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\(\left|x+\dfrac{1}{1.5}\right|+\left|x+\dfrac{1}{5.9}\right|+\left|x+\dfrac{1}{9.14}\right|+...+\left|x+\dfrac{1}{397.401}\right|\ge0\)
\(\Rightarrow101x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow x+\dfrac{1}{1.5}+x+\dfrac{1}{5.9}+...+x+\dfrac{1}{397.401}=101x\)
\(\Rightarrow101x+\left(\dfrac{1}{1.5}+\dfrac{1}{5.9}+...+\dfrac{1}{397.401}\right)=x\)
\(\Rightarrow\dfrac{1}{4}\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+...+\dfrac{4}{397.401}\right)=x\)
\(\Rightarrow x=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+....+\dfrac{1}{397}-\dfrac{1}{401}\right)\)
\(\Rightarrow x=\dfrac{1}{4}\left(1-\dfrac{1}{401}\right)\)
\(\Rightarrow x=\dfrac{1}{4}.\dfrac{400}{401}\)
\(\Rightarrow x=\dfrac{100}{401}\)
1: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^6\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{18}\)
=>4x=18
hay x=9/2
2: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^{36}\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{108}\)
=>4x=108
hay x=27
3: \(\left(\dfrac{1}{81}\right)^x=\left(\dfrac{1}{27}\right)^4\)
\(\Leftrightarrow\left(\dfrac{1}{3}\right)^{4x}=\left(\dfrac{1}{3}\right)^{12}\)
=>4x=12
hay x=3
a)
\(\left(3x+\dfrac{1}{3}\right)\left(x-\dfrac{1}{2}\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x+\dfrac{1}{3}=0\\x-\dfrac{1}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{9}\\x=\dfrac{1}{2}\end{matrix}\right.\)
b)
\(\left(x-\dfrac{3}{2}\right)\left(2x+1\right)>0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-\dfrac{3}{2}>0\\2x+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-\dfrac{3}{2}< 0\\2x+1< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\dfrac{3}{2}\\x>-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< \dfrac{3}{2}\\x< -\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>\dfrac{3}{2}\\x< -\dfrac{1}{2}\end{matrix}\right.\)
\(\left(\dfrac{3}{4}x+5\right)-\left(\dfrac{2}{3}x-4\right)-\left(\dfrac{1}{6}x+1\right)=\left(\dfrac{1}{3}x+4\right)-\left(\dfrac{1}{3}x-3\right)\)
\(\Rightarrow\dfrac{3}{4}x+5-\dfrac{2}{3}x+4-\dfrac{1}{6}x-1=\dfrac{1}{3}x+4-\dfrac{1}{3}x+3\)
\(\Rightarrow\left(\dfrac{3}{4}x-\dfrac{2}{3}x-\dfrac{1}{6}x\right)+\left(5+4-1\right)=\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(4+3\right)\)
\(\Rightarrow\left(\dfrac{9}{12}x-\dfrac{8}{12}x-\dfrac{2}{12}x\right)+8=7\)
\(\Rightarrow-\dfrac{1}{12}x+8=7\)
\(\Rightarrow-\dfrac{1}{12}x=-1\)
\(\Rightarrow x=12\)
mình làm lại câu b) nha
b) |x-3|=-4
th1: x-3=-4
x=3+(-4)
x=-1
th2: x-3=4
x=3+4
x=7
b) \(\left|x-3\right|=-4\)
t/h1:\(x-3=-4\)
\(x=3-\left(-4\right)\)
\(x=7\)
t/h2:\(x-3=4\)
\(x=3-4\)
\(x=-1\)
\(ĐKXĐ:x\ne1,x\ne3,x\ne8,x\ne20\)
\(\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}-\dfrac{1}{x-20}=-\dfrac{3}{4}\)
\(\Rightarrow\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}-\dfrac{1}{x-20}=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{2\left(x-8\right)\cdot\left(x-20\right)+5\left(x-1\right)\cdot\left(x-20\right)+12\left(x-1\right)\cdot\left(x-3\right)-\left(x-1\right)\cdot\left(x-3\right)\cdot\left(x-8\right)}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{\left(2x-16\right)\cdot\left(x-20\right)+\left(5x-5\right)\cdot\left(x-20\right)+\left(12x-12\right)\cdot\left(x-3\right)-\left(x^2-3x-x+3\right)\cdot\left(x-8\right)}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{2x^2-40x-16x+320+5x^2-100x-5x+100+12x^2-36x-12x+36-\left(x^2-4x+3\right)\cdot\left(x-8\right)}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{2x^2-40x-16x+320+5x^2-100x-5x+100+12x^2-36x-\left(x^3-8x^2-4x^2+32x+3x-24\right)}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}\)
\(\Leftrightarrow\dfrac{2x^2-40x-16x+320+5x^2-100x-5x+100+12x^2-36x-12x+36-\left(x^3-12x^2+35x-24\right)}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{2x^2-40x-16x+320+5x^2-100x-5x+100+12x^2-36x-12x+36-x^3+12x^2-35x+24}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{31x^2-244x+480-x^3}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{-x^3+31x^2-244x+480}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{-x^3+3x^2+28x^2-84x-160x+480}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{-x^2\cdot\left(x-3\right)+28x\cdot\left(x-3\right)-160\left(x-3\right)}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{-\left(x-3\right)\left(x^2-28x+160\right)}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{-1\left(x^2-8x-20x+160\right)}{\left(x-1\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{-1\left(x^2-8x-20x+160\right)}{\left(x-1\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{-1\left(x\cdot\left(x-8\right)-20\left(x-8\right)\right)}{\left(x-1\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{-1\left(x-20\right)\left(x-8\right)}{\left(x-1\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{-1}{x-1}=-\dfrac{3}{4}\)
\(\Leftrightarrow-\dfrac{1}{x-1}=-\dfrac{3}{4}\)
\(\Leftrightarrow-4=-3\left(x-1\right)\)
\(\Leftrightarrow-4=-3\left(x-1\right)\)
\(\Leftrightarrow-4=-3x+3\)
\(\Leftrightarrow3x=3+4\)
\(\Leftrightarrow3x=7\)
\(\Rightarrow x=\dfrac{7}{3}\)
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
\(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}=\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-1\right)\left(x-4\right)}\) Đk: \(x\ne1;x\ne2;x\ne3;x\ne4\)
\(\Leftrightarrow\dfrac{2x-4}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=\dfrac{2x-4}{\left(x-1\right)\left(x-3\right)\left(x-4\right)}\)
\(\Leftrightarrow\dfrac{2x-4}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}-\dfrac{2x-4}{\left(x-1\right)\left(x-3\right)\left(x-4\right)}=0\)
\(\Leftrightarrow\dfrac{\left(x-4\right)\left(2x-4\right)-\left(2x-4\right)\left(x-2\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)}=0\)
\(\Leftrightarrow\dfrac{2x^2-4x-8x+16-2x^2+4x+4x-8}{\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)}=0\)
\(\Leftrightarrow-4x+8=0\)
\(\Rightarrow x=2\) (KTM )
=> Pt vô nghiệm