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\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)
\(\Rightarrow xy+yz+xz=0\)
\(\Rightarrow\left\{{}\begin{matrix}xy=-yz--xz\\yz=-xy-xz\\xz=-xy-xz\end{matrix}\right.\)
\(\dfrac{yz}{x^2+2yz}=\dfrac{yz}{x^2+yz-xy-xz}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)
CMTT:
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(x-y\right)\left(x-z\right)}\\\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(x-y\right)\left(x-z\right)}\\\dfrac{yz}{x^2+2yz}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\end{matrix}\right.\)
A=\(\dfrac{xz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xy}{\left(x-y\right)\left(x-z\right)}+\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)
\(A=\dfrac{xz+xy+yz}{\left(x-y\right)\left(x-z\right)}\left(1\right)\)
mà \(xy+yz+xz=0\)
Từ \(\Rightarrow\dfrac{xz+xy+yz}{\left(x-y\right)\left(x-z\right)}=0\)
Vậy A=0
Dễ dàng chứng minh được : nếu \(a+b+c=0\) thì \(a^3+b^3+c^3=3abc\)
Ta có \(\frac{xy}{z^2}+\frac{yz}{x^2}+\frac{zx}{y^2}=xyz\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)=xyz.\frac{3}{xyz}=3\)( Vì \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\))
1) VT= \(\frac{1}{1+x+xy}+\frac{x}{x+xy+1}+\frac{xyz}{xyz+z+zx}\)
\(=\frac{1}{1+x+xy}+\frac{xy}{1+x+xy}+\frac{xyz}{z\left(x+xy+1\right)}\)
\(=\frac{1}{1+x+xy}+\frac{x}{1+x+xy}+\frac{xy}{1+x+xy}\)
\(=\frac{1+x+xy}{1+x+xy}=1\)
Bài 2 giả thiết trên tử làm mell gì có bình phương, nếu có thì tính làm gì nữa :D, kết quả là 2016(x+y+z)
a, Chứng minh \(x^3+y^3+z^3=\left(x+y\right)^3-3xy.\left(x+y\right)+z^3\)
Biến đổi vế phải thì ta phải suy ra điều phải chứng minh
b, Ta có: \(a+b+c=0\)thì
\(a^3+b^3+c^3==\left(a+b\right)^3-3ab\left(a+b\right)+c^3=-c^3-3ab\left(-c\right)+c^3=3abc\)
( Vì \(a+b+c=0\)nên \(a+b=-c\))
Theo giả thuyết \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Rightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\)
Khi đó \(A=\frac{yz}{x^2}+\frac{xz}{y^2}+\frac{xy}{z^2}\)
\(=\frac{xyz}{x^3}+\frac{xyz}{y^3}+\frac{xyz}{z^3}\)
\(=xyz\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)\)
\(=xyz.\frac{3}{xyz}=3\)
Đặt bài toán phụ : Chứng minh nếu \(a+b+c=0\Rightarrow a^3+b^3+c^3=3abc\)
Thật vậy :
\(a^3+b^3+c^3=\left(a+b+c\right)^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
\(a+b+c=0\Rightarrow\left(a+b+c\right)^3=0\)
\(a+b=-c\)
\(b+c=-a\)
\(c+a=-b\)
\(\Rightarrow\left(a+b+c\right)^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
\(=-3\left(-c\right)\left(-b\right)\left(-a\right)\)
\(=3abc\)
Trở lại bài toán chính :
Ta có:
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Rightarrow\frac{yz}{xyz}+\frac{xz}{xyz}+\frac{xy}{xyz}=0\)
\(\Rightarrow\frac{yz+xz+xy}{xyz}=0\)
\(\Rightarrow xy+xz+yz=0\)
\(\Rightarrow\left(xy\right)^3+\left(xz\right)^3+\left(yz^3\right)=3\left(xy\right)\left(xz\right)\left(yz\right)=3x^2y^2z^2\)
Lại có:
\(P=\frac{xy.y^2x^2}{x^2y^2z^2}+\frac{xz.z^2.x^2}{x^2y^2z^2}+\frac{z^2.y^2.yz}{x^2y^2z^2}\)
\(=\frac{\left(xy\right)^3}{x^2y^2z^2}+\frac{\left(xz\right)^3}{x^2y^2z^2}+\frac{\left(yz\right)^3}{x^2y^2z^2}\)
\(=\frac{\left(xy\right)^3+\left(xz\right)^3+\left(yz^3\right)}{x^2y^2z^2}\)
Thay \(\left(xy\right)^3+\left(xz\right)^3+\left(yz^3\right)=3x^2y^2z^2;\)ta có:
\(P=\frac{3x^2y^2z^2}{x^2y^2z^2}\)
\(=3\)
Vậy \(P=3.\)
Ta có \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\)
\(\Rightarrow\left(\frac{1}{x}+\frac{1}{y}\right)^3=\frac{1}{z}^3\)
\(\Rightarrow\frac{1}{x^3}+\frac{1}{y^3}+3\cdot\frac{1}{x}\cdot\frac{1}{y}\cdot\left(\frac{1}{x}+\frac{1}{y}\right)=\frac{1}{z^3}\)
\(\Rightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=-3\cdot\frac{1}{x}\cdot\frac{1}{y}\cdot\left(\frac{1}{x}+\frac{1}{y}\right)=\frac{3}{xyz}.\)Vì \(\frac{1}{x}+\frac{1}{y}=\frac{-1}{z}\)
Mặt khác : \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Rightarrow\frac{xy+yz+zx}{xyz}=0\)
\(\Rightarrow xy+yz+zx=0\)`
\(A=\frac{yz}{x^2}+2yz+\frac{xz}{y^2}+2xz+\frac{xy}{z^2}+2xy\)
\(=\frac{xyz}{x^3}+\frac{xyz}{y^3}+\frac{xyz}{z^3}+2\left(xy+yz+xz\right)\)Vì x , y , z khác 0 .
\(=xyz\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)\)Vì \(xy+yz+xz=0\)
\(=xyz\cdot\frac{3}{xyz}\)Vì \(\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)=\frac{3}{xyz}\)
\(=3\)
Vậy \(A=3\)
mk tưởng chố \(\left(\frac{1}{x}+\frac{1}{y}\right)^3\)phải bằng\(\left(\frac{-1}{z}\right)^3\)chứ