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3: \(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
\(\Leftrightarrow\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)
\(\left[{}\begin{matrix}\dfrac{x}{3}-\dfrac{12}{x}=0\\5:x-1=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{x^2-36}{3x}=0\\\dfrac{5}{x}=1\end{matrix}\right.\)
\(\left[{}\begin{matrix}=>x=6và\left(-6\right)\\=>x=5\end{matrix}\right.\)
\(ĐKXĐ:x\ne1,x\ne3,x\ne8,x\ne20\)
\(\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}-\dfrac{1}{x-20}=-\dfrac{3}{4}\)
\(\Rightarrow\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}-\dfrac{1}{x-20}=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{2\left(x-8\right)\cdot\left(x-20\right)+5\left(x-1\right)\cdot\left(x-20\right)+12\left(x-1\right)\cdot\left(x-3\right)-\left(x-1\right)\cdot\left(x-3\right)\cdot\left(x-8\right)}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{\left(2x-16\right)\cdot\left(x-20\right)+\left(5x-5\right)\cdot\left(x-20\right)+\left(12x-12\right)\cdot\left(x-3\right)-\left(x^2-3x-x+3\right)\cdot\left(x-8\right)}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{2x^2-40x-16x+320+5x^2-100x-5x+100+12x^2-36x-12x+36-\left(x^2-4x+3\right)\cdot\left(x-8\right)}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{2x^2-40x-16x+320+5x^2-100x-5x+100+12x^2-36x-\left(x^3-8x^2-4x^2+32x+3x-24\right)}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}\)
\(\Leftrightarrow\dfrac{2x^2-40x-16x+320+5x^2-100x-5x+100+12x^2-36x-12x+36-\left(x^3-12x^2+35x-24\right)}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{2x^2-40x-16x+320+5x^2-100x-5x+100+12x^2-36x-12x+36-x^3+12x^2-35x+24}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{31x^2-244x+480-x^3}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{-x^3+31x^2-244x+480}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{-x^3+3x^2+28x^2-84x-160x+480}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{-x^2\cdot\left(x-3\right)+28x\cdot\left(x-3\right)-160\left(x-3\right)}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{-\left(x-3\right)\left(x^2-28x+160\right)}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{-1\left(x^2-8x-20x+160\right)}{\left(x-1\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{-1\left(x^2-8x-20x+160\right)}{\left(x-1\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{-1\left(x\cdot\left(x-8\right)-20\left(x-8\right)\right)}{\left(x-1\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{-1\left(x-20\right)\left(x-8\right)}{\left(x-1\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{-1}{x-1}=-\dfrac{3}{4}\)
\(\Leftrightarrow-\dfrac{1}{x-1}=-\dfrac{3}{4}\)
\(\Leftrightarrow-4=-3\left(x-1\right)\)
\(\Leftrightarrow-4=-3\left(x-1\right)\)
\(\Leftrightarrow-4=-3x+3\)
\(\Leftrightarrow3x=3+4\)
\(\Leftrightarrow3x=7\)
\(\Rightarrow x=\dfrac{7}{3}\)
Vậy \(x=\dfrac{7}{3}\)
cho ngu ké với bài này lớp 5 dư sức làm áp dụng 1/n(n+1)=1/n-1/n+1
\(\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}+\dfrac{1}{x-20}=\dfrac{3}{4}\)
\(\Rightarrow\dfrac{1}{x-1}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-8}+\dfrac{1}{x-8}-\dfrac{1}{x-20}+\dfrac{1}{x-20}=\dfrac{3}{4}\)
\(\Rightarrow\dfrac{1}{x-1}=\dfrac{3}{4}\Rightarrow3x-3=4\Rightarrow x=\dfrac{7}{3}\)
Vậy...
a: \(x=\left(-\dfrac{2}{3}\right)^5:\left(-\dfrac{2}{3}\right)^2=\left(-\dfrac{2}{3}\right)^3=-\dfrac{8}{27}\)
b: =>x-1/2=1/3
=>x=5/6
c: =>2/3x-1=0 hoặc 3/4x+1/2=0
=>x=3/2 hoặc x=-1/2:3/4=-1/2*4/3=-4/6=-2/3
d =>4/9:x=10/3:9/4=10/3*4/9=40/27
=>x=4/9:40/27=4/9*27/40=108/360=3/10
a/dễ --> tự lm
b/ \(\left(x-\dfrac{4}{7}\right)\left(1\dfrac{3}{5}+2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{5}=0\\1\dfrac{3}{5}+2x=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\2x=\dfrac{8}{5}\Rightarrow x=\dfrac{4}{5}\end{matrix}\right.\)
Vậy...............
c/ \(\left(x-\dfrac{4}{7}\right):\left(x+\dfrac{1}{2}\right)>0\)
TH1: \(\left\{{}\begin{matrix}x-\dfrac{4}{7}>0\\x+\dfrac{1}{2}>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>\dfrac{4}{7}\\x>-\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow x>\dfrac{4}{7}\)
TH2: \(\left\{{}\begin{matrix}x-\dfrac{4}{7}< 0\\x+\dfrac{1}{2}< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x< \dfrac{4}{7}\\x< -\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow x< -\dfrac{1}{2}\)
Vậy \(x>\dfrac{4}{7}\) hoặc \(x< -\dfrac{1}{2}\) thì thỏa mãn đề
d/ \(\left(2x-3\right):\left(x+1\dfrac{3}{4}\right)< 0\)
TH1: \(\left\{{}\begin{matrix}2x-3>0\\x+1\dfrac{3}{4}< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>1,5\\x< -\dfrac{7}{4}\end{matrix}\right.\)(vô lý)
TH2: \(\left\{{}\begin{matrix}2x-3< 0\\x+1\dfrac{3}{4}>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x< 1,5\\x>-\dfrac{7}{4}\end{matrix}\right.\)\(\Rightarrow-\dfrac{7}{4}< x< 1,5\)
Vậy...................
Sửa đề:
\(\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}-\dfrac{1}{x-20}=-\dfrac{3}{4}\)
ĐKXĐ: \(x\notin\left\{1;3;8;20\right\}\)
PT=>\(-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-8}-\dfrac{1}{x-8}+\dfrac{1}{x-20}-\dfrac{1}{x-20}=-\dfrac{3}{4}\)
=>\(-\dfrac{1}{x-4}=-\dfrac{3}{4}\)
=>\(x-1=\dfrac{4}{3}\)
=>\(x=\dfrac{4}{3}+1=\dfrac{7}{3}\)(nhận)