Hình như đề bài của bạn bị lỗi rồi.

bạn ơi ghi lại đầu bài đi rồi mình làm chho nha

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: \(=2\cdot\dfrac{5}{4}-3\cdot\dfrac{7}{6}+4\cdot\dfrac{9}{8}=\dfrac{5}{2}-\dfrac{7}{2}+\dfrac{9}{2}=\dfrac{7}{2}\)

b: \(=18-16\cdot\dfrac{1}{2}+\dfrac{1}{16}\cdot\dfrac{3}{4}\)

=10+3/64

=643/64

c: \(=\dfrac{2}{3}\cdot\dfrac{9}{4}-\dfrac{3}{4}\cdot\dfrac{8}{3}+\dfrac{7}{5}\cdot\dfrac{5}{14}=\dfrac{3}{2}-2+\dfrac{1}{2}=2-2=0\)

\(\left[-\sqrt{2,25}+4\sqrt{\left(-2,15\right)^2}-\left(3\sqrt{\dfrac{7}{6}}\right)^2\right]\sqrt{1\dfrac{9}{16}}\)

\(=\left[-1,5+4\sqrt{2,15^2}-9\cdot\dfrac{7}{6}\right]\sqrt{\dfrac{25}{16}}\)

\(=\left[4\cdot\dfrac{43}{20}-10,5-1,5\right]\cdot\dfrac{5}{4}\)

\(=\left[\dfrac{43}{5}-12\right]\cdot\dfrac{5}{4}\)

\(=\dfrac{43}{5}\cdot\dfrac{5}{4}-12\cdot\dfrac{5}{4}\)

\(=\dfrac{43}{4}-15=\dfrac{-17}{4}\)

a: \(=7\cdot\dfrac{6}{7}-5+\dfrac{3\sqrt{2}}{2}=1+\dfrac{3}{2}\sqrt{2}\)

b: \(=-\dfrac{8}{7}-\dfrac{3}{5}\cdot\dfrac{5}{8}+\dfrac{1}{2}=\dfrac{-16+7}{14}-\dfrac{3}{8}=\dfrac{-9}{14}-\dfrac{3}{8}\)

\(=\dfrac{-72-42}{112}=\dfrac{-114}{112}=-\dfrac{57}{56}\)

c: \(=20\sqrt{5}-\dfrac{1}{4}\cdot\dfrac{4}{3}+\dfrac{3}{2}=20\sqrt{5}+\dfrac{3}{2}-\dfrac{1}{3}=20\sqrt{5}+\dfrac{7}{6}\)

a) \(4.\left(-\dfrac{1}{2}\right)^3-2.\left(-\dfrac{1}{2}\right)^2+3.\left(-\dfrac{1}{2}\right)+1\)

\(=4.\left(-\dfrac{1}{8}\right)-2.\dfrac{1}{4}+3.\left(-\dfrac{1}{2}\right)+1\)

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

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

b) \(8.\sqrt{9}-\sqrt{64}\)

\(=8.3-8\)

\(=24-8\)

\(=16\)

c) \(\sqrt{\dfrac{9}{16}}+\dfrac{25}{46}:\dfrac{5}{23}-\dfrac{7}{4}\)

\(=\dfrac{3}{4}+\dfrac{5}{2}-\dfrac{7}{4}\)

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

\(=\dfrac{3}{2}\)

56:54=

a,\(\left(\sqrt{1\dfrac{9}{16}}-\sqrt{\dfrac{9}{16}}\right):5=\left(\sqrt{\dfrac{25}{16}}-\dfrac{3}{4}\right):5=\left(\dfrac{5}{4}-\dfrac{3}{4}\right):5\)

\(=\dfrac{1}{2}:5=\dfrac{1}{10}\)

b,\(\left(\sqrt{3}-2\right)^2\left(\sqrt{3}+2\right)^2=\left[\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)\right]^2\)

\(=\left[3-4\right]^2=1\)

c,\(\left(11-4\sqrt{3}\right)\left(11+4\sqrt{3}\right)=11^2-\left(4\sqrt{3}\right)^2\)

\(=121-48=73\)

d,\(\left(\sqrt{2}-1\right)^2-\dfrac{3}{2}\sqrt{\left(-2\right)^2}+\dfrac{4\sqrt{2}}{5}+\sqrt{1\dfrac{11}{25}}.\sqrt{2}\)

\(=2-2\sqrt{2}+1-3+\dfrac{4\sqrt{2}}{5}+\sqrt{\dfrac{36}{25}.2}\)

\(=-2\sqrt{2}+\dfrac{4\sqrt{2}+6\sqrt{2}}{5}\)

\(=-2\sqrt{2}+\dfrac{10\sqrt{2}}{5}=-2\sqrt{2}+2\sqrt{2}=0\)

e,\(\left(1+\sqrt{2021}\right)\sqrt{2022-2\sqrt{2021}}\)

\(=\left(1+\sqrt{2021}\right)\sqrt{2021-2\sqrt{2021}.1+1}\)

\(=\left(1+\sqrt{2021}\right)\sqrt{\left(\sqrt{2021}-1\right)^2}\)

\(=\left(1+\sqrt{2021}\right)\left(\sqrt{2021}-1\right)\)

\(=\sqrt{2021}-1+\sqrt{2021^2}-\sqrt{2021}=2020\)