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Bài 1: Chỉ cần chú ý đẳng thức \(a^5+b^5=\left(a^2+b^2\right)\left(a^3+b^3\right)-a^2b^2\left(a+b\right)\) là ok!
Làm như sau: Từ \(x^2+\frac{1}{x^2}=14\Rightarrow x^2+2.x.\frac{1}{x}+\frac{1}{x^2}=16\)
\(\Rightarrow\left(x+\frac{1}{x}\right)^2=16\). Do \(x>0\Rightarrow x+\frac{1}{x}>0\Rightarrow x+\frac{1}{x}=4\)
: \(x^5+\frac{1}{x^5}=\left(x^2+\frac{1}{x^2}\right)\left(x^3+\frac{1}{x^3}\right)-\left(x+\frac{1}{x}\right)\)
\(=14\left(x^3+\frac{1}{x^3}\right)-\left(x+\frac{1}{x}\right)\)
\(=14\left(x+\frac{1}{x}\right)\left(x^2+\frac{1}{x^2}-1\right)-4\)
\(=14.4.\left(14-1\right)-4=724\) là một số nguyên (đpcm)
P/s: Lâu ko làm nên cũng ko chắc đâu nhé!
\(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\Leftrightarrow\frac{x^2+y^2+z^2}{a^2+b^2+c^2}-\frac{x^2}{a^2}-\frac{y^2}{b^2}-\frac{z^2}{c^2}=0\)
\(\Leftrightarrow\left(\frac{x^2}{a^2+b^2+c^2}-\frac{x^2}{a^2}\right)+\left(\frac{y^2}{a^2+b^2+c^2}-\frac{y^2}{b^2}\right)+\left(\frac{z^2}{a^2+b^2+c^2}-\frac{z^2}{c^2}\right)=0\)
\(\Leftrightarrow x^2.\left(\frac{1}{a^2+b^2+c^2}-\frac{1}{a^2}\right)+y^2.\left(\frac{1}{a^2+b^2+c^2}-\frac{1}{b^2}\right)+z^2.\left(\frac{1}{a^2+b^2+c^2}-\frac{1}{c^2}\right)=0\)
vì \(a,b,c\ne0\Rightarrow\hept{\begin{cases}\left(\frac{1}{a^2+b^2+c^2}-\frac{1}{a^2}\right)\ne0\\\left(\frac{1}{a^2+b^2+c^2}-\frac{1}{b^2}\right)\ne0\\\left(\frac{1}{a^2+b^2+c^2}-\frac{1}{c^2}\right)\ne0\end{cases}}\Rightarrow\hept{\begin{cases}x^2=0\\y^2=0\\z^2=0\end{cases}}\Rightarrow x=y=z=0\Rightarrow P=0+\frac{11}{2011}=\frac{11}{2011}\)
theo đầu bài ta có\(\dfrac{x^2+y^2}{xy}=\dfrac{10}{3}\)=>\(3x^2+3y^2=10xy\)
A=\(\dfrac{x-y}{x+y}\)
=>\(A^2=\left(\dfrac{x-y}{x+y}\right)^2=\dfrac{x^2-2xy+y^2}{x^2+2xy+y^2}=\dfrac{3x^2-6xy+3y^2}{3x^2+6xy+3y^2}=\dfrac{10xy-6xy}{10xy+6xy}=\dfrac{4xy}{16xy}=\dfrac{1}{4}\)
=>A=\(\sqrt{\dfrac{1}{4}}=\dfrac{-1}{2}hoặc\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}\) (cộng trừ căn 1/4 nhé)
vì y>x>0=> A=-1/2
ta có \(\frac{x^2}{a^2}\)+ \(\frac{y^2}{b^2}\)+\(\frac{z^2}{c^2}\)= \(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}\)
=> ( \(\frac{x^2}{a^2}\)+ \(\frac{y^2}{b^2}\)+ \(\frac{z^2}{c^2}\))( \(a^2+b^2+c^2\))= \(x^2+y^2+z^2\)
=> \(x^2\)+ \(\frac{\left(b^2+c^2\right)x^2}{a^2}\)+ \(y^2\)+ \(\frac{\left(a^2+c^2\right)y^2}{b^2}\)+ \(z^2\)+ \(\frac{\left(a^2+b^2\right)z^2}{c^2}\)= \(x^2+y^2+z^2\)
=> \(\frac{\left(b^2+c^2\right)x^2}{a^2}\)+ \(\frac{\left(a^2+c^2\right)y^2}{b^2}\)+ \(\frac{\left(a^2+b^2\right)z^2}{c^2}\)= 0
nhận xét ...... ( tát cả đều lớn hơn hoặc = 0 nên cả tổng sẽ lớn hơn hoặc = 0)
dấu = xảy ra khi và chi khi x=y = z = 0 ( vì a,b,c khác 0)
vậy \(x^{2011}+y^{2011}+z^{2011}\)= 0 +0+0 = 0
\(\text{Có: }x^2+y^2+z^2=xy+yz+xz\)
\(\Leftrightarrow2\left(x^2+y^2+z^2\right)=2\left(xy+yz+xz\right)\)
\(\Leftrightarrow x^2+x^2+y^2+y^2+z^2+z^2=2xy+2yz+2xz\)
\(\Leftrightarrow x^2+x^2+y^2+y^2+z^2+z^2-2xy-2yz-2xz=0\)
\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(x^2-2xz+z^2\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2=0\)
\(\text{Vì }\left(x-y\right)^2\ge0;\left(y-z\right)^2\ge0\text{ và }\left(x-z\right)^2\ge0\)
\(\text{Nên để }\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2=0\)
\(\text{thì }\hept{\begin{cases}\left(x-y\right)^2=0\\\left(y-z\right)^2=0\\\left(x-z\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x-y=0\\y-z=0\\x-z=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=y\\y=z\\x=z\end{cases}\Leftrightarrow}x=y=z}\)
\(\text{Khi đó: }x^{2011}+y^{2011}+z^{2011}=3^{2012}\)
\(\Leftrightarrow x^{2011}+x^{2011}+x^{2011}=3^{2012}\left(\text{Vì x = y = z}\right)\)
\(\Leftrightarrow3x^{2011}=3^{2012}\)
\(\Leftrightarrow x^{2011}=3^{2011}\)
\(\Leftrightarrow x=3\)
\(\text{Vậy }x=y=z=3\)
a/ \(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+ac+bc\right)=0\)
\(\Rightarrow ab+ac+bc=-7\Rightarrow\left(ab+ac+bc\right)^2=49\)
\(\Rightarrow\left(ab\right)^2+\left(ac\right)^2+\left(bc\right)^2+2a^2bc+2ab^2c+2abc^2=49\)
\(\Rightarrow\left(ab\right)^2+\left(ac\right)^2+\left(bc\right)^2+2abc\left(a+b+c\right)=49\)
\(\Rightarrow\left(ab\right)^2+\left(ac\right)^2+\left(bc\right)^2=49\)
Ta có:
\(a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2\left(\left(ac\right)^2+\left(ac\right)^2+\left(bc\right)^2\right)=14^2-2.49=98\)
b/ \(\frac{x^2}{a^2}-\frac{x^2}{a^2+b^2+c^2}+\frac{y^2}{b^2}-\frac{y^2}{a^2+b^2+c^2}-\frac{z^2}{c^2}-\frac{z^2}{a^2+b^2+c^2}=0\)
\(\Leftrightarrow x^2\left(\frac{b^2+c^2}{\left(a^2+b^2+c^2\right)a^2}\right)+y^2\left(\frac{a^2+c^2}{\left(a^2+b^2+c^2\right)b^2}\right)+z^2\left(\frac{a^2+b^2}{\left(a^2+b^2+c^2\right)c^2}\right)=0\)
\(\Leftrightarrow x^2=y^2=z^2=0\) (do \(a;b;c\ne0\))
\(\Rightarrow x=y=z=0\Rightarrow P=0\)
+ \(\left(x^{2011}+y^{2011}\right)\left(x+y\right)\)
\(=x^{2012}+y^{2012}+xy\left(x^{2010}+y^{2010}\right)\)
\(=\left(x^{2011}+y^{2011}\right)+xy\left(x^{2011}+y^{2011}\right)\)
\(=\left(xy+1\right)\left(x^{2011}+y^{2011}\right)\)
+ Vì x, y dương nên \(x^{2011}+y^{2011}>0\)
=> x + y = xy + 1
=> x + y - xy - 1 = 0
=> ( y - 1 ) - x( y - 1 ) = 0
=> ( 1 - x ) ( y - 1 ) = 0
\(\Rightarrow\left[{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
+ x = 1 => \(1+y^{2010}=1+y^{2011}=1+y^{2012}\)
\(\Rightarrow y^{2010}=y^{2011}\) \(\Rightarrow y^{2010}-y^{2011}=0\)
\(\Rightarrow y^{2010}\left(1-y\right)=0\)
\(\Rightarrow y=1\left(doy>0\right)\)
+ Tương tự nếu y = 1 ta cùng tìm được x = 1
Do đó : A = 2
Lời giải khác:
Ta có:
\(x^{2011}+y^{2011}=x^{2010}+y^{2010}\)
\(\Rightarrow x^{2011}-x^{2010}+y^{2011}-y^{2010}=0\)
\(\Leftrightarrow x^{2010}(x-1)+y^{2010}(y-1)=0(1)\)
Và: \(x^{2011}+y^{2011}=x^{2012}+y^{2012}\)
\(\Rightarrow x^{2012}-x^{2011}+y^{2012}-y^{2011}=0\)
\(\Leftrightarrow x^{2011}(x-1)+y^{2011}(y-1)=0(2)\)
Lấy (2)-(1) ta có:
\(x^{2011}(x-1)-x^{2010}(x-1)+y^{2011}(y-1)-y^{2010}(y-1)=0\)
\(\Leftrightarrow x^{2010}(x-1)^2+y^{2010}(y-1)^2=0\)
Dễ thấy \(x^{2010}(x-1)^2\geq 0; y^{2010}(y-1)^2\geq 0, \forall x,y>0\)
Do đó để tổng của chúng bằng $0$ thì \(x^{2010}(x-1)^2=y^{2010}(y-1)^2=0\)
Mà $x,y$ đều dương nên $x=y=1$
Khi đó ta dễ tính ra $A=2$