\(\dfrac{-2}{45}:\dfrac{8}{3}=0,01x:4\)
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\(\left(85\dfrac{7}{30}-83\dfrac{5}{18}\right):2\dfrac{2}{3}=0,01x:4\)
\(\Leftrightarrow[\left(85-83\right)+\left(\dfrac{7}{30}-\dfrac{5}{18}\right)]:2\dfrac{2}{3}=0,01x:4\)
\(\Leftrightarrow1\dfrac{43}{45}:2\dfrac{2}{3}=0,01x:4\)
\(\Leftrightarrow\dfrac{88}{45}:\dfrac{8}{3}=0,01x:4\)
\(\Leftrightarrow\dfrac{88}{45}.\dfrac{3}{8}=0,01x:4\)
\(\Leftrightarrow\dfrac{11}{15}=0,01x:4\)
\(\Leftrightarrow0,01x:4=\dfrac{11}{15}\)
\(\Leftrightarrow x=\dfrac{11}{15}.4:0,01\)
\(\Leftrightarrow x=\dfrac{880}{3}\)
Vậy x = \(\dfrac{880}{3}\)
Ta có:
\(\left(85\dfrac{7}{30}-83\dfrac{5}{18}\right):2\dfrac{2}{3}=0.01x:4\\ \left(\left(85-83\right)+\left(\dfrac{7}{30}-\dfrac{5}{18}\right)\right):\dfrac{8}{3}=0.01x:4\\ \dfrac{88}{45}:\dfrac{8}{3}=0.01x:4\\ \dfrac{11}{15}=0.01x:4\\ \dfrac{44}{15}=0.01x\\ x=\dfrac{44}{15}:0.01\\ x=\dfrac{880}{3}\)
Vậy \(x=\dfrac{880}{3}\)
a: \(\dfrac{8}{18}+\dfrac{5}{3}=\dfrac{4}{9}+\dfrac{5}{3}=\dfrac{4}{9}+\dfrac{15}{9}=\dfrac{4+15}{9}=\dfrac{19}{9}\)
b: \(\dfrac{8}{24}+\dfrac{4}{48}=\dfrac{1}{3}+\dfrac{1}{12}=\dfrac{4}{12}+\dfrac{1}{12}=\dfrac{4+1}{12}=\dfrac{5}{12}\)
c: \(\dfrac{20}{15}-\dfrac{4}{45}=\dfrac{4}{3}-\dfrac{4}{45}=\dfrac{60}{45}-\dfrac{4}{45}=\dfrac{60-4}{45}=\dfrac{56}{45}\)
d: \(\dfrac{40}{32}-\dfrac{1}{2}=\dfrac{5}{4}-\dfrac{1}{2}=\dfrac{5-2}{4}=\dfrac{3}{4}\)
Bài 1:
a)
\(\dfrac{1}{2}=\dfrac{1\times6}{2\times6}=\dfrac{6}{12}\)
\(\dfrac{2}{3}=\dfrac{2\times4}{3\times4}=\dfrac{8}{12}\)
\(\dfrac{3}{4}=\dfrac{3\times3}{4\times3}=\dfrac{9}{12}\)
b)
\(\dfrac{1}{3}=\dfrac{1\times15}{3\times15}=\dfrac{15}{45}\)
\(\dfrac{2}{15}=\dfrac{2\times3}{15\times3}=\dfrac{6}{45}\)
\(\dfrac{4}{45}\) (giữ nguyên)
c)
\(\dfrac{1}{8}=\dfrac{1\times3}{8\times3}=\dfrac{3}{24}\)
\(\dfrac{2}{3}=\dfrac{2\times8}{3\times8}=\dfrac{16}{24}\)
\(\dfrac{5}{2}=\dfrac{5\times12}{2\times12}=\dfrac{60}{24}\)
d)
\(\dfrac{2}{7}=\dfrac{2\times4}{7\times4}=\dfrac{8}{28}\)
\(\dfrac{9}{4}=\dfrac{9\times7}{4\times7}=\dfrac{63}{28}\)
\(\dfrac{5}{28}\) (giữ nguyên)
Bài 2:
a)
\(4=\dfrac{4}{1}=\dfrac{4\times12}{1\times12}=\dfrac{48}{12}\)
\(\dfrac{9}{4}=\dfrac{9\times3}{4\times3}=\dfrac{27}{12}\)
b)
\(\dfrac{5}{8}=\dfrac{5\times30}{8\times30}=\dfrac{150}{240}\)
\(\dfrac{25}{30}=\dfrac{5}{6}=\dfrac{5\times40}{6\times40}=\dfrac{200}{240}\)
\(2=\dfrac{2}{1}=\dfrac{2\times240}{1\times240}=\dfrac{480}{240}\).
\(a,=\dfrac{5}{7}-\dfrac{33}{8}=-\dfrac{191}{56}\\ b,=\left(\dfrac{12}{17}+\dfrac{5}{17}\right)+\left(\dfrac{19}{7}+\dfrac{3}{7}\right)=1+3=4\\ c,=\left(0,125\cdot8\right)^{12}-\left(\dfrac{45}{15}\right)^3=1-3^3=-26\\ d,=\left(-\dfrac{1}{3}\right)\left(5\dfrac{2}{7}-2\dfrac{2}{7}\right)=-\dfrac{1}{3}\cdot3=-1\\ e,=\dfrac{3^4\cdot3^6}{3^9}=3\)
\(a)\) Mẫu số chung: 45.
\(\dfrac{2}{5}=\dfrac{2\times7}{5\times7}=\dfrac{14}{45}\\ \dfrac{13}{15}=\dfrac{13\times3}{15\times3}=\dfrac{39}{45}\\ \dfrac{19}{45}\)
b) Mẫu số chung: 16.
\(\dfrac{1}{2}=\dfrac{1\times8}{2\times8}=\dfrac{8}{16}\\ \dfrac{3}{8}=\dfrac{3\times2}{8\times2}=\dfrac{6}{16}\\ \dfrac{5}{4}=\dfrac{5\times4}{4\times4}=\dfrac{\dfrac{20}{16}\\ 7}{16}\)
3)\(\dfrac{-41}{32}\left(\dfrac{15}{8}-\dfrac{16}{41}\right)+\dfrac{15}{8}\left(\dfrac{41}{32}-\dfrac{8}{3}\right)\)
=\(\dfrac{-41}{32}.\dfrac{15}{8}-\dfrac{-41}{32}.\dfrac{16}{41}+\dfrac{15}{8}.\dfrac{41}{32}-\dfrac{15}{8}.\dfrac{8}{3}\)
=\(\left(\dfrac{-41}{32}.\dfrac{15}{8}+\dfrac{15}{8}.\dfrac{41}{32}\right)+\dfrac{-16}{41}.\dfrac{-41}{32}-\dfrac{15}{8}.\dfrac{8}{3}\)
=\(0+\dfrac{1}{2}-5=\dfrac{-9}{2}\)
4)\(\dfrac{13}{29}\left(\dfrac{29}{5}-\dfrac{45}{8}\right)-\dfrac{45}{8}\left(\dfrac{9}{8}-\dfrac{13}{29}\right)\)
=\(\dfrac{13}{29}.\dfrac{29}{5}-\dfrac{45}{8}.\dfrac{13}{29}-\dfrac{45}{8}.\dfrac{9}{8}-\dfrac{45}{8}.\dfrac{13}{29}\)
=\(\left(\dfrac{45}{8}.\dfrac{13}{29}-\dfrac{45}{8}.\dfrac{13}{29}\right)-\dfrac{13}{29}.\dfrac{29}{5}-\dfrac{45}{8}.\dfrac{9}{8}\)
=\(0-\dfrac{13}{5}-\dfrac{405}{64}=\dfrac{-2857}{320}\)
\(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}\)
\(=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\sqrt{3}+2\sqrt{7}}\)
\(=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}\)
\(=\dfrac{\sqrt{2}}{2}\)
___________
\(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}+\sqrt{8}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)
__________
\(\dfrac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{3\sqrt{18}-2\sqrt{27}+\sqrt{45}}\)
\(=\dfrac{3\cdot2\sqrt{2}-2\cdot2\sqrt{3}+2\sqrt{5}}{3\cdot3\sqrt{2}-2\cdot3\sqrt{3}+3\sqrt{5}}\)
\(=\dfrac{6\sqrt{2}-4\sqrt{3}+2\sqrt{5}}{9\sqrt{2}-6\sqrt{3}+3\sqrt{5}}\)
\(=\dfrac{2\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}{3\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}\)
\(=\dfrac{2}{3}\)
a: \(=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}=\dfrac{\sqrt{2}}{2}\)
b: \(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+2\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+2}=1+\sqrt{2}\)
c: \(=\dfrac{6\sqrt{2}-4\sqrt{3}+2\sqrt{5}}{9\sqrt{2}-6\sqrt{3}+3\sqrt{5}}=\dfrac{2}{3}\)
\(\dfrac{-2}{45}:\dfrac{8}{3}=0,01x:4\)
\(0,01x:4=\dfrac{-2}{45}\times\dfrac{3}{8}\)
\(0,01x:4=\dfrac{-6}{360}\)
\(0,01x:4=\dfrac{-1}{60}\)
\(0,01x=\dfrac{-1}{60}\times\dfrac{1}{4}\)
\(0,01x=\dfrac{-1}{240}\)
\(x=\dfrac{-1}{240}:0,01\)
\(x=\dfrac{-1}{240}\times100\)
\(x=\dfrac{-100}{240}=-2,4\)
\(\dfrac{-2}{45}:\dfrac{8}{3}=0,01x:4\)
\(\dfrac{-2}{45}\times\dfrac{3}{8}=0,01x:4\)
\(\dfrac{-6}{360}=0,01x:4\)
\(\dfrac{-1}{60}=0,01x:4\)
\(0,01x=\dfrac{-1}{60}\times\dfrac{4}{1}\)
\(0,01x=\dfrac{-4}{60}\)
\(0,01x=\dfrac{-1}{15}\)
\(x=\dfrac{-1}{15}:0,01\)
\(x=\dfrac{-1}{15}\times100\)
\(x=\dfrac{-100}{15}=\dfrac{-20}{3}\)
Vậy \(x=\dfrac{-20}{3}\).