Tính:a) 8x5 : 4x3b) 120x7 : (-24x5)c) \(\dfrac{3}{4}{( - x)^3}:\dfrac{1}{8}x\)d) -3,72x4 :...
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Tính:
a) 8x5 : 4x3
b) 120x7 : (-24x5)
c) \(\dfrac{3}{4}{( - x)^3}:\dfrac{1}{8}x\)
d) -3,72x4 : (-4x2)
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O
10 tháng 5 2022
a) \(\dfrac{11}{10}+\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{11}{10}+\dfrac{3}{5}\times\dfrac{3}{2}=\dfrac{11}{10}+\dfrac{9}{10}=\dfrac{20}{10}=2\)
b) \(\dfrac{4}{3}+5\times\dfrac{5}{8}=\dfrac{4}{3}+\dfrac{25}{8}=\dfrac{32}{24}+\dfrac{75}{24}=\dfrac{107}{24}\)
c) \(\left(\dfrac{2}{5}+\dfrac{3}{7}\right)\times\dfrac{25}{29}=\left(\dfrac{14}{35}+\dfrac{15}{35}\right)\times\dfrac{25}{39}=\dfrac{29}{35}\times\dfrac{25}{39}=\dfrac{145}{274}\)
d) \(\dfrac{1}{4}\times\dfrac{5}{12}+\dfrac{5}{12}\times\dfrac{4}{5}=\dfrac{5}{12}\times\left(\dfrac{1}{4}+\dfrac{4}{5}\right)=\dfrac{5}{12}\times\dfrac{21}{20}=\dfrac{105}{240}=\dfrac{7}{16}\)
C
10 tháng 5 2022
a) \(\dfrac{11}{10}+\dfrac{3}{5}x\dfrac{3}{2}=\dfrac{11}{10}+\dfrac{9}{10}=\dfrac{20}{10}=2\)
b) \(\dfrac{4}{3}+\dfrac{25}{8}=\dfrac{32}{24}+\dfrac{75}{24}=\dfrac{107}{24}\)
c) \(\dfrac{29}{35}x\dfrac{25}{29}=\dfrac{5}{7}\)
\(=\dfrac{5}{12}x\left(\dfrac{1}{4}+\dfrac{4}{5}\right)=\dfrac{5}{12}x\dfrac{21}{20}=\dfrac{7}{16}\)
14 tháng 5 2022
a: \(=\dfrac{2x-2x+y}{2\left(2x-y\right)}=\dfrac{y}{2\left(2x-y\right)}\)
b: \(=\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{6x+2-x^2-x}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x^2+5x+2}{2\left(x-1\right)\left(x+1\right)}\)
c: \(=\dfrac{1}{x+2}+\dfrac{x+8}{3x\left(x+2\right)}\)
\(=\dfrac{3x+x+8}{3x\left(x+2\right)}=\dfrac{4x+8}{3x\left(x+2\right)}=\dfrac{4}{3x}\)
d: \(=\dfrac{4x+6-2x^2+3x+2x+1}{\left(2x-3\right)\left(2x+3\right)}\)
\(=\dfrac{-2x^2+9x+7}{\left(2x-3\right)\left(2x+3\right)}\)
14 tháng 9 2023
a: \(=\dfrac{13}{3}+\dfrac{17}{6}=\dfrac{26}{6}+\dfrac{17}{6}=\dfrac{43}{6}\)
b: \(=7-\dfrac{8}{3}=\dfrac{21-8}{3}=\dfrac{13}{3}\)
c: \(=\dfrac{17}{7}\cdot\dfrac{7}{4}=\dfrac{17}{4}\)
d: \(=\dfrac{16}{3}:\dfrac{16}{5}=\dfrac{16}{3}\cdot\dfrac{5}{16}=\dfrac{5}{3}\)
F
26 tháng 12 2021
a)\(\dfrac{x^2}{x-1}+\dfrac{1-2x}{x-1}\)
=\(\dfrac{x^2+1-2x}{x-1}\)
=\(\dfrac{x^2-2x+1}{x-1}\)
=\(\dfrac{\left(x-1\right)^2}{x-1}\)
= x - 1
F
26 tháng 12 2021
b) \(\dfrac{x}{x-3}\) + \(\dfrac{-9}{x^2-3x}\)
=\(\dfrac{x}{x-3}\)+ \(\dfrac{-9}{x\left(x-3\right)}\)
=\(\dfrac{x.x}{x\left(x-3\right)}\) + \(\dfrac{-9}{x\left(x-3\right)}\)
=\(\dfrac{x^2+3^2}{x\left(x-3\right)}\)
=\(\dfrac{\left(x+3\right)\left(x-3\right)}{x\left(x-3\right)}\)
=\(\dfrac{x+3}{x}\)
#Fiona
HQ
Hà Quang Minh
Giáo viên
27 tháng 11 2023
a) $\frac{2}{5} \times \frac{3}{8} \times \frac{3}{4} = \frac{{2 \times 3 \times 3}}{{5 \times 8 \times 4}} = \frac{{18}}{{160}} = \frac{9}{{80}}$
b) $\frac{1}{3} \times \frac{1}{6} \times \frac{1}{9} = \frac{{1 \times 1 \times 1}}{{3 \times 6 \times 9}} = \frac{1}{{162}}$
c) $\frac{3}{4}:\frac{1}{5}:\frac{7}{8} = \frac{3}{4} \times \frac{5}{1} \times \frac{8}{7} = \frac{{3 \times 5 \times 8}}{{4 \times 1 \times 7}} = \frac{{120}}{{28}} = \frac{{30}}{7}$
d) $\frac{3}{5}:\frac{1}{5}:\frac{3}{8} = \frac{3}{5} \times \frac{5}{1} \times \frac{8}{3} = \frac{{3 \times 5 \times 8}}{{5 \times 1 \times 3}} = 8$
31 tháng 8 2021
a: \(\dfrac{-1}{2}+\dfrac{5}{6}+\dfrac{1}{3}\)
\(=\dfrac{-3}{6}+\dfrac{5}{6}+\dfrac{2}{6}\)
\(=\dfrac{4}{6}=\dfrac{2}{3}\)
b: \(\dfrac{-3}{8}+\dfrac{7}{4}-\dfrac{1}{12}\)
\(=\dfrac{-9}{24}+\dfrac{42}{24}-\dfrac{2}{24}\)
\(=\dfrac{31}{24}\)
c: \(\dfrac{3}{5}:\left(\dfrac{1}{4}\cdot\dfrac{7}{5}\right)=\dfrac{3}{4}:\dfrac{7}{20}=\dfrac{3}{4}\cdot\dfrac{20}{7}=\dfrac{15}{7}\)
d: \(\dfrac{10}{11}+\dfrac{4}{11}:4-\dfrac{1}{8}\)
\(=\dfrac{10}{11}+\dfrac{1}{11}-\dfrac{1}{8}=\dfrac{7}{8}\)
10 tháng 7 2021
a) Ta có: \(\left(x-1\right)\left(x-2\right)\left(x^2+x+1\right)\left(x^2+2x+4\right)-x^6+9x^3\)
\(=\left(x-1\right)\left(x^2+x+1\right)\left(x-2\right)\left(x^2+2x+4\right)-x^6+9x^3\)
\(=\left(x^3-1\right)\left(x^3-8\right)-x^6+9x^3\)
\(=x^6-9x^3+8-x^6+9x^3=8\)
b) Ta có: \(\left(\dfrac{1}{3}+2x\right)\left(\dfrac{1}{9}-\dfrac{2}{3}x+4x^2\right)-\left(2x-\dfrac{1}{3}\right)\left(4x^2+\dfrac{2}{3}x+\dfrac{1}{4}\right)\)
\(=\dfrac{1}{27}+8x^3-8x^3+\dfrac{1}{27}\)
\(=\dfrac{2}{27}\)
c) Ta có: \(\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3x\left(1-x\right)\)
\(=x^3-3x^2+3x-1-x^3+1-3x+3x^2\)
=0
d) Ta có: \(\left(x^2-y^2\right)\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)-x^6+y^6\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)-x^6+y^6\)
\(=\left(x^3-y^3\right)\left(x^3+y^3\right)-x^6+y^6\)
\(=x^6-y^6-x^6+y^6=0\)
em c.ơn trước
a) 8x5 : 4x3 = (8 : 4) . (x5 : x3) = 2.x2
b) 120x7 : (-24x5) = [120 : (-24)] . (x7 : x5) = -5.x2
c) \(\dfrac{3}{4}{( - x)^3}:\dfrac{1}{8}x = \dfrac{{ - 3}}{4}{x^3}:\dfrac{1}{8}x = \left( {\dfrac{{ - 3}}{4}:\dfrac{1}{8}} \right).({x^3}:x) = - 6{x^2}\)
d) -3,72x4 : (-4x2) = [(-3,72) : (-4)] . (x4 : x2) = 0,93x2