3x-2y/4 = 2z-4x/3 = 4y-3z/2 và x ²+ z ²= 2(yz-8) tìm x,y,z 14/08/2021 Bởi Reagan 3x-2y/4 = 2z-4x/3 = 4y-3z/2 và x ²+ z ²= 2(yz-8) tìm x,y,z
Giải thích các bước giải: Ta có :$\dfrac{3x-2y}{4}=\dfrac{2z-4x}3\to 9x-6y=8z-16x\to 25x=6y+8z $ $\dfrac{3x-2y}4=\dfrac{4y-3z}2\to 3x-2y=8y-6z\to 3x=10y-6z$ $\to 75x+12x=3(6y+8z)+4(10y-6z)\to 87x=58y\to y=\dfrac{87x}{58}=\dfrac{3x}{2}$ $\to z=\dfrac{10y-3x}{6}=\dfrac{10\dfrac{3x}{2}-3x}{6}=2x$ $\to x^2+(2x)^2=2(2.\dfrac{3x}{2}.2x-8)\to 5x^2=12x^2-16\to 7x^2=16\to x^2=\dfrac{16}7$ $\to x=\dfrac{4}{\sqrt{7}}\to y=\dfrac{6}{\sqrt{7}},z=\dfrac{8}{\sqrt{7}}$ Hoặc $ x=\dfrac{-4}{\sqrt{7}}\to y=\dfrac{-6}{\sqrt{7}},z=\dfrac{-8}{\sqrt{7}}$ Bình luận
Giải thích các bước giải:
Ta có :
$\dfrac{3x-2y}{4}=\dfrac{2z-4x}3\to 9x-6y=8z-16x\to 25x=6y+8z $
$\dfrac{3x-2y}4=\dfrac{4y-3z}2\to 3x-2y=8y-6z\to 3x=10y-6z$
$\to 75x+12x=3(6y+8z)+4(10y-6z)\to 87x=58y\to y=\dfrac{87x}{58}=\dfrac{3x}{2}$
$\to z=\dfrac{10y-3x}{6}=\dfrac{10\dfrac{3x}{2}-3x}{6}=2x$
$\to x^2+(2x)^2=2(2.\dfrac{3x}{2}.2x-8)\to 5x^2=12x^2-16\to 7x^2=16\to x^2=\dfrac{16}7$
$\to x=\dfrac{4}{\sqrt{7}}\to y=\dfrac{6}{\sqrt{7}},z=\dfrac{8}{\sqrt{7}}$
Hoặc $ x=\dfrac{-4}{\sqrt{7}}\to y=\dfrac{-6}{\sqrt{7}},z=\dfrac{-8}{\sqrt{7}}$