Tìm x : (x-3) + (x-2) +(x-1)+...+9+10+2011 = 2011
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Câu 1 bị sai đề bài.
Câu 2:
\(\frac{2011}{2012}+\frac{2012}{2013}+\frac{2013}{2011}=\frac{2012-1}{2012}+\frac{2013-1}{2013}+\frac{2011+1+1}{2011}\)
\(=1-\frac{1}{2012}+1-\frac{1}{2013}+1+\frac{1}{2011}+\frac{1}{2011}\)
Vì:
\(\frac{1}{2011}>\frac{1}{2012};\frac{1}{2011}>\frac{1}{2013}\Rightarrow\frac{1}{2011}+\frac{1}{2011}-\frac{1}{2012}-\frac{1}{2013}>0\)
\(\Rightarrow\)\(\frac{2012-1}{2012}+\frac{2013-1}{2013}+\frac{2011+1+1}{2011}>3\)
\(\Rightarrow\frac{2011}{2012}+\frac{2012}{2013}+\frac{2013}{2011}>3\)
\(\Leftrightarrow\left(\frac{x-1}{2012}-1\right)+\left(\frac{x-2}{2011}-1\right)+...+\left(\frac{x-2012}{1}-1\right)=0\)
\(\Leftrightarrow\frac{x-2013}{2012}+\frac{x-2013}{2011}+...+\frac{x-2013}{1}=0\)
\(\Leftrightarrow\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+....+1\right)=0\)
\(\Leftrightarrow x-2013=0\)(because 1/2012 +1/2011+...+1 luôn lớn hơn 0
\(\Leftrightarrow x=2013\)
Vậy ........
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\times\left(x+1\right):2}=\frac{2011}{2013}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\times\left(x+1\right)}\times\frac{1}{2}=\frac{2011}{2013}\)
\(\Rightarrow2\times\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\times\left(x+1\right)}\right)=\frac{2011}{2013}\)
\(\Rightarrow2\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{x\times\left(x+1\right)}\right)=\frac{2011}{2013}\)
\(\Rightarrow2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
\(\Rightarrow2\times\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{2013}:2\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2011}{4016}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2013}\)
\(\Rightarrow x+1=2013\)
\(\Rightarrow x=2012\)
Vậy x = 2012
Ta có : \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+......+\frac{2}{n\left(n+1\right)}\)
\(=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.....+\frac{2}{n\left(n+1\right)}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+......+\frac{2}{n\left(n+1\right)}\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+......+\frac{1}{n\left(n+1\right)}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{n}-\frac{1}{n+1}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{n+1}\right)\)
\(=1-\frac{2}{n+1}\)
\(=\frac{n+1}{n+1}-\frac{2}{n+1}\)
\(=\frac{n-1}{n+1}\)
a. 26-|x+9|=-13
=> |x+9|=26-(-13)
=> |x+9|=39
=> x+9 = 39 hoặc x+9 = -39
=> x=30 hoặc x = -48
Vậy, x thuộc {30, -48}
b. 2011-|x|=2009
=> |x| = 2011-2009
=> |x|=2
=> x = 2 hoặc x = -2
Vậy, x thuộc {2, -2}
c.|x+2010|=2011
=> x + 2010 = 2011 hoặc x + 2010 = -2011
=> x = 2011 - 2010 hoặc x= -2011 - 2010
=> x = 1 hoặc x= -4021
Vậy, x thuộc {1, -4021}
a) Kết quả là: \(x^2-\frac{7}{2}x+3\)
b) Ta có: \(x^2-2x+1-y^2=\left(x-1\right)^2-y^2=\left(x-1-y\right)\left(x-1+y\right)\)
c) Ta có: \(\frac{\left(3^{10}+x^9\right)}{x^9}=2011\)
\(\Leftrightarrow\frac{3^{10}}{x^9}+\frac{x^9}{x^9}=2011\)
\(\Leftrightarrow\frac{3}{x}+1=2011\)
\(\Leftrightarrow\frac{3}{x}=2011-1\Leftrightarrow x=\frac{2011-1}{3}=670\)
Vậy: x=670
d) Ta có: \(A=x^2+4x+20\)
\(=x^2+4x+4+16=\left(x+2\right)^2+16\)
Ta có: \(\left(x+2\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+2\right)^2+16\ge16\forall x\)
Dấu '=' xảy ra khi
\(\left(x+2\right)^2=0\Leftrightarrow x+2=0\Leftrightarrow x=-2\)
Vậy: GTNN của biểu thức \(A=x^2+4x+20\) là 16 khi x=-2
đáp án:ko muốn giải
Con nào nhanh nhất ko đc k à chế