Tính tổng : A=\(\dfrac{2010}{2}+\dfrac{2010}{6}+\dfrac{2010}{12}+.....+\dfrac{2010}{9900}\)
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8 tháng 10 2021 lúc 13:48
a) Cho các số a,b,c,d khác 0 . Tính :T x^{2011}+y^{2011}+z^{2011}+t^{2011} Biết x,y,z,t thoả mãn dfrac{x^{2010}+y^{2010}+z^{2010}+t^{2010}}{a^2+b^2+c^2+d^2}dfrac{x^{2010}}{a^2}+dfrac{y^{2010}}{b^2}+dfrac{z^{2010}}{c^2}+dfrac{t^{2010}}{d^2}b) Tìm số tự nhiên M nhỏ nhất có 4 chữ số thoả mãn điều kiện Ma+bc+de+f Nếu câu b thiếu j thì các bạn cứ bỏ qua nha
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a) Cho các số a,b,c,d khác 0 . Tính :
T = \(x^{2011}+y^{2011}+z^{2011}+t^{2011}\)
Biết x,y,z,t thoả mãn \(\dfrac{x^{2010}+y^{2010}+z^{2010}+t^{2010}}{a^2+b^2+c^2+d^2}=\dfrac{x^{2010}}{a^2}+\dfrac{y^{2010}}{b^2}+\dfrac{z^{2010}}{c^2}+\dfrac{t^{2010}}{d^2}\)
b) Tìm số tự nhiên M nhỏ nhất có 4 chữ số thoả mãn điều kiện
M=a+b=c+d=e+f
Nếu câu b thiếu j thì các bạn cứ bỏ qua nha
tính A biết A=(\(\dfrac{\sqrt[4]{2010^2}-\sqrt[4]{2010}}{1-\sqrt[4]{2010}}+\dfrac{1+\sqrt{2010}}{\sqrt[4]{2010}}\)) -\(\dfrac{\sqrt{1+\dfrac{2}{2010}+\dfrac{1}{2010}}}{1+\sqrt{2010}}\)
Tính giá trị của biểu thức :
\(A=\left(\dfrac{\sqrt[4]{2010^2}-\sqrt[4]{2010}}{1-\sqrt[4]{2010}}+\dfrac{1+\sqrt{2010}}{\sqrt[4]{2010}}\right)^2-\dfrac{\sqrt{1+\dfrac{2}{\sqrt{2010}}+\dfrac{1}{2010}}}{1+\sqrt{2010}}\)
Cho a2+b2+c2=ab+bc+ca.Tính C=\(\dfrac{a^{2010}+b^{2010}}{c^{2010}}+\dfrac{b^{2010}+c^{2010}}{a^{2010}}+\dfrac{c^{2010}+a^{2010}}{b^{2010}}\)
ta có : \(a^2+b^2+c^2=ab+bc+ca\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)
\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\) \(\Leftrightarrow a=b=c\)
\(\Rightarrow C=\dfrac{a^{2010}+b^{2010}}{c^{2010}}+\dfrac{b^{2010}+c^{2010}}{a^{2010}}+\dfrac{c^{2010}+a^{2010}}{b^{2010}}=3\dfrac{a^{2010}+a^{2010}}{a^{2010}}\)
\(=3\dfrac{2a^{2010}}{a^{2010}}=3.2=6\)
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cho các số a,b,c,d khác 0 và các số x,y,z,t thỏa mãn \(\dfrac{x^{2010}+y^{2010}+z^{2010}+t^{2010}}{a^2+b^2+c^2+d^2}=\dfrac{x^{2010}}{a^2}+\dfrac{y^{2010}}{b^2}+\dfrac{z^{2010}}{c^2}+\dfrac{t^{2010}}{d^2}\)
Bài 2 : So sánh
\(A=\dfrac{2008}{2009}+\dfrac{2009}{2010}+\dfrac{2010}{2011}vàB=\dfrac{2008+2009+2010}{2009+2010+2011}\)
Ta có :
\(B=\frac{2008+2009+2010}{2009+2010+2011}=\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{2010}{2009+2010+2011}\)
Vì :
\(\frac{2008}{2009}>\frac{2008}{2009+2010+2011}\)
\(\frac{2009}{2010}>\frac{2009}{2009+2010+2011}\)
\(\frac{2010}{2011}>\frac{2010}{2009+2010+2011}\)
Nên \(\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}>\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{2010}{2009+2010+2011}\)
\(\Rightarrow\)\(A>B\)
Vậy \(A>B\)
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Ta có: \(B=\frac{2008+2009+2010}{2009+2010+2011}\)
\(=\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{2010}{2009+2010+2011}\)
Vì \(\frac{2008}{2009}>\frac{2008}{2009+2010+2011}\)
\(\frac{2009}{2010}>\frac{2009}{2009+2010+2011}\)
\(\frac{2010}{2011}>\frac{2010}{2009+2010+2011}\)
nên \(\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}>\frac{2008+2009+2010}{2009+2010+2011}\)
hay A > B
Vậy A > B
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So sánh A và B biết: A= \(\dfrac{2008+2008+2010}{2009+2010+2011}\) và B= \(\dfrac{2008}{2009}\)+ \(\dfrac{2009}{2010}\)+ \(\dfrac{2010}{2011}\)
A = \(\dfrac{2008}{2009+2010+2011}+\dfrac{2009}{2009+2010+2011}+\dfrac{2010}{2009+2010+2011}\)
Ta có:
\(\dfrac{2008}{2009}>\dfrac{2008}{2009+2010+2011}\)
\(\dfrac{2009}{2010}>\dfrac{2009}{2009+2010+2011}\)
\(\dfrac{2010}{2011}>\dfrac{2010}{2009+2010+2011}\)
Từ 3 điều trên suy ra : A < B
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\(\dfrac{x+1}{2010}+\dfrac{x+2}{2009}+\dfrac{x-3}{2008}+...+\dfrac{x-2009}{2}+\dfrac{x-2010}{1}=-2010\)
\(\Leftrightarrow\dfrac{x+1}{2010}+1+\dfrac{x+2}{2009}+1+...+\dfrac{x+2009}{2}+1+\dfrac{x+2010}{1}+1=0\)
=>x+2011=0
hay x=-2011
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Bài 2 : So sánh
\(A=\dfrac{2008}{2009}+\dfrac{2009}{2010}+\dfrac{2010}{2011}vàB=\dfrac{2008+2009+2010}{2009+2010+2011}\)
\(B=\dfrac{2008+2009+2010}{2009+2010+2011}=\dfrac{2008}{2009+2010+2011}+\dfrac{2009}{2009+2010+2011}+\dfrac{2010}{2009+2010+2011}\)Ta có : \(\dfrac{2008}{2009}>\dfrac{2008}{2009+2010+2011}\)
\(\dfrac{2009}{2010}>\dfrac{2009}{2009+2010+2011}\)
\(\dfrac{2010}{2011}>\dfrac{2010}{2009+2010+2011}\)\(=>\dfrac{2008}{2009}+\dfrac{2009}{2010}+\dfrac{2010}{2011}>\dfrac{2008+2009+2010}{2009+2010+2011}\)
Hay A > B
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