\(x^3+3x^2+3x+1+y^3+3y^3+3y+1+x+y+2=0\)

\(\Leftrightarrow\left(x+1\right)^3+\left(y+1\right)^3+x+y+2=0\)

\(\Leftrightarrow\left(x+y+2\right)\left(\left(x+1\right)^2+\left(y+1\right)^2-\left(x+1\right)\left(y+1\right)\right)+\left(x+y+2\right)=0\)

\(\Leftrightarrow\left(x+y+2\right)\left(\left(x+1\right)^2+\left(y+1\right)^2-\left(x+1\right)\left(y+1\right)+1\right)=0\)

\(\Leftrightarrow x+y+2=0\)

(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

3) áp dụng đẳng thức \(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)

<=>\(1-3xyz=1\left(1-xy-yz-zx\right)\)

<=>\(3xyz=xy+yz+zx\)

mặt khác ta có \(\left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2yz+2zx=1\)

<=>\(1+2xy+2yz+2zx=1\)

<=> \(xy+yz+zx=0\)

do đó 3xyz=0<=> \(\hept{\begin{cases}x=0\\y=0\\z=0\end{cases}}\)

lần lượt thay x;y;z vào hệ ta có các cặp nghiệm (x;y;z)=(0;0;1),(0;1;0),(1;0;0)

do đó x^2017+y^2017+z^2017=1 

bạn ơi bài 3 thì x^3+y^3+z^3 bằng mấy đấy 

     \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\left(x;y;z,x+y+z\ne0\right)\)

\(\Rightarrow\frac{xy+yz+xz}{xyz}=\frac{1}{x+y+z}\)

\(\Rightarrow\left(xy+yz+xz\right)\left(x+y+z\right)=xyz\)

\(\Leftrightarrow\left(xy+yz+xz\right)\left(x+y+z\right)-xyz=0\)

\(\Leftrightarrow\left(xy+yz\right)\left(x+y+z\right)+xz\left(x+z\right)=0\)

\(\Leftrightarrow y\left(x+z\right)\left(x+y+z\right)+xz\left(x+z\right)=0\)

\(\Leftrightarrow\left(x+z\right)\left(xy+y^2+yz\right)+xz\left(x+z\right)=0\)

\(\Leftrightarrow\left(x+z\right)\left(xy+y^2+yz+xz\right)=0\)

\(\Leftrightarrow\left(x+z\right)\left[y\left(x+y\right)+z\left(x+y\right)\right]=0\)

\(\Leftrightarrow\left(x+z\right)\left(x+y\right)\left(y+z\right)=0\)

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 \(M=\frac{19}{4}\)

tự túc

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

Nhanh k cho nè

làm lần lượt nhá,dài dòng quá khó coi.ahihihi!

\(\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{7\left(\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)

\(=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}=\frac{1}{4}\)

b

Tổng quát:\(1-\frac{1}{1+2+3+....+n}=1-\frac{1}{\frac{n\left(n+1\right)}{2}}=1-\frac{2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n^2+2n\right)-\left(n+2\right)}{n\left(n+1\right)}\)

\(=\frac{n\left(n+2\right)-\left(n+2\right)}{n\left(n+1\right)}=\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)

Thay số vào,ta được:

\(\frac{\left(2-1\right)\left(2+2\right)}{2\left(2+1\right)}\cdot\frac{\left(3-1\right)\left(3+2\right)}{3\left(3+1\right)}\cdot.....\cdot\frac{\left(2017-1\right)\left(2017+2\right)}{2017\left(2017+1\right)}\)

\(=\frac{1\cdot4}{2\cdot3}\cdot\frac{2\cdot5}{3\cdot4}\cdot...\cdot\frac{2016\cdot2019}{2017\cdot2018}\)

\(=\frac{1\cdot2\cdot3\cdot...\cdot2016}{2\cdot3\cdot4\cdot...\cdot2017}\cdot\frac{4\cdot5\cdot6\cdot...\cdot2019}{3\cdot4\cdot5\cdot...\cdot2018}\)

\(=\frac{1}{2017}\cdot\frac{2019}{3}=\frac{2019}{6051}\)