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\(3\cdot\left(\frac{-1}{3}\right)^3+\frac{1}{3}=3\cdot\left(-\frac{1}{27}\right)+\frac{1}{3}\)
\(=-\frac{1}{9}+\frac{1}{3}\)
\(=-\frac{1}{9}+\frac{3}{9}\)
\(=\frac{2}{9}\)
\(\frac{x^2}{2}+\frac{x^2}{3}+\frac{x^2}{4}\)
\(=\frac{6x^2}{12}+\frac{4x^2}{12}+\frac{3x^2}{12}\)
\(=\frac{6x^2+4x^2+3x^2}{12}\)
\(=\frac{13x^2}{12}\)
\(\frac{x^2}{2}+\frac{x^2}{3}+\frac{x^2}{4}\)
\(=x^2.\frac{1}{2}+x^2.\frac{1}{3}+x^2.\frac{1}{4}\)
\(=x^2.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)\)
\(=x^2.\frac{13}{12}\)
a)\(\frac{1}{8}.\frac{3}{5} = \frac{{1.3}}{{8.5}} = \frac{3}{{40}}\)
b)\(\frac{{ - 6}}{7}:\left( { - \frac{5}{3}} \right) = \frac{{ - 6}}{7}.\frac{{ - 3}}{5} = \frac{{18}}{{35}}\)
c)\(0,6.\left( { - 0,15} \right) = \frac{6}{{10}}.\frac{{ - 15}}{{100}} = \frac{{ - 90}}{{1000}} = \frac{{ - 9}}{{100}}\).
\(\frac{-3}{11}:\frac{17}{15}+\frac{-3}{11}:\frac{17}{2}+\frac{8}{11}\)
\(=\frac{-3}{11}.\frac{15}{17}+\frac{-3}{11}.\frac{2}{17}+\frac{8}{11}\)
\(=\frac{-3}{11}.\left(\frac{15}{17}+\frac{2}{17}\right)+\frac{8}{11}\)
\(=\frac{-3}{11}.1+\frac{8}{11}\)
\(=\frac{-3}{11}+\frac{8}{11}\Rightarrow\frac{5}{11}\)
K mik nhe
\(\frac{-3}{11}:\frac{17}{15}+\frac{-3}{11}:\frac{17}{2}+\frac{8}{11}\)
\(=\frac{-3}{11}.\left(\frac{15}{17}+\frac{2}{17}\right)+\frac{8}{11}\)
\(=\frac{-3}{11}.1+\frac{8}{11}\)
=\(\frac{-3}{11}+\frac{8}{11}\)
\(=\frac{5}{11}\)
(\(x\) - 3).(3 + \(x\))
= 3\(x\) + \(x^2\) - 9 - 3\(x\)
= \(x^2\) - 9
\(\left(x-3\right)\left(3+x\right)\\ =\left(x-3\right).3+\left(x-3\right).x\\ =3x-9+2x-3x\\ =3x-9-x\\ =3x-x+9\\ =2x+9.\)
Bài 1:
a) \(=\dfrac{8}{15}\left(\dfrac{7}{13}+\dfrac{6}{13}\right)=\dfrac{8}{15}.1=\dfrac{8}{15}\)
b) \(=\dfrac{3.3-7-2.4}{12}=-\dfrac{6}{12}=-\dfrac{1}{2}\)
Bài 2:
\(\dfrac{x}{2,7}=-\dfrac{2}{3,6}\Rightarrow x=\dfrac{\left(-2\right).2,7}{3,6}\Rightarrow x=-\dfrac{3}{2}\)
Bài 3:
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=-\dfrac{21}{7}=-3\)
\(\Rightarrow\left\{{}\begin{matrix}x=\left(-3\right).2=-6\\y=\left(-3\right).5=-10\end{matrix}\right.\)
a) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Mà \(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)\ne0\)
nên x + 1 = 0 => x = -1
Vậy x = -1
b) \(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(1+\frac{x+4}{2000}+1+\frac{x+3}{2001}=1+\frac{x+2}{2002}+1+\frac{x+1}{2003}\)
\(\frac{2004+x}{2000}+\frac{2004+x}{2001}=\frac{2004+x}{2002}+\frac{2004+x}{2003}\)
\(\frac{2004+x}{2000}+\frac{2004+x}{2001}-\frac{2004+x}{2002}-\frac{2004+x}{2003}=0\)
\(\left(2004+x\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
Mà \(\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)\ne0\)
nên 2004 + x = 0 => x = -2004
Vậy x = -2004
=))
1.(2515.415)/(517.2016)=(530.230)/(517.516.232)=1/(53.22)=1/500
2.a,-x/2=8/-x=>-x.(-x)=2*8 =>x^2=16=(-4)^2=4^2
=>x=4 hoặc x=-4
b,(3/4)2x/(2/5)10=(15/8)10
(3/4)2x=(2/5*15/8)10
(3/4)2x=(3/4)10
2x=10
x=5
1) \(\frac{25^{15}\cdot4^{15}}{5^{17}\cdot20^{16}}\)
\(=\frac{5^{30}\cdot2^{30}}{5^{33}\cdot2^{32}}\)
\(=\frac{1}{5^3\cdot2^2}\)
\(=\frac{1}{500}\)
2)
a) \(\frac{-x}{2}=\frac{8}{-x}\)
\(\Rightarrow\left(-x\right)\left(-x\right)=8\cdot2\)
\(\Rightarrow x^2=16\)
\(\Rightarrow x=\left\{\pm4\right\}\)
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