cho a,b,c>0 và abc=1. CM \(\frac{1}{a^{2010}+b^{2010}+1}+\frac{1}{b^{2010}+c^{2010}+1}+\frac{1}{c^{2010}+a^{2010}+1}\le1\)
a/ Cho x,y,z khác 0 thỏa mãn \(\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z}{z}\)
tính B=\(\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)\)
b/ Cho a,b,c,d khác 0. Tính
\(T=x^{2011}+y^{2011}+z^{2011}+t^{2011}\) biết x,y,z,t thỏa mãn :
\(\frac{x^{2010}+y^{2010}+z^{2010}+t^{2010}}{a^2+b^2+c^2+=d^2}=\frac{x^{2010}}{a^2}+\frac{y^{2010}}{b^2}+\frac{z^{2010}}{c^2}+\frac{t^{2010}}{d^2}\)
Từ \(\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z}{z}\)
\(\Rightarrow\frac{y+z-x}{x}+2=\frac{z+x-y}{y}+2=\frac{x+y-z}{z}+2\)
\(\Rightarrow\frac{x+y+z}{x}=\frac{x+y+z}{y}=\frac{x+y+z}{z}\left(1\right)\)
*)Xét \(x+y+z\ne0\left(2\right)\). Từ (1) và (2)
\(\Rightarrow x=y=z\). Khi đó \(B=\frac{x+y}{y}\cdot\frac{y+z}{z}\cdot\frac{x+z}{x}=2\cdot2\cdot2=8\)
*)Xét \(x+y+z=0\)\(\Rightarrow\left\{\begin{matrix}x+y=-z\\y+z=-x\\x+z=-y\end{matrix}\right.\)
Khi đó \(B=\frac{x+y}{y}\cdot\frac{y+z}{z}\cdot\frac{x+z}{x}=\frac{-z}{y}\cdot\frac{-x}{z}\cdot\frac{-y}{x}=-1\)
a)
Ta có \(\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z}{z}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có
\(\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z}{z}=\frac{y+z-x+z+x-y+x+y-z}{x+y+z}=\frac{x+y+z}{x+y+z}=1\)
\(\Rightarrow\left\{\begin{matrix}\frac{y+z-x}{x}=1\\\frac{z+x-y}{y}=1\\\frac{x+y-z}{z}=1\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}y+z-x=x\\z+x-y=y\\x+y-z=z\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}y+z=2x\\z+x=2y\\x+y=2z\end{matrix}\right.\) (1)
Ta có \(B=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)\)
\(\Rightarrow B=\frac{x+y}{y}.\frac{y+z}{z}.\frac{x+z}{x}\)
Thế (1) vào biểu thức B
\(\Rightarrow B=\frac{2z}{y}.\frac{2x}{z}.\frac{2y}{x}\)
\(\Rightarrow B=2.2.2=8\)
Vậy biểu thức \(B=8\)
Cho 3 số a,b,c thỏa mãn abc=2010. Tính giá trị biểu thức:
M= \(\frac{2010a}{ab+2010a+2010}+\frac{b}{bc+b+2010}+\frac{c}{ac+c+1}\)
Ta có: \(M=\frac{2010a}{ab+2010a+2010}+\frac{b}{bc+b+2010}+\frac{c}{ac+c+1}\)
Thế: abc = 2010 ta được:
\(M=\frac{a^2bc}{ab+a^2bc+abc}+\frac{b}{bc+b+abc}+\frac{c}{ac+c+1}\)
\(\Leftrightarrow\frac{a^2bc}{ab\left(1+ac+c\right)}+\frac{b}{b\left(c+1+ac\right)}+\frac{c}{ac+c+1}\)
\(\Leftrightarrow\frac{a^2bc}{ab\left(1+ac+c\right)}+\frac{ab}{ab\left(c+1+ac\right)}+\frac{abc}{ab\left(ac+c+1\right)}\)
\(\Leftrightarrow\frac{a^2bc+ab+abc}{ab\left(1+ac+c\right)}=\frac{ab\left(ac+1+c\right)}{ab\left(1+ac+c\right)}=1\)
Vậy \(M=1\)
cho A=\(\frac{2010^{2011}+1}{2010^{2012}+1}\) và B=\(\frac{2010^{2010}+1}{2010^{2011}+1}\)
So sánh A và B
so sánh : cho A\(\frac{2010^{2011}+1}{2010^{2012}+1}\)
cho B =\(\frac{2010^{2010}+1}{2010^{2011}+1}\)
Ta có:
\(A=\frac{2010^{2011}+1}{2010^{2012}+1}\)
\(2010A=\frac{2010^{2012}+2010}{2010^{2012}+1}\)
\(2010A=1+\frac{2009}{2010^{2012}+1}\)
Lại có:
\(B=\frac{2010^{2010}+1}{2010^{2011}+1}\)
\(2010B=\frac{2010^{2011}+2010}{2010^{2011}+1}\)
\(2010B=1+\frac{2009}{2010^{2011}+1}\)
Vì \(1+\frac{2009}{2010^{2012}+1}< 1+\frac{2009}{2010^{2011}+1}\)
nên 2010A < 2010B
hay A < B
Vậy A < B
Bài 1: So sánh:
A = \(\frac{10^{1992}+1}{10^{1991}+1}\) và B = \(\frac{10^{1993}+1}{10^{1992}+1}\)
C = \(\frac{2010^{2008}+1}{2010^{2009}+1}\) và C = \(\frac{2010^{2007}+1}{2010^{2008}+1}\)
Cho:
\(A=\frac{2010^{2011}+1}{2010^{2012}+1}\) Và \(B=\frac{2010^{2010}+1}{2010^{2011}+1}\)
So sánh A và B
\(1-A=1-\frac{2010^{2011}+1}{2010^{2012}+1}=\frac{2010^{2012}+1}{2010^{2012}+1}-\frac{2010^{2011}+1}{2010^{2012}+1}\)=\(\frac{2010}{2010^{2012}+1}\)
\(1-A=1-\frac{2010^{2011}+1}{2010^{2012}+1}=\frac{2010^{2012}+1}{2010^{2012}+1}-\frac{2010^{2011}+1}{2010^{2012}+1}=\frac{2010}{2010^{2012}+1}\)
\(1-B=1-\frac{2010^{2010}+1}{2010^{2011}+1}=\frac{2010^{2011}+1}{2010^{2011}+1}-\frac{2010^{2010}+1}{2010^{2011}+1}=\frac{2010}{2010^{2011}+1}\)
\(\frac{2010}{2010^{2012}+1}<\frac{2010}{2010^{2011}+1}\Rightarrow A>B\)
Ta có:
\(2010A=\frac{2010^{2012}+2010}{2010^{2012}+1}=\frac{2010^{2012}+1+2009}{2010^{2012}+1}=1+\frac{2009}{2010^{2012}+1}\)
\(2010B=\frac{2010^{2011}+2010}{2010^{2011}+1}=\frac{2010^{2011}+1+2009}{2010^{2011}+1}=1+\frac{2009}{2010^{2011}+1}\)
Do \(2010^{2012}+1>2010^{2011}+1\) => \(\frac{2009}{2010^{2012}+1}<\frac{2009}{2010^{2011}+1}\)
Nên \(1+\frac{2009}{2010^{2012}+1}<1+\frac{2009}{2010^{2011}+1}\) hay 2010A < 2010B
Vậy A<B
cho các số a,b,c khác 0.thỏa mãn:
\(\frac{x^{2010}+y^{2010}+z^{2010}+t^{2010}}{a^2+b^2+c^2+d^2}=\frac{x^{2010}}{a^2}+\frac{y^{2010}}{b^2}+\frac{z^{2010}}{c^2}+\frac{t^{2010}}{d^2}\)
1. Cho các số a,b,c,d khác 0. Tính T = x2011 + y2011 + z2011 + t2011
Biết x,y,z,t thoả mãn:
\(\frac{x^{2010}+y^{2010}+z^{2010}+t^{2010}}{a^2+b^2+c^2+d^2}=\frac{x^{2010}}{a^2}+\frac{y^{2010}}{b^2}+\frac{z^{2010}}{c^2}+\frac{t^{2010}}{d^2}\)
2. Tìm số tự nhiên M nhỏ nhất có 4 chữ số thoả mãn điều kiện:
M = a+b = c+d = e+f
Biết a,b,c,d,e,f thuộc tập hợp N* và \(\frac{a}{b}=\frac{14}{22};\frac{c}{d}=\frac{11}{13};\frac{e}{f}=\frac{13}{17}\)
M=a+b=c+d=e+f.M=a+b=c+d=e+f.
⇒⎧⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪⎩a7=b11=a+b7+11=M18(1)c11=d13=c+d11+13=M24(2)e13=f17=e+f13+17=M30(3)⇒{a7=b11=a+b7+11=M18(1)c11=d13=c+d11+13=M24(2)e13=f17=e+f13+17=M30(3)
Kết hợp (1),(2)và(3)(1),(2)và(3)
⇒M∈BCNN(18;24;30).⇒M∈BCNN(18;24;30).
⇒M∈{0;360;720;1080;...}⇒M∈{0;360;720;1080;...}
Mà MM là số tự nhiên nhỏ nhất có 4 chữ số.
⇒M=1080.⇒M=1080.
Vậy M=1080.
nhớ cho mình 1 k nhé chúc bạn học tốt
cho A=\(\frac{1}{2010}+\frac{2}{2009}+\frac{3}{2008}+...+\frac{2009}{2}+\frac{2010}{1}\)
B=\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2010}+\frac{1}{2011}\)
tính\(\frac{a}{b}\)
b.giả sử 2^2010 có m chữ số và 5^2010 có n chữ số.tính m+n
a) A= 1/2010+1+2/2009+1+3/2008+1+...+2009/2+1+1
= 2011/2010+20011/2009+2011/2008+...+2011/2+2011/2011
= 2011(1/2+1/3+1/4+...+1/2011)
Ta có: B= 1/2+1/3+1/4+...+1/2011
suy ra A/B= 2011
\(\frac{A}{B}\)=2011