a, \(1+\dfrac{3}{4}=\dfrac{7}{4}\)

b, \(\dfrac{4}{5}-\dfrac{3}{8}=\dfrac{32-15}{40}=\dfrac{17}{40}\)

c, \(1:\dfrac{2}{3}=\dfrac{1.3}{2}=\dfrac{3}{2}\)

d, \(\dfrac{2}{5}.\dfrac{5}{2}=1\)

a)1 + 3/4=7/4

   b)4/5 -3/8=17/40

  c)1 : 2/3=3/2

  d)2/5 x 5/2=1

Đáp án:

a)1 + 3/4=7/4

b)4/5 -3/8=17/40

c)1 : 2/3=3/2

d)2/5 x 5/2=1

a) \(\dfrac{2}{3}+\dfrac{4}{5}=\dfrac{10}{15}+\dfrac{12}{15}=\dfrac{22}{15}\)

b) \(\dfrac{5}{6}-\dfrac{3}{4}=\dfrac{20}{24}-\dfrac{18}{24}=\dfrac{2}{24}=\dfrac{1}{12}\)

\(\dfrac{10}{15}+\dfrac{12}{15}=\dfrac{22}{15}\)

\(\dfrac{20}{24}-\dfrac{18}{24}=\dfrac{1}{12}\)

a) \(\dfrac{2}{3}+\dfrac{4}{5}=\dfrac{10}{15}+\dfrac{12}{15}=\dfrac{22}{15}\)

b) \(\dfrac{5}{6}-\dfrac{3}{4}=\dfrac{10}{12}-\dfrac{9}{12}=\dfrac{1}{12}\)

\(a,\dfrac{4}{7}+\dfrac{7}{2}\\ =\dfrac{8}{14}+\dfrac{49}{14}\\ =\dfrac{8+49}{14}\\ =\dfrac{57}{14}\)

\(b,\dfrac{5}{8}\times\dfrac{3}{2}\\ =\dfrac{5\times3}{8\times2}\\ =\dfrac{15}{16}\)

\(c,\dfrac{3}{2}\times\dfrac{5}{6}-\dfrac{2}{3}\\ =\dfrac{3\times5}{2\times6}-\dfrac{2}{3}\\ =\dfrac{15}{12}-\dfrac{2}{3}\\ =\dfrac{15}{12}-\dfrac{8}{12}\\ =\dfrac{15-8}{12}\\ =\dfrac{7}{12}\)

\(d,\dfrac{13}{15}+\dfrac{2}{5}:\dfrac{3}{4}\\ =\dfrac{13}{15}+\dfrac{2}{5}\times\dfrac{4}{3}\\ =\dfrac{13}{15}+\dfrac{2\times4}{5\times3}\\ =\dfrac{13}{15}+\dfrac{8}{15}\\ =\dfrac{13+8}{15}\\ =\dfrac{21}{15}\\ =\dfrac{7}{5}\)

a \(\dfrac{53}{84}\)

b \(\dfrac{20}{63}\)

c \(\dfrac{1}{60}\)

a: \(\dfrac{2}{3}+\dfrac{5}{7}-\dfrac{3}{4}=\dfrac{56}{84}+\dfrac{60}{84}-\dfrac{63}{84}=\dfrac{53}{84}\)

b: \(=\dfrac{3}{7}-\dfrac{1}{9}=\dfrac{27}{63}-\dfrac{7}{63}=\dfrac{20}{63}\)

c: \(=\dfrac{2}{5}\cdot\dfrac{1}{8}\cdot\dfrac{1}{3}=\dfrac{2}{8}\cdot\dfrac{1}{15}=\dfrac{1}{60}\)

a.5/14+1/2=6/7

b.5/3-1=2/3

c.4/3-1/3=1

a) 5/14+ 1/2= 6/7

b) 5/3 - 1= 2/3

c) 4/3- 1/3 =1

a.5/14+1/2=6/7

b.5/3-1=2/3

c.4/3-1/3=1

a: Ta có: \(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)

\(=\dfrac{\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{3}-1-\sqrt{3}-1}{\sqrt{2}}=-\sqrt{2}\)

b: Ta có: \(\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)

\(=\dfrac{\left(\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}-2\right)}{\sqrt{2}}\)

\(=\dfrac{\sqrt{5}+1+3-\sqrt{5}-2}{\sqrt{2}}=\sqrt{2}\)

a: Ta có: \(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)

\(=\dfrac{\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{3}-1-\sqrt{3}-1}{\sqrt{2}}=-\sqrt{2}\)

b: Ta có: \(\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)

\(=\dfrac{\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}-2}{\sqrt{2}}\)

\(=\dfrac{\sqrt{5}+1+3-\sqrt{5}-2}{\sqrt{2}}\)

\(=\sqrt{2}\)

c: \(\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)=5-4=1\)

\(A=\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}-\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)

\(=\dfrac{3^4\left(3-1\right)}{3^5\left(3+1\right)}-\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\cdot9}\)

\(=\dfrac{1}{3\cdot2}-\dfrac{1}{5}\cdot\dfrac{-6}{9}=\dfrac{1}{6}+\dfrac{6}{45}=\dfrac{45+36}{270}=\dfrac{81}{270}=\dfrac{3}{10}\)

a.4/5+3/15 = 4/5+ 3/15:3=4/5+1/5=5/5

b.2/3+32/24=2/3+32/24:8=2/3+4/3=6/3

c.5/6+15/18=5/6x3+15/18=15/18+15/18=30/18

a: \(\sqrt{\left(3-\sqrt{5}\right)^2}-\sqrt{5}\)

\(=\left|3-\sqrt{5}\right|-\sqrt{5}\)

\(=3-\sqrt{5}-\sqrt{5}=3-2\sqrt{5}\)

b: \(\sqrt{\left(4-2\sqrt{3}\right)^2}\)

\(=\left|4-2\sqrt{3}\right|\)

\(=4-2\sqrt{3}\)