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\(\dfrac{x+24}{1996}+\dfrac{x+25}{1995}+\dfrac{x+26}{1994}+\dfrac{x+27}{1993}+\dfrac{x+2036}{4}=0\)
\(\Rightarrow\left(\dfrac{x+24}{1996}+1\right)+\left(\dfrac{x+25}{1995}+1\right)+\left(\dfrac{x+26}{1994}+1\right)+\left(\dfrac{x+27}{1993}+1\right)+\left(\dfrac{x+2036}{4}-4\right)=0\)\(\Rightarrow\dfrac{x+2020}{1996}+\dfrac{x+2020}{1995}+\dfrac{x+2020}{1994}+\dfrac{x+2020}{1993}+\dfrac{x+2020}{4}=0\)\(\Rightarrow\left(x+2020\right)\left(\dfrac{1}{9996}+\dfrac{1}{1995}+\dfrac{1}{1994}+\dfrac{1}{1993}+\dfrac{1}{4}\right)=0\)
\(\Rightarrow x+2020=0\Rightarrow x=-2020\)
\(\frac{x+24}{1996}+\frac{x+25}{1995}+\frac{x+26}{1994}+\frac{x+27}{1993}+\frac{x+2036}{4}=0\)
\(\Rightarrow\left(\frac{x+24}{1996}+1\right)+\left(\frac{x+25}{1995}+1\right)+\left(\frac{x+26}{1994}+1\right)+\left(\frac{x+27}{1993}+1\right)+\left(\frac{x+2036}{4}-4\right)=0\)
\(\Rightarrow\frac{x+2020}{1996}+\frac{x+2020}{1995}+\frac{x+2020}{1994}+\frac{x+2020}{1993}+\frac{x+2020}{4}=0\)
\(\Rightarrow\left(x+2020\right)\left(\frac{1}{4}+\frac{1}{1993}+\frac{1}{1994}+\frac{1}{1995}+\frac{1}{1996}\right)=0\)
Vì \(\left(\frac{1}{4}+\frac{1}{1993}+\frac{1}{1994}+\frac{1}{1995}+\frac{1}{1996}\right)\ne0\)nên \(x+2020=0\Rightarrow x=-2020\)
Vậy x = -2020
Ta có \(\frac{x+24}{1996}+\frac{x+25}{1995}+\frac{x+26}{1994}+\frac{x+27}{1993}+\frac{x+2036}{4}\)
\(\Leftrightarrow\left(\frac{x+24}{1996}+1\right)+\left(\frac{x+25}{1995}+1\right)+\left(\frac{x+26}{1994}\right)+\left(\frac{x+27}{1993}\right)+\left(\frac{x+2036}{4}-4\right)=0\)
\(\Leftrightarrow\frac{x+2020}{1996}+\frac{x+2020}{1995}+\frac{x+2020}{1994}+\frac{x+2020}{1993}+\frac{x+2020}{4}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{1996}+\frac{1}{1995}+\frac{1}{1994}+\frac{1}{1993}\right)=0\)
\(V\text{ì}\) \(\frac{1}{1996}+\frac{1}{1995}+\frac{1}{1994}+\frac{1}{1993}+\frac{1}{4}\ne0\)
\(\Rightarrow x+2020=0\Leftrightarrow x=-2020\)
Vậy phương trình có tập nghiệm \(S=\left\{-2020\right\}\)
a) 2 x 3 x 4 x 8 x 50 x 25 x 125
= (2 x 50) x ( 4 x 25 ) x (8 x 125) x 3 = 100 x 100 x 1000 x 3
= 30 000 000
a/ 1 . 3 . 5 . ... . 59
= 5 . [số lẻ] = [...5] . Vậy tận cùng là 5
b/32 . 44 . 75.69 - 21 . 49 . 65 . 55
= [75 . 32] . 44 . 69 - [21 . 49 . 65] . 55
=[...0] . 44 . 69 - [số lẻ] . 55
=[...0] - [...5] = [...5]. vậy tận cùng bằng 5
c/1991 . 1992 . 1993 . 1994 + 1995 . 1996 . 1997 . 1998 . 1999
= [1991 . 1992 . 1993] . 1994 + [1995 . 1996] . 1997 . 1998 . 1999
=...6 . 1994 + ...0 . 1997 . 1998 . 1999
=...4 + ...0 = ...4 . Vậy tận cùng là 4
Ta có:
(x+2)/2012+(x+2)/2011+...+(x+2)/1996 = (x+2)/1995+(x+2)/1994+ ... + (x+ 2) / 1002
=> (x+2)(1/2012+1/2011+...+1/1996) = (x+2)(1/1995+1/1994+...+1/1002)
=> [(x+2)(1/2012+1/2011+...+1/1996)] - [(x+2)(1/1995+1/1994+...+1/1002)]=0
=> (x+2)[(1/2012+1/2011+...+1/1996)+(1/1995+1/1994+...+1/1002)]=0
=> x+2=0 (Vì (1/2012+1/2011+...+1/1996)+(1/1995+1/1994+...+1/1002) không thể bằng 0 được)
x=0-2=-2
d) Ta có: \(32\%-0.25:x=-\dfrac{17}{5}\)
\(\Leftrightarrow0.25:x=\dfrac{8}{25}+\dfrac{17}{5}=\dfrac{93}{25}\)
hay \(x=\dfrac{25}{372}\)
Vậy: \(x=\dfrac{25}{372}\)
e) Ta có: \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)
\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{2}{5};-\dfrac{4}{5}\right\}\)
f) Ta có: \(-\dfrac{32}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)
\(\Leftrightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-8}{27}\)
\(\Leftrightarrow3x-\dfrac{7}{9}=-\dfrac{2}{3}\)
\(\Leftrightarrow3x=\dfrac{1}{9}\)
hay \(x=\dfrac{1}{27}\)
g) Ta có: \(60\%\cdot x+0.4x+x:3=2\)
\(\Leftrightarrow\dfrac{4}{3}x=2\)
hay \(x=\dfrac{3}{2}\)
Vậy: \(x=\dfrac{3}{2}\)
h) PT \(\Leftrightarrow\left|\dfrac{20}{9}-x\right|=\dfrac{2}{9}\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{20}{9}-x=\dfrac{2}{9}\\x-\dfrac{20}{9}=\dfrac{2}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{22}{9}\end{matrix}\right.\)
Vậy ...
i) PT \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}x=\dfrac{16}{5}\) \(\Leftrightarrow\dfrac{2}{5}x=\dfrac{8}{5}\) \(\Leftrightarrow x=4\)
Vậy ...
