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ta có;
x/5 - 5/2=-31
=>x/5= -31+5/2=-28 1/2
=> x = -57/2* 5
=> x=-142 1/2
GTLN:A=11
GTNN:B=2
CÒN GTLN CÂU B KO TIM ĐƯỢC
GTNN CÂU A KO TÌM ĐƯỢC
\(\frac{1}{x\left(x+1\right)}=\frac{\left(x+1\right)-x}{x\left(x+1\right)}=\frac{x+1}{x\left(x+1\right)}-\frac{x}{x\left(x+1\right)}=\frac{1}{x}-\frac{1}{x+1}\)
=>\(\frac{1}{x}-\frac{1}{x+1}=\frac{1}{x}+\frac{1}{2011}\)
=>\(\frac{1}{x}-\frac{1}{x+1}-\frac{1}{x}=\frac{1}{2011}\)
=>\(\frac{1}{x}-\frac{1}{x}-\frac{1}{x+1}=\frac{1}{2011}\)
=>\(0-\frac{1}{x+1}=\frac{1}{2011}\)
=>\(-\frac{1}{x+1}=\frac{1}{2011}\)
=>-x+1=2011
=>-x=2011-1
=>-x=2010
=>x=-2010
Vậy x=-2010
\(\frac{1}{x\left(x+1\right)}=\frac{1}{x}+\frac{1}{2011}\)
<=>\(\frac{1}{x}-\frac{1}{x+1}=\frac{1}{x}+\frac{1}{2011}\)
<=>\(-\frac{1}{x+1}=\frac{1}{2011}\)
<=>-x-1=2011
<=>x=-2012
Đáp số: \(x=-2012\)
\(\Rightarrow\left(x-3\right)\left[\left(x-3\right)^x-\left(x-3\right)^{10}\right]=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x-3=0\\\left(x-3\right)^x-\left(x-3\right)^{10}=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=3\\\left(x-3\right)^x=\left(x-3\right)^{10}\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=3\\x=10\end{array}\right.\)
Vậy \(x\in\left\{3;10\right\}\)
\(\Rightarrow\left(x-3\right)\left[\left(x-3\right)^x-\left(x-3\right)^9\right]=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x-3=0\\\left(x-3\right)^x-\left(x-3\right)^9=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=3\\\left(x-3\right)^x=\left(x-3\right)^9\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=3\\x=9\end{array}\right.\)
Vậy \(x\in\left\{3;9\right\}\)
Ta có :
\(\frac{x}{2^2}+\frac{x}{2^3}+\frac{x}{2^4}=\frac{x}{3^2}+\frac{x}{3^3}+\frac{x}{3^4}\)
\(\frac{x}{2^2}+\frac{x}{2^3}+\frac{x}{2^4}-\frac{x}{3^2}-\frac{x}{3^3}-\frac{x}{3^4}=0\)
\(x^2\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{3^2}-\frac{1}{3^3}-\frac{1}{3^4}\right)=0\)
\(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{3^2}-\frac{1}{3^3}-\frac{1}{3^4}\ne0\Rightarrow x^2=0\)
\(x=0\)