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\(A=\dfrac{bcy^2+bcz^2+caz^2+cax^2+abx^2+aby^2-2bcyz-2cazx-2abxy}{ax^2+by^2+cz^2}=\dfrac{\left(bcy^2+bcz^2+caz^2+cax^2+abx^2+aby^2+a^2x^2+b^2y^2+c^2z^2\right)-\left(ax+by+cz\right)^2}{ax^2+by^2+cz^2}=\dfrac{\left(ax^2+by^2+cz^2\right)\left(a+b+c\right)}{ax^2+by^2+cz^2}=a+b+c\)
Vì \(x=by+cz\)
\(\Rightarrow by=x-cz\)
Mà \(z=ax+by\)
\(\Rightarrow by=z-ax\)
\(\Rightarrow x-cz=z-ax\left(=by\right)\)
\(\Rightarrow x+ax=z+cz\)
\(\Rightarrow x\left(a+1\right)=z\left(c+1\right)\)
Cũng có :
\(z=ax+by\)
\(\Rightarrow ax=z-by\)
\(y=ax+cz\)
\(\Rightarrow ax=y-cz\)
\(\Rightarrow z-by=y-cz\left(=ax\right)\)
\(\Rightarrow z+cz=y+by\)
\(\Rightarrow z\left(c+1\right)=y\left(b+1\right)\)
\(\Rightarrow x\left(a+1\right)=y\left(b+1\right)=z\left(c+1\right)\)
Đặt \(x\left(a+1\right)=y\left(b+1\right)=z\left(c+1\right)=k\)
\(\Rightarrow3k=x\left(a+1\right)+y\left(b+1\right)+z\left(c+1\right)\)
Có :
\(Q=\frac{1}{a+1}+\frac{1}{1+b}+\frac{1}{c+1}\)
\(=\frac{x}{x\left(a+1\right)}+\frac{y}{y\left(b+1\right)}+\frac{z}{z\left(c+1\right)}\)
\(=\frac{x}{k}+\frac{y}{k}+\frac{z}{k}\)
\(=\frac{x+y+z}{k}\)
\(=\frac{3\left(x+y+z\right)}{3k}\)
Mà \(3k=x\left(a+1\right)+y\left(b+1\right)+z\left(c+1\right)\)
\(\Rightarrow Q=\frac{3\left(x+y+z\right)}{x\left(a+1\right)+y\left(b+1\right)+z\left(c+1\right)}\)
\(=\frac{3\left(x+y+z\right)}{xa+x+by+y+zc+z}\)
\(=\frac{3\left(x+y+z\right)}{\left(x+y+z\right)+\left(xa+by+zc\right)}\)
\(=\frac{3\left(x+y+z\right)}{\left(x+y+z\right)+\frac{1}{2}\left[\left(xa+by\right)+\left(xa+zc\right)+\left(by+zc\right)\right]}\)
Có \(x+y+z=\left(ax+by\right)+\left(by+cz\right)+\left(ax+cz\right)\)
\(\Rightarrow Q=\frac{3\left(x+y+z\right)}{\left(x+y+z\right)+\frac{1}{2}\left(x+y+z\right)}\)
\(=\frac{3\left(x+y+z\right)}{\frac{3}{2}\left(x+y+z\right)}\)
\(=\frac{3}{\frac{3}{2}}\)
\(=2\)
Vậy \(Q=2.\)
Ta có:
\(ax+by+cz=0\Rightarrow\left(ax+by+cz\right)^2=0\)
\(\Rightarrow a^2x^2+b^2y^2+c^2z^2+2axby+2bycz+2axcz=0\)
\(\Rightarrow a^2x^2+b^2y^2+c^2z^2=-2axby-2bycz-2axcz\)
Ta có:
\(bc\left(y-z\right)^2+ca\left(z-x\right)^2+ab\left(x-y\right)^2=bc\left(y^2-2yz+z^2\right)+ca\left(z^2-2xz+x^2\right)+ab\left(x^2-2xy+y^2\right)\)
\(=bcy^2-2bcyz+bcz^2+acz^2-2acxz+acx^2+abx^2-2abxy+aby^2\)
\(=bcy^2+bcz^2+acz^2+acx^2+abx^2+aby^2-2axby-2bycz-2axcz\)
\(=bcy^2+bcz^2+acz^2+acx^2+abx^2+aby^2+a^2x^2+b^2y^2+c^2z^2\)
\(=\left(abx^2+a^2x^2+acx^2\right)+\left(bcy^2+aby^2+b^2y^2\right)+\left(bcz^2+acz^2+c^2z^2\right)\)
\(=ax^2\left(b+a+c\right)+by^2\left(c+a+b\right)+cz^2\left(b+a+c\right)\)
\(=\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)\)
Thay vào A ta được:
\(A=\dfrac{\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)}{ax^2+by^2+cz^2}=a+b+c\)
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x+y+z=0 sao tính được. sửa đề: x+y+z khác 0
Ta có: \(x+y=by+cz+ax+cz=2cz+z\Leftrightarrow2cz=x+y-z\Leftrightarrow c=\frac{x+y-z}{2z}\Leftrightarrow c+1=\frac{x+y+z}{2z}\Leftrightarrow\frac{1}{c+1}=\frac{2z}{x+y+z}\left(1\right)\)
Tương tự, ta có: \(\frac{1}{a+1}=\frac{2x}{x+y+z}\left(2\right);\frac{1}{b+1}=\frac{2y}{x+y+z}\left(3\right)\)
Cộng (1),(2),(3) vế với vế ta được:
\(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=\frac{2\left(x+y+z\right)}{x+y+z}=2\) hay Q = 2
Vậy Q=2
Với a, b, c khác -1 thì x + y + z khác 0.
Từ đề bài ta có: y + z = ax + cz + ax + by
<=> 2ax = y + z - x
--> a = (y + z - x)/(2x) --> a + 1 = (x + y + z)/(2x)
--> 1/(1 + a) = 2x/(x + y + z)
tương tự: 1/(1 + b) = 2y/(x + y + z)
1/(1 + c) = 2z/(x + y + z)
--> 1/(1 + a) + 1/(1 + b) + 1/(1 + c) = (2x + 2y + 2z)/(x + y + z) = 2
B1:Cong 7heo ve cac gia 7hie7: \(x+y+z=2\left(ax+by+cz\right)\)
Ma` \(x=by+cz\Leftrightarrow x\left(a+1\right)=ax+by+cz=\frac{x+y+z}{2}\)
\(\Leftrightarrow\frac{1}{a+1}=\frac{2x}{x+y+z}\).7uong 7u cho 2 dang 7huc con lai roi cong 7heo ve:
\(V7=\frac{2\left(x+y+z\right)}{x+y+z}=2=VP\) (DPCM)
B2: chu y \(a^5+b^5=\left(a+b\right)\left(a^4-a^3b+a^2b^2-ab^3+b^4\right)\)
\(=\left(a+b\right)\left(a^3\left(a-b\right)+a^2b^2-b^3\left(a-b\right)\right)\)
\(=\left(a+b\right)\left(\left(a-b\right)\left(a^3-b^3\right)+a^2b^2\right)\)
\(=\left(a+b\right)\left(\left(a-b\right)^2\left(a^2+b^2-ab\right)+a^2b^2\right)\)
\(\ge ab\left(a+b\right)\left(a^2+b^2-ab\right)\)\(\ge a^2b^2\left(a+b\right)\)
\(\Leftrightarrow a^5+b^5+ab\ge ab\left(ab\left(a+b\right)+abc\right)=a^2b^2\left(a+b+c\right)\)
\(\Leftrightarrow\frac{ab}{a^5+b^5+ab}\le\frac{abc}{ab\left(a+b+c\right)}=\frac{c}{a+b+c}\)
7uong 7u cho 2 BD7 con lai roi cong 7heo ve
\(V7\le\frac{a+b+c}{a+b+c}=1=VP\)
Dau "=" khi \(a=b=c=1\)