\sqrt{2}=3+1-2=2\)
c. \(\left(\sqrt{\dfrac{8}{3}}-\sqrt{24}+\sqrt{\dfrac{50}{3}}\right).\sqrt{6}=4-12+10=2\)
d. \(\left(\sqrt{\dfrac{2}{3}}-\sqrt{\dfrac{3}{2}}\right)^2=\dfrac{2}{3}+\dfrac{3}{2}-2\sqrt{\dfrac{2}{3}.\dfrac{3}{2}}=\dfrac{1}{6}\)
2)
a. \(\sqrt{4+2\sqrt{3}}=\sqrt{3+2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
b. \(\sqrt{8-2\sqrt{7}}=\sqrt{7-2\sqrt{7}+1}=\sqrt{\left(\sqrt{7}-1\right)^2}=\sqrt{7}-1\)
c. \(1+\sqrt{6-2\sqrt{5}}=1+\sqrt{5-2\sqrt{5}+1}=1-\sqrt{\left(\sqrt{5}-1\right)^2}=1-\sqrt{5}+1=2-\sqrt{5}\)
d. \(\sqrt{7-2\sqrt{10}}+\sqrt{2}=\sqrt{5-2.\sqrt{5}.\sqrt{2}+2}+\sqrt{2}=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}+\sqrt{2}=\sqrt{5}-\sqrt{2}+\sqrt{2}=\sqrt{5}\)
3. \(a.A=x^2+2x+16=\left(\sqrt{2}-1\right)^2+2.\left(\sqrt{2}-1\right)+16=2-2\sqrt{2}+1+2\sqrt{2}-2+16=17\)
\(b.B=x^2+12x-14=\left(5\sqrt{2}-6\right)^2+12.\left(5\sqrt{2}-6\right)-14=50+36-60\sqrt{2}+60\sqrt{2}-72-14=0\)
Help me nha @Phùng Khánh Linh@Nhã Doanh@Liana@Yukru Cảm ơn trước nhé
a) ; b) ;
c) ; d) .
a) \(\sqrt{\dfrac{289}{225}}=\dfrac{\sqrt{289}}{\sqrt{225}}=\dfrac{17}{15}\)
b) \(\sqrt{2\dfrac{14}{25}}=\dfrac{\sqrt{64}}{\sqrt{25}}=\dfrac{8}{5}\)
c) \(\sqrt{\dfrac{0,25}{9}}=\dfrac{\sqrt{0,25}}{\sqrt{9}}=\dfrac{0,5}{3}=\dfrac{1}{6}\)
d) \(\sqrt{\dfrac{8,1}{1,6}}=\dfrac{\sqrt{81}}{\sqrt{16}}=\dfrac{9}{4}\)