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thay x = -0,3 vào f(x) ta có
f(-0,3) = (-0,3)3+0,027.(-0,3)2-2019
f(-0,3) = -0,9+0,027.0,6-2019
f(-0,3) = -0,9+0,0162-2019
f(-0,3) = 0,9162-2019
f(-0,3) = -2018,0838
`Answer:`
\(f\left(x\right)=x^3+0,027x^2-2019\)
\(\Rightarrow f\left(-0,3\right)=\left(-0,3\right)^3+0,027.\left(-0,3\right)^2-2019\)
\(\Rightarrow f\left(-0,3\right)=-0,027+0,027.0,09-2019\)
\(\Rightarrow f\left(-0,3\right)=-0,027+0,00243-2019\)
\(\Rightarrow f\left(-0,3\right)=-0,02457-2019\)
\(\Rightarrow f\left(-0,3\right)=-2019,02457\)
\(=\left(\dfrac{1}{2}+\dfrac{1}{2}\right)-\left(\dfrac{5}{41}+\dfrac{36}{41}\right)=1-1=0\)
\(A=9-\frac{3}{5}+\frac{2}{3}-7-\frac{7}{5}+\frac{3}{2}-3+\frac{9}{5}-\frac{5}{2}\)
\(=\left(9-7-3\right)+\left(\frac{9}{5}-\frac{7}{5}-\frac{3}{5}\right)+\left(\frac{3}{2}-\frac{5}{2}\right)\)
\(=-2-\frac{1}{5}=-\frac{11}{5}\)
b) = \(\frac{3}{4}\div\)\(\left(-\frac{1}{3}+\frac{2}{3}+\frac{1}{2}\right)\)
= \(\frac{3}{4}\div\frac{5}{6}\)
= \(\frac{9}{10}\)
c) \(\frac{16.2^3}{4}\)
\(=4.8=32\)
\(a)\left|-\frac{1}{2}\right|+3^0+\frac{1}{4}+4+2021^0.\)
\(=\frac{1}{2}+1+\frac{1}{4}+4+1\)
\(=\left(\frac{1}{2}+\frac{1}{4}\right)+\left(1+4+1\right)\)
\(=\frac{3}{4}+6=\frac{27}{4}\)
\(b)\frac{3}{4}\div\left(-\frac{1}{3}\right)+\frac{3}{4}\div\frac{2}{3}+\frac{3}{4}\div\frac{1}{2}\)
\(=\frac{3}{4}\div\left(-\frac{1}{3}+\frac{2}{3}+\frac{1}{2}\right)\)
\(=\frac{3}{4}\div\frac{5}{6}=\frac{9}{10}\)
Bài 1:
$0,2-\frac{4}{7}+\frac{-6}{5}=\frac{1}{5}+\frac{-6}{5}-\frac{4}{7}$
$=\frac{-5}{5}-\frac{4}{7}=-1-\frac{4}{7}=\frac{-11}{7}$
b.
$=(\frac{-2}{3})^2+\frac{5}{6}+(-1)=\frac{4}{9}+\frac{5}{6}-1$
$=\frac{8}{18}+\frac{15}{18}-1=\frac{23}{18}-1=\frac{5}{18}$
c.
$=1+3+5+7+9=25$
Bài 2:
a. $-(0,5+x)-\frac{1}{3}=\frac{1}{6}$
$-(0,5+x)=\frac{1}{3}+\frac{1}{6}=\frac{1}{2}$
$0,5+x=\frac{-1}{2}$
$x=\frac{-1}{2}-0,5=-1$
b.
$(x+\frac{4}{9})(x-\frac{11}{5})=0$
$\Rightarrow x+\frac{4}{9}=0$ hoặc $x-\frac{11}{5}=0$
$\Rightarrow x=\frac{-4}{9}$ hoặc $x=\frac{11}{5}$
c.
$\frac{1}{3}-|\frac{5}{4}-2x|=\frac{1}{4}$
$|\frac{5}{4}-2x|=\frac{1}{3}-\frac{1}{4}=\frac{1}{12}$
$\Rightarrow \frac{5}{4}-2x=\frac{1}{12}$ hoặc $\frac{5}{4}-2x=\frac{-1}{12}$
$\Rightarrow x=\frac{1}{2}(\frac{5}{4}-\frac{1}{12})$ hoặc $x=\frac{1}{2}(\frac{5}{4}+\frac{1}{12})$
$\Rightarrow x=\frac{7}{12}$ hoặc $x=\frac{2}{3}$
\(\frac{3}{5}.\left(\frac{5}{3}-\frac{2}{7}\right)-\left(\frac{7}{3}-\frac{3}{7}\right).\frac{3}{5}\)
\(=\frac{3}{5}.\text{[}\left(\frac{5}{3}-\frac{2}{7}\right)-\left(\frac{7}{3}-\frac{3}{7}\right)\text{]}\)
\(=\frac{3}{5}.\text{[}\frac{5}{3}-\frac{2}{7}-\frac{7}{3}+\frac{3}{7}\text{]}\)
\(=\frac{3}{5}.\text{[}\left(\frac{5}{3}-\frac{7}{3}\right)-\left(\frac{2}{7}-\frac{3}{7}\right)\text{]}\)
\(=\frac{3}{5}.\text{[}\frac{-2}{3}-\frac{-1}{7}\text{]}\)
\(=\frac{3}{5}.\left(\frac{-2}{3}+\frac{1}{7}\right)\)
\(=\frac{3}{5}.\left(\frac{-14}{21}+\frac{3}{21}\right)\)
\(=\frac{3}{5}.\frac{-11}{21}\)
\(=\frac{3.\left(-11\right)}{5.21}\)
\(=\frac{-11}{5.7}=\frac{-11}{35}\)
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ta thấy
6,3 x 12-21 x 3,6 = 3,6 x 7/4 x 12 - 21 x 3,6 = 3,6 x 21 - 21 x 3,6 = 0
Vậy (1+2+..+100)X(6,3 x 12-21 x 3,6)x(1/3-1/5-1/7-1/9)=0