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17 tháng 10 2021
a: Ta có: \(\dfrac{x-2}{x-1}=\dfrac{x+4}{x+7}\)
\(\Leftrightarrow x^2+5x-10=x^2+3x-4\)
\(\Leftrightarrow2x=6\)
hay x=3
12 tháng 4 2022
a.Giả sử: \(A\left(x\right)=0\)
\(\Rightarrow9-3x=0\)
\(-3x=-9\)
\(x=3\)
b. Giả sử \(B\left(x\right)=0\)
\(\Rightarrow x^3+x=0\)
\(x\left(x^2+1\right)=0\)
\(x=0\) ( vì \(x^2+1\ge1>0\) )
c.Giả sử: \(C\left(x\right)=0\)
\(\Rightarrow x^2+5=0\) ( vô lí ) ( vì \(x^2+5\ge5>0\) )
d.Giả sử: \(D\left(x\right)=0\)
\(\Rightarrow\left(x+5\right)\left(\left|x\right|-1\right)=0\)
\(\left[{}\begin{matrix}x+5=0\\\left|x\right|-1=0\end{matrix}\right.\) \(\left[{}\begin{matrix}x=5\\x=\pm1\end{matrix}\right.\)
17 tháng 9 2021
a) \(\left|x-17\right|=2,3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-17=2,3\\x-17=-2,3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=19,3\\x=14,7\end{matrix}\right.\)
b) \(\left|x+\dfrac{3}{4}\right|=0\)
\(\Leftrightarrow x+\dfrac{3}{4}=0\Leftrightarrow x=-\dfrac{3}{4}\)
c) \(\left|x+\dfrac{3}{4}\right|+\dfrac{1}{3}=0\)
\(\Leftrightarrow\left|x+\dfrac{3}{4}\right|=-\dfrac{1}{3}\)( vô lý do \(\left|x+\dfrac{3}{4}\right|\ge0\forall x\))
Vậy \(S=\varnothing\)
15 tháng 10 2021
b: \(\left|x+\dfrac{1}{3}\right|-4=-2\)
\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=2\\x+\dfrac{1}{3}=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{7}{3}\end{matrix}\right.\)
CM
1 tháng 7 2017
Thay x=-4 vào f(x) ta có
5/4.(-4)a + a + 1 = 0 ⇒ -5a + a + 1 = 0
⇒ -4a + 1 = 0 ⇒ a = 1/4. Chọn C
HV
2
16 tháng 12 2021
a: \(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{5}\\x-\dfrac{3}{4}=-\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{19}{20}\\x=\dfrac{11}{20}\end{matrix}\right.\)
16 tháng 12 2021
\(a,\left|x-\dfrac{3}{4}\right|=\dfrac{1}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{5}\\x-\dfrac{3}{4}=\dfrac{-1}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{19}{20}\\x=\dfrac{11}{20}\end{matrix}\right.\)
\(b,\dfrac{-1}{3}+\left|x\right|=0,5\)
\(\Leftrightarrow\left|x\right|=\dfrac{5}{6}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{6}\\x=\dfrac{-5}{6}\end{matrix}\right.\)
17 tháng 7 2021
a) Ta có: \(\dfrac{1}{4}-\left|x+\dfrac{1}{2}\right|=\dfrac{1}{8}\)
\(\Leftrightarrow\left|x+\dfrac{1}{2}\right|=\dfrac{1}{8}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{8}\\x+\dfrac{1}{2}=-\dfrac{1}{8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{8}\\x=\dfrac{-5}{8}\end{matrix}\right.\)
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) \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Leftrightarrow\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)\)
\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Leftrightarrow x+2004=0\) \(\left(do\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\right)\)
\(\Leftrightarrow x=-2004\)
