.......x3........-4=11
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PQ
2
23 tháng 4 2023
a, = \(\dfrac{3}{4}\)x\(\dfrac{41}{24}\) = \(\dfrac{41}{32}\)
b, = \(\dfrac{1}{11}\) : \(\dfrac{1}{42}\) - \(\dfrac{20}{11}\) = \(\dfrac{42}{11}\) - \(\dfrac{20}{11}\) = \(\dfrac{22}{11}\) = 2
c, = \(\dfrac{26}{7}\) x 3 : \(\dfrac{1}{3}\) = \(\dfrac{78}{7}\) : \(\dfrac{1}{3}\) = \(\dfrac{234}{7}\)
Học tốt nhá
29 tháng 1 2023
\(\dfrac{22}{5}\times\dfrac{6}{121}\times\dfrac{11}{4}\times\dfrac{3}{5}\times\dfrac{1}{3}\times\dfrac{5}{4}\)
\(=\left(\dfrac{22}{5}\times\dfrac{5}{4}\right)\times\left(\dfrac{6}{121}\times\dfrac{11}{4}\right)\times\left(\dfrac{3}{5}\times\dfrac{1}{3}\right)\)
\(=\dfrac{11}{2}\times\dfrac{3}{22}\times\dfrac{1}{5}\)
\(=\dfrac{3}{20}\)
TD
29 tháng 1 2023
\(=\dfrac{22\times6\times11\times3\times1\times5}{5\times121\times4\times5\times3\times4}=\dfrac{11\times2\times6\times11\times1}{11\times11\times4\times5\times4}=\dfrac{2\times6\times1}{4\times5\times4}=\dfrac{18}{100}=\dfrac{9}{50}\)
30 tháng 10 2021
\(A=\left(-\dfrac{5}{9}\right).\dfrac{3}{11}+\left(-\dfrac{13}{18}\right).\dfrac{3}{11}=\dfrac{3}{11}\left(-\dfrac{5}{9}-\dfrac{13}{18}\right)\)
\(=\dfrac{3}{11}.\dfrac{-23}{18}=-\dfrac{23}{66}\)
1 tháng 6 2021
\(3\dfrac{1}{3}.x+11\dfrac{3}{4}=\left(-13,25\right)\)
\(\dfrac{10}{3}.x+\dfrac{47}{4}=\dfrac{-53}{4}\)
\(\dfrac{10}{3}.x=\dfrac{-53}{4}-\dfrac{47}{4}\)
\(\dfrac{10}{3}.x=-25\)
\(x=-25:\dfrac{10}{3}\)
\(x=\dfrac{-15}{2}\)
CM
31 tháng 1 2019
a) Đa thức thương x 2 – 6x + 9.
b) Đa thức thương 2 x 2 – 5.
c) Đa thức thương x 2 + 4x + 3 và đa thức dư -12.
d) Đa thức x + 5 và đa thức dư x – 4.
24 tháng 4 2022
(2xy-4/9)3/2=6/11
<=>3xy-12/18=6/11
<=>99xy/33-22/33=18/33
<=>99xy-22=18
<=>99xy=18+22
<=>99xy=40
<=>xy=40/99
VQ
24 tháng 4 2022
(2xy-4/9)3/2=6/11
<=>3xy-12/18=6/11
<=>99xy/33-22/33=18/33
<=>99xy-22=18
<=>99xy=18+22
<=>99xy=40
<=>xy=40/99
H
30 tháng 9 2023
\(\dfrac{1}{x+2}+\dfrac{5}{2x^2+3x-2}\\ =\dfrac{1}{x+2}+\dfrac{5}{\left(2x-1\right)\left(x+2\right)}\\ =\dfrac{2x-1}{\left(2x-1\right)\left(x+2\right)}+\dfrac{5}{\left(2x-1\right)\left(x+2\right)}\\ =\dfrac{2x-1+5}{\left(2x-1\right)\left(x+2\right)}\\ =\dfrac{2x+4}{\left(2x-1\right)\left(x+2\right)}\\ =\dfrac{2\left(x+2\right)}{\left(2x-1\right)\left(x+2\right)}\\ =\dfrac{2}{2x-1}\)
__
`x^3+1` chứ cậu nhỉ?
\(\dfrac{-3x^2}{x^3+1}+\dfrac{1}{x^2-x+1}+\dfrac{1}{x+1}\\ =\dfrac{-3x^2}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{1}{x^2-x+1}+\dfrac{1}{x+1}\\ =\dfrac{-3x^2}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{x^2-x+1}{\left(x-1\right)\left(x^2-x+1\right)}\\ =\dfrac{-3x^2+x+1+x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\\ =\dfrac{-2x^2+2}{\left(x+1\right)\left(x^2-x+1\right)}\\ =\dfrac{-2\left(x^2-1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{-2\left(x-1\right)\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\\ =\dfrac{-2\left(x-1\right)}{x^2-x+1}\)
__
T
30 tháng 9 2023
a) \(\dfrac{1}{x+2}+\dfrac{5}{2x^2+3x-2}\)
\(=\dfrac{1}{x+2}+\dfrac{5}{2x^2+4x-x-2}\)
\(=\dfrac{2x-1}{\left(2x-1\right)\left(x+2\right)}+\dfrac{5}{2x\left(x+2\right)-\left(x+2\right)}\)
\(=\dfrac{2x-1+5}{\left(2x-1\right)\left(x+2\right)}\)
\(=\dfrac{2x+4}{\left(2x-1\right)\left(x+2\right)}\)
\(=\dfrac{2\left(x+2\right)}{\left(2x-1\right)\left(x+2\right)}\)
\(=\dfrac{2}{2x-1}\)
\(---\)
b) \(\dfrac{-3x^2}{x^3+1}+\dfrac{1}{x^2-x+1}+\dfrac{1}{x+1}\) (sửa đề)
\(=\dfrac{-3x^2}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{-3x^2+x+1+x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{-2x^2+2}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{-2\left(x^2-1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{-2\left(x-1\right)\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{-2x+2}{x^2-x+1}\)
\(---\)
c) \(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}\)
\(=\dfrac{1+x}{\left(1-x\right)\left(1+x\right)}+\dfrac{1-x}{\left(1-x\right)\left(1+x\right)}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}\)
\(=\dfrac{1+x+1-x}{1^2-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}\)
\(=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}\)
\(=\dfrac{2\left(1+x^2\right)}{\left(1-x^2\right)\left(1+x^2\right)}+\dfrac{2\left(1-x^2\right)}{\left(1-x^2\right)\left(1+x^2\right)}+\dfrac{4}{1+x^4}\)
\(=\dfrac{2+2x^2+2-2x^2}{1-x^4}+\dfrac{4}{1+x^4}\)
\(=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}\)
\(=\dfrac{4\left(1+x^4\right)}{\left(1-x^4\right)\left(1+x^4\right)}+\dfrac{4\left(1-x^4\right)}{\left(1-x^4\right)\left(1+x^4\right)}\)
\(=\dfrac{4+4x^4+4-4x^4}{1-x^8}\)
\(=\dfrac{8}{1-x^8}\)
#\(Toru\)
5x3-4=11 k cho minh nha