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a: \(=\dfrac{5^4\cdot5^4\cdot4^4}{5^{10}\cdot4^5}=\dfrac{1}{5^2}\cdot\dfrac{1}{4}=\dfrac{1}{100}\)
b: \(=\dfrac{\left[5^3\left(5-1\right)\right]^3}{5^{12}}=\dfrac{5^9}{5^{12}}\cdot\dfrac{4^3}{1}=\dfrac{4^3}{5^3}\)
c: \(=\sqrt{1.8^2}=1.8\)
a) \(\sqrt{\left(2,5+0,7\right)^2}=\left(2,5+0,7\right)=3,2\)
b) \(\frac{\sqrt{3^2}+\sqrt{39^2}}{\sqrt{7^2}+\sqrt{91^2}}=\frac{3+39}{7+91}=\frac{42}{98}=\frac{3}{7}\)
Ko có x nha bạn
a)\(\sqrt{\dfrac{3^2}{7^2}}=\sqrt{\dfrac{9}{49}}=\sqrt{\dfrac{3}{7}}\)
b)\(\dfrac{\sqrt{3^2}+\sqrt{39^2}}{\sqrt{7^2}+\sqrt{91^2}}=\dfrac{\sqrt{9}+\sqrt{1521}}{\sqrt{49}+\sqrt{8281}}=\dfrac{3+39}{7+91}=\dfrac{42}{98}\)
c)Tương tự câu b, ta đc:
\(\dfrac{\sqrt{3^2}-\sqrt{39^2}}{\sqrt{7^2}-\sqrt{91^2}}=\dfrac{3-39}{7-91}=\dfrac{-36}{86}=\dfrac{3}{7}\)
d)Tương tự câu a, ta đc:
\(\dfrac{\sqrt{39^2}}{\sqrt{91^2}}=\dfrac{39}{91}\)
Chúc Bạn Học Tốt!!!
a) \(\sqrt{\dfrac{3^2}{7^2}}=\sqrt{\left(\dfrac{3}{7}\right)^2}=\left|\dfrac{3}{7}\right|=\dfrac{3}{7}\)
b) \(\dfrac{\sqrt{3}^2+\sqrt{39}^2}{\sqrt{7}^2+\sqrt{91}^2}=\dfrac{\left|3\right|+\left|39\right|}{\left|7\right|+\left|91\right|}=\dfrac{3+39}{7+91}=\dfrac{42}{98}=\dfrac{3}{7}\)
c) \(\dfrac{\sqrt{3}^2-\sqrt{39}^2}{\sqrt{7}^2-\sqrt{91}^2}=\dfrac{\left|3\right|- \left|39\right|}{\left|7\right|-\left|91\right|}=\dfrac{3-39}{7-91}=\dfrac{-36}{-84}=\dfrac{3}{7}\)
d) \(\sqrt{\dfrac{39^2}{91^2}}=\sqrt{\left(\dfrac{39}{91}\right)^2}=\left|\dfrac{39}{91}\right|=\dfrac{39}{91}=\dfrac{3}{7}\)
a: \(\left(18\dfrac{1}{3}:\sqrt{225}+8\dfrac{2}{3}\cdot\sqrt{\dfrac{49}{4}}\right):\left[\left(12\dfrac{1}{3}+8\dfrac{6}{7}\right)-\dfrac{\left(\sqrt{7}\right)^2}{\left(3\sqrt{2}\right)^2}\right]:\dfrac{1704}{445}\)
\(=\left(\dfrac{55}{3}:15+\dfrac{26}{3}\cdot\dfrac{7}{4}\right):\left[\left(12+\dfrac{1}{3}+8+\dfrac{6}{7}\right)-\dfrac{7}{18}\right]\cdot\dfrac{445}{1704}\)
\(=\left(\dfrac{55}{45}+\dfrac{91}{6}\right):\left[20+\dfrac{101}{126}\right]\cdot\dfrac{445}{1704}\)
\(=\dfrac{295}{18}:\dfrac{2621}{126}\cdot\dfrac{445}{1704}\)
\(=\dfrac{295}{18}\cdot\dfrac{126}{2621}\cdot\dfrac{445}{1704}\simeq0,21\)
b: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)
c: \(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{n+1}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{n}{n+1}\)
\(=\dfrac{1}{n+1}\)
d: \(-66\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{11}\right)+124\cdot\left(-37\right)+63\cdot\left(-124\right)\)
\(=-66\cdot\dfrac{33-22+6}{66}+124\left(-37-63\right)\)
\(=-17-12400=-12417\)
e: \(\dfrac{7}{4}\left(\dfrac{33}{12}+\dfrac{3333}{2020}+\dfrac{333333}{303030}+\dfrac{33333333}{42424242}\right)\)
\(=\dfrac{7}{4}\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)
\(=\dfrac{7}{4}\cdot33\cdot\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\right)\)
\(=33\cdot\dfrac{7}{4}\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)\)
\(=33\cdot\dfrac{7}{4}\cdot\left(\dfrac{1}{3}-\dfrac{1}{7}\right)\)
\(=33\cdot\dfrac{7}{4}\cdot\dfrac{4}{21}=\dfrac{33\cdot1}{3}=11\)
a) \(\left(\frac{2^2}{5}\right)+5\frac{1}{2}.\left(4,5-2,5\right)+\frac{2^3}{-4}\)
\(=\frac{4}{5}+\frac{11}{2}.2+\frac{-8}{4}\)
\(=\frac{4}{5}+11-2\)
\(=\frac{4}{5}+9\)
\(=\frac{49}{9}\)
b) \(\left(-2^3\right)+\frac{1}{2}:\frac{1}{8}-\sqrt{25}+\left|-64\right|\)
\(=-8+4-5+64\)
= 55
c) \(\frac{\sqrt{3^2+\sqrt{39}^2}}{\sqrt{91^2}-\sqrt{\left(-7\right)^2}}\)
\(=\frac{\sqrt{9+39}}{91-\sqrt{49}}\)
\(=\frac{\sqrt{48}}{91-7}\)
\(=\frac{4\sqrt{3}}{84}\)
\(=\frac{\sqrt{3}}{41}\)
d) Xem lại đề nhé em!
e) \(\sqrt{25}-3\sqrt{\frac{4}{9}}\)
\(=5-3.\frac{2}{3}\)
= 5 - 2
= 3
h) \(\left(-3^2\right).\frac{1}{3}-\sqrt{49}+\left(5^3\right):\sqrt{25}\)
\(=-9.\frac{1}{3}-7+125:5\)
\(=-3-7+25\)
= 15
\(a,=\sqrt{6,25-0,49}=\sqrt{5,76}=2,4\)
\(b,=\sqrt{2,5-0,49}=\sqrt{2,01}\)
\(c,\sqrt{5,76}=2,4\)
a)\(\dfrac{3}{4}-\dfrac{5}{2}-\dfrac{3}{5}=\dfrac{15}{20}-\dfrac{50}{20}-\dfrac{12}{20}=-\dfrac{47}{20}\)
b) \(\sqrt{7^2}+\sqrt{\dfrac{25}{16}-\dfrac{3}{2}}=7+\sqrt{\dfrac{1}{16}}=7+\dfrac{1}{4}=\dfrac{29}{4}\)
c) \(\dfrac{1}{2}.\sqrt{100}-\sqrt{\dfrac{1}{16}+\left(\dfrac{1}{3}\right)^0}=\dfrac{1}{2}.10-\sqrt{\dfrac{1}{16}+1}=5-\sqrt{\dfrac{17}{16}}\)
\(B=\dfrac{\left(\dfrac{5}{70}-\dfrac{10\sqrt{2}}{70}+\dfrac{6\sqrt{2}}{70}\right)\cdot\dfrac{-4}{15}}{\left(\dfrac{5}{50}+\dfrac{6\sqrt{2}}{50}-\dfrac{10\sqrt{2}}{50}\right)\cdot\dfrac{5}{7}}=\dfrac{\dfrac{5-4\sqrt{2}}{70}\cdot\dfrac{-4}{15}}{\dfrac{5-4\sqrt{2}}{50}\cdot\dfrac{5}{7}}\)
\(=\dfrac{-4\left(5-4\sqrt{2}\right)}{70\cdot15}\cdot\dfrac{50\cdot7}{5\left(5-4\sqrt{2}\right)}=\dfrac{-4}{5}\cdot\dfrac{350}{70\cdot15}=\dfrac{-4}{5}\cdot\dfrac{1}{3}=\dfrac{-4}{15}\)
a)\(\frac{\sqrt{3^2}+\sqrt{39^2}}{\sqrt{7^2}+\sqrt{91^2}}.1\)
=\(\frac{3+39}{7+91}\)
=\(\frac{42}{98}\)
=\(\frac{3}{7}\)
b)\(\sqrt{\left(2,5-0,7\right)^2}\)
=\(|2,5-0,7|\)
=2,5-0,7
=1,8