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\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\Leftrightarrow\frac{x+y}{xy}-\frac{x+y}{z\left(x+y+z\right)}=0\Leftrightarrow\left(x+y\right)\left(z+x\right)\left(y+z\right)=0\)
<=> x=-y hoặc y=-z hoặc z=-x
=> B=0
( Các bước làm tóm tắt ):))
Lời giải:
Ta có:
\(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right):\left(\frac{1}{x+y+z}\right)=1\)
\(\Leftrightarrow \frac{xy+yz+xz}{xyz}.(x+y+z)=1\Leftrightarrow (xy+yz+xz)(x+y+z)=xyz\)
\(\Leftrightarrow xy(x+y)+yz(y+z)+xz(x+z)+2xyz=0\)
\(\Leftrightarrow xy(x+y+z)+yz(y+z+x)+xz(x+z)=0\)
\(\Leftrightarrow y(x+y+z)(x+z)+xz(x+z)=0\)
\(\Leftrightarrow (x+z)[y(x+y+z)+xz]=0\)
\(\Leftrightarrow (x+z)(y+x)(y+z)=0\)
Do đó:
\(B=(x+y)(x^{20}+....+y^{20})(y+z)(y^{10}+...+z^{10})(z+x)(z^{2016}+x^{2016})\)
\(=(x+y)(y+z)(x+z)(x^{20}+..+y^{20})(y^{10}+..+z^{10})(z^{2016}+x^{2016})=0\)
Lời giải:
Ta có:
\(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right):\left(\frac{1}{x+y+z}\right)=1\)
\(\Leftrightarrow \frac{xy+yz+xz}{xyz}.(x+y+z)=1\Leftrightarrow (xy+yz+xz)(x+y+z)=xyz\)
\(\Leftrightarrow xy(x+y)+yz(y+z)+xz(x+z)+2xyz=0\)
\(\Leftrightarrow xy(x+y+z)+yz(y+z+x)+xz(x+z)=0\)
\(\Leftrightarrow y(x+y+z)(x+z)+xz(x+z)=0\)
\(\Leftrightarrow (x+z)[y(x+y+z)+xz]=0\)
\(\Leftrightarrow (x+z)(y+x)(y+z)=0\)
Do đó:
\(B=(x+y)(x^{20}+....+y^{20})(y+z)(y^{10}+...+z^{10})(z+x)(z^{2016}+x^{2016})\)
\(=(x+y)(y+z)(x+z)(x^{20}+..+y^{20})(y^{10}+..+z^{10})(z^{2016}+x^{2016})=0\)
Từ giải thiết ta suy ra được: \(\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
Thay vào thì P=0
P/S: Tìm trên gg cũng có thể loại này :v
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\left(x;y;z,x+y+z\ne0\right)\)
\(\Rightarrow\frac{xy+yz+xz}{xyz}=\frac{1}{x+y+z}\)
\(\Rightarrow\left(xy+yz+xz\right)\left(x+y+z\right)=xyz\)
\(\Leftrightarrow\left(xy+yz+xz\right)\left(x+y+z\right)-xyz=0\)
\(\Leftrightarrow\left(xy+yz\right)\left(x+y+z\right)+xz\left(x+z\right)=0\)
\(\Leftrightarrow y\left(x+z\right)\left(x+y+z\right)+xz\left(x+z\right)=0\)
\(\Leftrightarrow\left(x+z\right)\left(xy+y^2+yz\right)+xz\left(x+z\right)=0\)
\(\Leftrightarrow\left(x+z\right)\left(xy+y^2+yz+xz\right)=0\)
\(\Leftrightarrow\left(x+z\right)\left[y\left(x+y\right)+z\left(x+y\right)\right]=0\)
\(\Leftrightarrow\left(x+z\right)\left(x+y\right)\left(y+z\right)=0\)
Từ đó \(x=-z\)hoặc \(x=-y\)hoặc \(y=-z\)
-Nếu \(x=-z\Rightarrow z^{2017}+x^{2017}=0\Rightarrow M=\frac{19}{4}+0=\frac{19}{4}\)
Tương tự với các trường hợp còn lại, ta cũng tính được \(M=\frac{19}{4}\)
\(x^3+3x^2+3x+1+y^3+3y^3+3y+1+x+y+2=0\)
\(\Leftrightarrow\left(x+1\right)^3+\left(y+1\right)^3+x+y+2=0\)
\(\Leftrightarrow\left(x+y+2\right)\left(\left(x+1\right)^2+\left(y+1\right)^2-\left(x+1\right)\left(y+1\right)\right)+\left(x+y+2\right)=0\)
\(\Leftrightarrow\left(x+y+2\right)\left(\left(x+1\right)^2+\left(y+1\right)^2-\left(x+1\right)\left(y+1\right)+1\right)=0\)
\(\Leftrightarrow x+y+2=0\)
(phần trong ngoặc \(\left(x+1\right)^2-\left(x+1\right)\left(y+1\right)+\frac{\left(y+1\right)^2}{4}+\frac{3\left(y+1\right)^2}{4}+1\)
\(=\left(x+1-\frac{y+1}{4}\right)^2+\frac{3\left(y+1\right)^2}{4}+1\) luôn dương)
\(\Rightarrow x+y=-2\)
Mà \(xy>0\Rightarrow\left\{{}\begin{matrix}x< 0\\y< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}-x>0\\-y>0\end{matrix}\right.\)
Ta có: \(\frac{1}{-x}+\frac{1}{-y}\ge\frac{4}{-\left(x+y\right)}=2\) \(\Leftrightarrow\frac{1}{x}+\frac{1}{y}\le-2\) (đpcm)
Dấu "=" xảy ra khi và chỉ khi \(x=y=-1\)
2/ \(x;y;z\ne0\)
\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
\(\Leftrightarrow\frac{x+y}{xy}+\frac{1}{z}-\frac{1}{x+y+z}=0\)
\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{xz+yz+z^2}=0\)
\(\Leftrightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{xz+yz+z^2}\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(\frac{xy+yz+xz+z^2}{xyz\left(x+y+z\right)}\right)=0\)
\(\Leftrightarrow\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz\left(x+y+z\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\) dù trường hợp nào thì thay vào ta đều có \(B=0\)
3/ \(\Leftrightarrow mx-2x+my-y-1=0\)
\(\Leftrightarrow m\left(x+y\right)-\left(2x+y+1\right)=0\)
Gọi \(A\left(x_0;y_0\right)\) là điểm cố định mà d đi qua
\(\Leftrightarrow\left\{{}\begin{matrix}x_0+y_0=0\\2x_0+y_0+1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_0=-1\\y_0=1\end{matrix}\right.\)
Vậy d luôn đi qua \(A\left(-1;1\right)\) với mọi m
\(\frac{\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)}{\frac{1}{x+y+x}}=1\Leftrightarrow\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{x}\right)=1\)
\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
\(\Rightarrow\left(\frac{1}{x}+\frac{1}{y}\right)+\left(\frac{1}{z}-\frac{1}{x+y+z}\right)=0\)
\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\)\(\Leftrightarrow\left(x+y\right)\left[z\left(x+y+z\right)+xy\right]=0\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(x+z\right)=0\)
B=\(\left(x+y\right)\left(y+z\right)\left(z+x\right).M=0\)
M ở đâu ra thế bạn