tìm x,y,z biết (x-1)/3=(y+1)/2=z/5=18/3x+2y+z+1 24/11/2021 Bởi Everleigh tìm x,y,z biết (x-1)/3=(y+1)/2=z/5=18/3x+2y+z+1
Đáp án: $(x,y,z)\in\{(1,-1,0),(\dfrac{11}{3}, \dfrac79,\dfrac{40}{9})\}$ Giải thích các bước giải: Ta có :$\dfrac{x-1}{3}=\dfrac{y+1}{2}=\dfrac{z}{5}=\dfrac{18}{3x+2y+z+1}=k$ $\to x=3k+1,y=2k-1,z=5k$ $\to 3x+2y+z+1=3(3k+1)+2(2k-1)+5k+1=18k+2$ Mà $\dfrac{18}{3x+2y+z+1}=k$ $\to\dfrac{18k}{18k+2}=k$ $\to 18k=k(18k+2)$ $\to 18k=18k^2+2k$ $\to 18k^2-16k=0$$\to 2k(9k-8)=0$ $\to k\in\{0,\dfrac89\}$ $+) k=0\to x=1,y=-1,z=0$ $+) k=\dfrac89\to x=\dfrac{11}{3}, y=\dfrac79,z=\dfrac{40}{9}$ Bình luận
Đáp án: $(x,y,z)\in\{(1,-1,0),(\dfrac{11}{3}, \dfrac79,\dfrac{40}{9})\}$
Giải thích các bước giải:
Ta có :
$\dfrac{x-1}{3}=\dfrac{y+1}{2}=\dfrac{z}{5}=\dfrac{18}{3x+2y+z+1}=k$
$\to x=3k+1,y=2k-1,z=5k$
$\to 3x+2y+z+1=3(3k+1)+2(2k-1)+5k+1=18k+2$
Mà $\dfrac{18}{3x+2y+z+1}=k$
$\to\dfrac{18k}{18k+2}=k$
$\to 18k=k(18k+2)$
$\to 18k=18k^2+2k$
$\to 18k^2-16k=0$
$\to 2k(9k-8)=0$
$\to k\in\{0,\dfrac89\}$
$+) k=0\to x=1,y=-1,z=0$
$+) k=\dfrac89\to x=\dfrac{11}{3}, y=\dfrac79,z=\dfrac{40}{9}$