chứng minh rằng,nếu \(x=by+cz,y=ax+cz,z=ax+by\)và x+y+z\(\ne\)0
thì \(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=2\)
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Ta có x + y = 2cz + ax + by = 2cz + z
hay 2cz = x + y - z, suy ra c = \(\frac{x+y-z}{2z}\)
do đó: \(1+c=\frac{x+y+z}{2z}\) hay \(\frac{1}{1+c}=\frac{2z}{z+y+z}\)
Tương tự \(1+a=\frac{x+y+z}{2x}\) hay \(\frac{1}{1+a}=\frac{2x}{x+y+z}\)
\(1+b=\frac{x+y+z}{2y}\) hay \(\frac{1}{1+b}=\frac{2y}{x+y+z}\)
Vậy \(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)
Ta có \(\left\{\begin{matrix}x=by+cz\\y=ax+cz\\z=ax+by\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}ax+x=ax+by+cz\\by+y=ax+by+cz\\cz+z=ax+by+cz\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}x\left(a+1\right)=ax+by+cz\\y\left(b+1\right)=ax+by+cz\\z\left(c+1\right)=ax+by+cz\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}a+1=\frac{ax+by+cz}{x}\\b+1=\frac{ax+by+cz}{y}\\c+1=\frac{ax+by+cz}{z}\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}\frac{1}{a+1}=\frac{x}{ax+by+cz}\\\frac{1}{b+1}=\frac{y}{ax+by+cz}\\\frac{1}{c+1}=\frac{z}{ax+by+cz}\end{matrix}\right.\)
\(\Rightarrow\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=\frac{x+y+z}{ax+by+cz}\)
Ta lại có \(\left\{\begin{matrix}x=by+cz\\y=ax+cz\\z=ax+by\end{matrix}\right.\)
\(\Rightarrow x+y+z=2\left(ax+by+cz\right)\)
\(\Rightarrow\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=\frac{x+y+z}{ax+by+cz}=\frac{2\left(ax+by+cz\right)}{ax+by+cz}=2\)
Vậy \(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=2\left(đpcm\right)\)
Ta có: \(x+y+z=\left(by+cz\right)+\left(ax+cz\right)+\left(ax+by\right)=2\left(ax+by+cz\right)\)
=> \(x+y+z=2\left(ax+by+cz\right)=2\left[\left(ax+by\right)+cz\right]=2\left[z+cz\right]=2\left(1+c\right)z\)
=> \(\frac{1}{1+c}=\frac{2z}{x+y+z}\) (1)
Tượng tự:
\(\frac{1}{1+a}=\frac{2x}{x+y+z}\) (2)
\(\frac{1}{1+b}=\frac{2y}{x+y+z}\) (3)
Cộng các vế của (1), (2), (3) ta có:
\(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=\frac{2\left(x+y+z\right)}{x+y+z}=2\) (ĐPCM)
Ta có x+y=ax+by+2cz=z+2cz
=> x+y-z=2cz
=> \(c=\frac{x+y-z}{2z}\Rightarrow c+1=\frac{x+y-z}{2z}+1=\frac{x+y+z}{2z}\)
\(\Rightarrow\frac{1}{c+1}=\frac{2z}{x+y+z}\left(1\right)\)
\(y+z=2ax+by+cz\Rightarrow y+z-x=2ax\Rightarrow a=\frac{y+z-x}{2x}\Rightarrow a+1=\frac{x+y+z}{2x}\)
\(\Rightarrow\frac{1}{a+1}=\frac{2x}{x+y+z}\left(2\right)\)
\(z+x=2by+ax+cz=2by+y\Rightarrow z+x-y=2by\)
\(\Rightarrow b=\frac{z+x-y}{2y}\Rightarrow b+1=\frac{z+x-y}{2y}+1=\frac{x+y+z}{2y}\)
\(\Rightarrow\frac{1}{b+1}=\frac{2y}{x+y+z}\left(3\right)\)
Cộng từng vế của (1)(2)(3) ta có
\(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=\frac{2x}{x+y+z}+\frac{2y}{x+y+z}+\frac{2z}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)
Cộng vế với vế của ba đẳng thức ta đc :
\(x+y+z=2\left(ax+by+cz\right)\Rightarrow ax+by+cz=\frac{x+y+z}{2}\) (*)
Lấy (*) - (1) ta có : \(ax+by+cz-\left(by+cz\right)=\frac{x+y+z}{2}-x\)
<=> \(ax=\frac{y+z-x}{2}\Leftrightarrow a=\frac{y+z-x}{2x}\Rightarrow a+1=\frac{y+z-x}{2x}+1=\frac{x+y+z}{2x}\)
=> \(\frac{1}{a+1}=\frac{2x}{x+y+z}\)
CMTT với 1/b+1 và 1/c+1
=> ĐPCM
Ta có : \(\begin{cases}x=by+cz\\y=ax+cz\\z=ax+by\end{cases}\) . Cộng các đẳng thức trên theo vế :
\(x+y+z=2\left(ax+by+cz\right)\Rightarrow\frac{x+y+z}{ax+by+cz}=2\)
Lại có : \(y=ax+cz\Rightarrow a=\frac{y-cz}{x}\Rightarrow a+1=\frac{x+y-cz}{x}\Rightarrow\frac{1}{a+1}=\frac{x}{x+y-cz}=\frac{x}{ax+by+cz}\)
Tương tự : \(\frac{1}{b+1}=\frac{y}{ax+by+cz};\frac{1}{c+1}=\frac{z}{ax+by+cz}\)
\(\Rightarrow P=\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=\frac{x}{ax+by+cz}+\frac{y}{ax+by+cz}+\frac{z}{ax+by+cz}\)
\(=\frac{x+y+z}{ax+by+cz}=2\)
Ta có : \(\begin{cases}x=by+cz\\y=ax+cz\\z=ax+by\end{cases}\) . Cộng các đẳng thức trên theo vế :
\(x+y+z=2\left(ax+by+cz\right)\)\(\Rightarrow\frac{x+y+z}{ax+by+cz}=2\)
Ta có : \(y=ax+cz\Rightarrow a=\frac{y-cz}{x}\Rightarrow a+1=\frac{x+y-cz}{x}\Rightarrow\frac{1}{a+1}=\frac{x}{x+y-cz}\)
\(\Rightarrow\frac{1}{a+1}=\frac{x}{ax+by+cz}\)
\(\Rightarrow P=\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=\frac{x+y+z}{ax+by+cz}=2\)
Tương tự : \(\frac{1}{b+1}=\frac{y}{ax+by+cz}\) ; \(\frac{1}{c+1}=\frac{z}{ax+by+cz}\)
Đặt \(ax^3=by^3=cz^3=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=\frac{k}{x^3}\\b=\frac{k}{y^3}\\c=\frac{k}{z^3}\end{matrix}\right.\)
Thay vào VT ta được :
\(VT=\sqrt[3]{x^2\cdot\frac{k}{x^3}+y^2\cdot\frac{k}{y^3}+z\cdot\frac{k}{z^3}}=\sqrt[3]{\frac{k}{x}+\frac{k}{y}+\frac{k}{z}}\)
\(=\sqrt[3]{k\cdot\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)}=\sqrt[3]{k}\) (1)
Thay vào VP ta được :
\(VP=\sqrt[3]{\frac{k}{x^3}}+\sqrt[3]{\frac{k}{y^3}}+\sqrt[3]{\frac{k}{z^3}}=\frac{\sqrt[3]{k}}{x}+\frac{\sqrt[3]{k}}{y}+\frac{\sqrt[3]{k}}{z}=\sqrt[3]{k}\cdot\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
\(=\sqrt[3]{k}\) (2)
Từ (1) và (2) \(\Rightarrow VT=VP\)
Ta có đpcm.
Ta có: \(ax^3+by^3+cz^3=\frac{ax^3}{x}+\frac{by^3}{y}+\frac{cz^3}{z}\)
mà \(ax^3=by^3=cz^3\)
\(\Rightarrow ax^3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=ax^3\)
\(\Rightarrow\sqrt[3]{ax^3+by^3+cz^3}=x\sqrt[3]{a}\\ \Leftrightarrow\frac{\sqrt[3]{ax^3+by^3+cz^3}}{x}=\sqrt[3]{a}\\ \Leftrightarrow\sqrt[3]{ax^3+by^3+cz^3}.\frac{1}{x}=\sqrt[3]{a}\)
Tương tự, ta có:
\(\sqrt[3]{ax^3+by^3+cz^3}.\frac{1}{y}=\sqrt[3]{b}\)
\(\sqrt[3]{ax^3+by^3+cz^3}.\frac{1}{z}=\sqrt[3]{c}\)
Cộng vế theo vế các đẳng thức, ta có:
\(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\sqrt[3]{ax^3+by^3+cz^3}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\\ =\sqrt[3]{ax^3+by^3+cz^3}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\left(đpcm\right)\)
Chúc bạn học tốt!
Làm biếng chép :'<
Link : Câu hỏi jj đó vào đây rồi biết :))
Đặt \(A=\frac{ax^2+by^2+cz^2}{ab\left(x-y\right)^2+bc\left(y-z\right)^2+cz\left(z-x\right)}\)
Từ ax+by+cz=0
=>(ax+by+cz)2=0
=>a2x2+b2y2+c2z2+2axby+2bycz+2czax=0
=>a2x2+b2y2+c2z2=-2(ax+by+byca+czax)
Xét mẫu thức: \(ab\left(x-y\right)^2+bc\left(y-z\right)^2+ca\left(z-x\right)^2\)
\(=ab\left(x^2-2xy+y^2\right)+bc\left(y^2-2yz+z^2\right)+ca\left(z^2-2zx+x^2\right)\)
\(=abx^2-2abxy+aby^2+bcy^2-2bcyz+bcz^2+caz^2-2cazx+cax^2\)
\(=\left(abx^2+bcz^2\right)+\left(aby^2+acz^2\right)+\left(acx^2+bcy^2\right)-2\left(abxy+bcyz+cazx\right)\)
\(=\left(aby^2+acz^2\right)+\left(abx^2+bcz^2\right)+\left(acx^2+bcy^2\right)+a^2x^2+b^2y^2+c^2z^2\)
\(=\left(a^2x^2+aby^2+acz^2\right)+\left(abx^2+b^2y^2+bcz^2\right)+\left(acx^2+bcy^2+c^2z^2\right)\)
\(=a\left(ax^2+by^2+cz^2\right)+b\left(ax^2+by^2+cz^2\right)+c\left(ax^2+by^2+cz^2\right)\)
\(=\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)\)
Do đó: \(A=\frac{ax^2+by^2+cz^2}{\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)}=\frac{1}{a+b+c}=\frac{1}{\frac{1}{2018}}=2018\) (dpcm)