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\(a,\left(\frac{1}{7}x-\frac{2}{7}\right)\left(-\frac{1}{5}x+\frac{3}{5}\right)\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
TH1 : \(\frac{1}{7}x-\frac{2}{7}=0\Rightarrow\frac{x-2}{7}=0\Rightarrow x-2=0\Leftrightarrow x=2\)
TH2 : \(-\frac{1}{5}x+\frac{3}{5}=0\Rightarrow\frac{-x+3}{5}=0\Rightarrow-x+3=0\Leftrightarrow x=3\)
TH3 : \(\frac{1}{3}x+\frac{4}{3}=0\Rightarrow\frac{x+4}{3}=0\Rightarrow x+4=0\Leftrightarrow x=-4\)
\(\Rightarrow x\in\left\{2;3;-4\right\}\)
\(b,\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)
\(\Rightarrow\frac{5}{30}x+\frac{3}{30}x-\frac{8}{30}x+1=0\)
\(\Rightarrow\frac{5x+3x-8x}{30}+1=0\)
\(\Rightarrow1=0\)( vô lý )\(\Rightarrow x\in\varnothing\)
Bài 1:
\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{101}\right|=101x\)
Ta thấy:
\(VT\ge0\Rightarrow VP\ge0\Rightarrow101x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{6}\right)+...+\left(x+\frac{1}{101}\right)=101x\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{101}\right)=0\)
\(\Rightarrow10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\right)=0\)
\(\Rightarrow10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)=0\)
\(\Rightarrow10x+\left(1-\frac{1}{11}\right)=0\)
\(\Rightarrow10x+\frac{10}{11}=0\)
\(\Rightarrow10x=-\frac{10}{11}\Rightarrow x=-\frac{1}{11}\)(loại,vì x\(\ge\)0)
Bài 2:
Ta thấy: \(\begin{cases}\left(2x+1\right)^{2008}\ge0\\\left(y-\frac{2}{5}\right)^{2008}\ge0\\\left|x+y+z\right|\ge0\end{cases}\)
\(\Rightarrow\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|\ge0\)
Mà \(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)
\(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)
\(\Rightarrow\begin{cases}\left(2x+1\right)^{2008}=0\\\left(y-\frac{2}{5}\right)^{2008}=0\\\left|x+y+z\right|=0\end{cases}\)\(\Rightarrow\begin{cases}2x+1=0\\y-\frac{2}{5}=0\\x+y+z=0\end{cases}\)
\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\x+y+z=0\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{2}+\frac{2}{5}+z=0\end{cases}\)
\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{10}=-z\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{1}{10}\end{cases}\)
a)
\(\begin{array}{l}\frac{2}{9}:x + \frac{5}{6} = 0,5\\\frac{2}{9}:x = \frac{1}{2} - \frac{5}{6}\\\frac{2}{9}:x = \frac{3}{6} - \frac{5}{6}\\\frac{2}{9}:x = \frac{{ - 2}}{6}\\x = \frac{2}{9}:\frac{{ - 2}}{6}\\x = \frac{2}{9}.\frac{{ - 6}}{2}\\x = \frac{{ - 2}}{3}\end{array}\)
Vậy \(x = \frac{{ - 2}}{3}\).
b)
\(\begin{array}{l}\frac{3}{4} - \left( {x - \frac{2}{3}} \right) = 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - \frac{4}{3}\\x - \frac{2}{3} = \frac{9}{{12}} - \frac{{16}}{{12}}\\x - \frac{2}{3} = \frac{{ - 7}}{{12}}\\x = \frac{{ - 7}}{{12}} + \frac{2}{3}\\x = \frac{{ - 7}}{{12}} + \frac{8}{{12}}\\x = \frac{1}{12}\end{array}\)
Vậy\(x = \frac{1}{12}\).
c)
\(\begin{array}{l}1\frac{1}{4}:\left( {x - \frac{2}{3}} \right) = 0,75\\\frac{5}{4}:\left( {x - \frac{2}{3}} \right) = \frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}:\frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}.\frac{4}{3}\\x - \frac{2}{3} = \frac{5}{3}\\x = \frac{5}{3} + \frac{2}{3}\\x = \frac{7}{3}\end{array}\)
Vậy \(x = \frac{7}{3}\).
d)
\(\begin{array}{l}\left( { - \frac{5}{6}x + \frac{5}{4}} \right):\frac{3}{2} = \frac{4}{3}\\ - \frac{5}{6}x + \frac{5}{4} = \frac{4}{3}.\frac{3}{2}\\ - \frac{5}{6}x + \frac{5}{4} = 2\\ - \frac{5}{6}x = 2 - \frac{5}{4}\\ - \frac{5}{6}x = \frac{8}{4} - \frac{5}{4}\\ - \frac{5}{6}x = \frac{3}{4}\\x = \frac{3}{4}:\left( { - \frac{5}{6}} \right)\\x = \frac{3}{4}.\frac{{ - 6}}{5}\\x = \frac{{ - 9}}{{10}}\end{array}\)
Vậy \(x = \frac{{ - 9}}{{10}}\).
\(\Rightarrow\frac{3}{4}x+5-\frac{2}{3}x+4-\frac{1}{6}x-1=\frac{1}{3}x+4-\frac{1}{3}+3\)+3
\(\Rightarrow\left(\frac{3}{4}x-\frac{2}{3}x-\frac{1}{6}x\right)+\left(5+4-1\right)=\frac{1}{3}x+\left(4-\frac{1}{3}+3\right)\)
=>\(\frac{-1}{12}x+8=\frac{1}{3}x+\frac{20}{3}\)\(\Rightarrow\frac{-1}{12}x+8-\frac{1}{3}x=\frac{20}{3}\)
\(\Rightarrow\left(\frac{-1}{12}-\frac{1}{3}\right)x+8=\frac{20}{3}\)
\(\Rightarrow\frac{-5}{12}x+8=\frac{20}{3}\Rightarrow\frac{-5}{12}x=\frac{20}{3}-8\)
\(\Rightarrow\frac{-5}{12}x=\frac{-4}{3}\Rightarrow x=\frac{-4}{3}:\frac{-5}{12}=\frac{16}{5}\)
\(\left(3-\frac{1}{2}x\right)\left(\left|x+\frac{3}{4}\right|-\frac{5}{6}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3-\frac{1}{2}x=0\\\left|x+\frac{3}{4}\right|-\frac{5}{6}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x=3\\\left|x+\frac{3}{4}\right|=\frac{5}{6}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=6\\x=\frac{1}{12}\\x=\frac{-19}{12}\end{cases}}\)
\(\left(3-\frac{1}{2}x\right)\cdot\left(\left|x+\frac{3}{4}\right|-\frac{5}{6}\right)=0\)
\(\Rightarrow\hept{\begin{cases}3-\frac{1}{2}x=0\\\left|x+\frac{3}{4}\right|-\frac{5}{6}=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=6\\x+\frac{3}{4}=\pm\frac{5}{6}\end{cases}}\)
Ta có
\(x+\frac{3}{4}=\pm\frac{5}{6}\)
\(\hept{\begin{cases}x+\frac{3}{4}=\frac{5}{6}\\x+\frac{3}{4}=-\frac{5}{6}\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{1}{12}\\x=-\frac{19}{12}\end{cases}}}\)
Vậy \(x\in\left\{3;\frac{1}{2};-\frac{19}{12}\right\}\)
b, \(x\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)+1=0\)
\(0+1=0\)
=> x thuoc rong
a) Dễ thấy VT > 0;mà VT=VP
=>VP > 0 => 4x > 0=> x > 0
=>\(\left|x+\frac{1}{2}\right|=x+\frac{1}{2};\left|x+\frac{1}{3}\right|=x+\frac{1}{3};\left|x+\frac{1}{6}\right|=x+\frac{1}{6}\)
=>BT đầu tương đương \(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{6}\right)=4x\)
\(=>3x+1=4x=>x=1\)
a) Để đẳng thức xảy ra thì: x>0 (vì: \(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{6}\right|>0\) )
Khi đó: \(\left|x+\frac{1}{2}\right|=x+\frac{1}{2};\left|x+\frac{1}{3}\right|=x+\frac{1}{3};\left|x+\frac{1}{6}\right|=x+\frac{1}{6}\)
=>\(x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{6}=4x\)
<=>x=1
Vậy x=1
b)Điều kiện: \(x\ne-3;-10;-21;-34\)
\(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
<=>\(\frac{1}{x+3}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+21}+\frac{1}{x+21}-\frac{1}{x+34}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
<=>\(\frac{1}{x+3}-\frac{1}{x+34}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
=>x+34-x-3=x
<=>x=31 (nhận)
Vậy x=31
Ta có : \(P=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}-2\right)}{\left(x+\frac{1}{x}\right)^3+x^3+\frac{1}{x^3}}=\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)\)
\(=3\left(x+\frac{1}{x}\right)\ge6\) \(\left(x>0\right)\).
Vậy \(P_{Min}=6\) khi \(x=1.\)
Happy New year :)