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T
5
18 tháng 9 2021
Bài 1:
a) \(\left|3x-5\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-5=4\\3x-5=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)
c) \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Leftrightarrow x=-2004\)( do \(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\))
Bài 2:
a) \(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)
\(=\dfrac{1}{4}+\dfrac{3}{4}=1\)
b) \(=-\left(\dfrac{1}{99.100}+\dfrac{1}{98.99}+\dfrac{1}{97.98}+...+\dfrac{1}{2.3}+\dfrac{1}{1.2}\right)\)
\(=-\left(\dfrac{1}{99}-\dfrac{1}{100}+\dfrac{1}{98}-\dfrac{1}{99}+...+1-\dfrac{1}{2}\right)\)
\(=-\left(1-\dfrac{1}{100}\right)=-\dfrac{99}{100}\)
18 tháng 9 2021
Bài 1:
a) \(\left|3x-5\right|=4\) (1)
\(\Leftrightarrow\left[{}\begin{matrix}3x-5=4\\3x-5=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=9\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)
b) \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\) \(\left(do\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\right)\)
\(\Leftrightarrow x=-1\)
c) \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Leftrightarrow\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)\)
\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Leftrightarrow x+2004=0\) \(\left(do\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\right)\)
\(\Leftrightarrow x=-2004\)
HT
31 tháng 8 2015
|x - 5/3| < 1/3
=> x - 5/3 < 1/3 và x - 5/3 > -1/3
=> x < 2 và x > 4/3
=> 4/3 < x < 2
|x + 11/2| > |-5,5|
|x + 11/2| > 5,5
=> x + 11/2 > 5,5 hoặc x + 11/2 < -5,5
=> x > 0 hoặc x < -11
YP
2
TH
17 tháng 11 2021
a,Ta có:
\(\dfrac{x}{y}=\dfrac{7}{4}=\dfrac{x}{7}=\dfrac{y}{4}\)
ÁP dụng tcdtsbn , ta có:
\(\dfrac{x}{7}=\dfrac{y}{4}=\dfrac{x+y}{7+4}=\dfrac{33}{11}=3\)
\(\Rightarrow\left\{{}\begin{matrix}x=21\\y=12\end{matrix}\right.\)
b,
\(\Rightarrow3.\left(x-1\right)=-24\)
\(\Rightarrow x-1=-8\)
\(\Rightarrow x=-7\)
17 tháng 11 2021
A)\(\dfrac{x}{y}=\dfrac{7}{4}\Rightarrow\dfrac{x}{7}=\dfrac{y}{4}\)
Áp dụng t/c dtsbn ta có:
\(\dfrac{x}{7}=\dfrac{y}{4}=\dfrac{x+y}{7+4}=\dfrac{33}{11}=3\)
\(\dfrac{x}{7}=3\Rightarrow x=21\\ \dfrac{y}{4}=3\Rightarrow y=12\)
B) \(3\left(x-1\right)+5=-19\\ \Rightarrow3\left(x-1\right)=-24\\ \Rightarrow x-1=-8\\ \Rightarrow x=-7\)
NT
0
HT
3
27 tháng 5 2018
b) | x + 11/2 | > |-5,5| hay 5,5
Xét :
+) x + 11/2 > 5,5
<=> x > 0
+) -x - 11/2 > 5.5
<=> -x > 11 hay x > -11
vậy....
27 tháng 5 2018
a) | x - 5/3 | > 1/3
Xét :
+) x - 5/3 > 1/3
<=> x > 2 ( tm )
+) -x + 5/3 > 1/3
<=> -x > -4/3 => x > 4/3 (tm)
Vậy,.....