p=1 rồi kìa

Xl n mk viết nhầm

Theo bài ra, ta có:  \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=k\) \(\Rightarrow\hept{\begin{cases}x=3k\\y=4k\\z=5k\end{cases}}\)

Ta có: \(P=\frac{2017x+2018y-2019z}{2017x-2018y+2019z}=\frac{2017.3k+2018.4k-2019.5k}{2017.3k-2018.4k+2019.5k}\)

\(P=\frac{6051k+8072k-10095k}{6051k-8072k+10095k}=\frac{k\left(6051+8072-10095\right)}{k\left(6051-8072+10095\right)}=\frac{4028}{8074}=\frac{2014}{4037}\)

Ta có:Đặt\(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=k\)

\(\Rightarrow\hept{\begin{cases}x=3k\\y=4k\\z=5k\end{cases}}\)

Thay vào đề bài

\(\Rightarrow P=\frac{2017x+2018y-2019z}{2017x-2018y+2019z}=\frac{2017.3.k+2018.4.k-2019.5.k}{2017.3.k-2018.4.k+2019.5.k}=\frac{4028k}{8074k}=\frac{2014}{4037}\)

                                                   Vậy\(P=\frac{2014}{4037}\)

Ta có:Vì x,y,z tỉ lệ với 3,4,5 nên

\(\Rightarrow\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)

Do đó đặt:\(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=3k\\y=4k\\z=5k\end{matrix}\right.\)

Thay vào P

\(\Rightarrow P=\frac{2017.3.k+2018.4.k-2019.5.k}{2017.3.k-2018.4.k+2019.5.k}=\frac{4028.k}{8074.k}=\frac{2014}{4037}\)

Vậy\(P=\frac{2014}{4037}\)

Kí hiệu  x + 2 m y − z = 1     ( 1 ) 2 x − m y − 2 z = 2     ( 2 ) x − ( m + 4 ) y − z = 1     ( 3 )

Lấy (1) – (3) vế với vế ta được 3 m + 4 y = 0 ⇔ y = 0    ( d o   m ≠ 0 ; − 4 3 )

Khi đó x − z = 1 y = 0

Ta có T = 2017 x − 2018 y − 2017 z = 2017 x − z = 2017

Đáp án cần chọn là: C

Lời giải:

Ta có: \(x^2+y^2+z^2=xy+yz+xz\)

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

\(\Leftrightarrow \frac{(x-y)^2+(y-z)^2+(z-x)^2}{2}=0\)

\(\Leftrightarrow (x-y)^2+(y-z)^2+(z-x)^2=0\)

Vì \((x-y)^2; (y-z)^2;(z-x)^2\geq 0\), do đó để tổng của chúng bằng $0$ thì:

\((x-y)^2=(y-z)^2=(z-x)^2=0\Rightarrow x=y=z\)

\(\Rightarrow 3x^{2017}=3y^{2017}=3z^{2017}=x^{2017}+y^{2017}+z^{2017}=9\)

\(\Rightarrow x=y=z=\sqrt[2017]{3}\)

\(\Rightarrow \left(\frac{2017x+2018y-4023z}{3}\right)^{2017}=\left(\frac{12x}{3}\right)^{2017}=(4x)^{2017}=3.4^{2017}\)

Từ \(x^2+y^2+z^2=xy+yz+zx\)

\(\Rightarrow2x^2+2y^2+2z^2=2xy+2yz+2zx\)

\(\Rightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=0\)

\(\Rightarrow\left(x^2-2xy-y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)=0\)

\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

Mà \(\left(x-y\right)^2;\left(y-z\right)^2;\left(z-x\right)^2\ge0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(y-z\right)^2=0\\\left(z-x\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-z=0\\z-x=0\end{matrix}\right.\Leftrightarrow x=y=z\)

Với \(x^{2017}+y^{2017}+z^{2017}=9\)

\(\Leftrightarrow3x^{2017}=9\Leftrightarrow x^{2017}=3\Leftrightarrow x=\sqrt[2017]{3}=y=z\)

\(\Rightarrow\left(\dfrac{2017x+2018y-4023z}{3}\right)^{2017}=\left(\dfrac{2017x+2018x-4032x}{3}\right)^{2017}=\left(\dfrac{9x}{3}\right)^{2017}=\left(3x\right)^{2017}=\left(3\sqrt[2017]{3}\right)^{2017}=3^{2017}\cdot3=3^{2018}\)

\(x^2+xy-2017x-2018y-2019=0\)

\(x^2+xy+x-2018x-2018y-2018-1=0\)

\(x\left(x+y+1\right)-2018\left(x+y+1\right)=1\)

\(\left(x-2018\right)\left(x+y+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2018\\x=-y-1\end{matrix}\right.\)

Đến đây rồi e thay vào phương trình dùng delta giải phương trình bậc 2 nha