\(\left(2x-1\right)^2=2\)
\(\Rightarrow\left(2x-1\right)^2=\pm\sqrt{2}^2\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=\sqrt{2}\Rightarrow2x=\sqrt{2}+1\Rightarrow x=\dfrac{\sqrt{2}+1}{2}\\2x-1=-\sqrt{2}\Rightarrow2x=-\sqrt{2}+1\Rightarrow x=\dfrac{-\sqrt{2}+1}{2}\end{matrix}\right.\)
Từ \(3x=2y\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}\)\(\Rightarrow\dfrac{-x}{4}=\dfrac{-y}{6}\)
\(-y=2z\Rightarrow\dfrac{-y}{2}=\dfrac{z}{1}\)\(\Rightarrow\dfrac{-y}{6}=\dfrac{z}{3}\)
\(\Rightarrow-\dfrac{x}{4}=-\dfrac{y}{6}=\dfrac{z}{3}=k\Rightarrow\left\{{}\begin{matrix}x=-4k\\y=-6k\\z=3k\end{matrix}\right.\)
Khi đó \(xyz=576\Leftrightarrow\left(-4\right)k\cdot\left(-6\right)k\cdot3k=576\)
\(\Leftrightarrow72k^3=576\Leftrightarrow k^3=8\Leftrightarrow k=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=-4k=-4\cdot2=-8\\y=-6k=-6\cdot2=-12\\z=3k=3\cdot2=6\end{matrix}\right.\)
\(3x=2y;-y=2z\)
\(\Rightarrow\dfrac{x}{3}=\dfrac{y}{2};\dfrac{y}{-1}=\dfrac{z}{2}\)
\(\Rightarrow\dfrac{x}{3}=\dfrac{y}{2};\dfrac{y}{2}=\dfrac{z}{-4}\)
\(\Rightarrow\dfrac{x}{3}=\dfrac{y}{2}=\dfrac{z}{-4}\)
Đặt:
\(\dfrac{x}{3}=\dfrac{y}{2}=\dfrac{z}{-4}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=3k\\y=2k\\z=-4k\end{matrix}\right.\)
\(\Rightarrow3k.2k.-4k=576\)
\(-24k^3=576\)
\(\Rightarrow k^3=-24\)
Tính tiếp ạ,ở đây em không có máy tính,bấm cái này không được
Bài 1
a, ( 2x - 1 ) 2 = 2
4x2 - 1 = 2
4x2 = 3
x2 = \(\dfrac{3}{4}\)
x = \(\pm\sqrt{\dfrac{3}{4}}\)
b, 3x = 2y ; -y = 2z và xyz=576
Ta có : 3x = 2y => \(\dfrac{x}{2}=\dfrac{y}{3}\)
- y = 2z => \(\dfrac{y}{2}=\dfrac{z}{-1}\)
=> \(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{2}=\dfrac{z}{-1}\)
=> \(\dfrac{x}{4}=\dfrac{y}{6};\dfrac{y}{6}=\dfrac{z}{-3}\)
=> \(\dfrac{x}{4}=\dfrac{y}{6}=\dfrac{z}{-3}\)
Đặt \(\dfrac{x}{4}=k;\dfrac{y}{6}=k;\dfrac{z}{-3}=k\)
=> \(x=4k;y=6k;z=-3k\)
Mà x . y . z = 576
=> 4k . 6k . (-3)k = 576
=> (-72) . k3 = 576
=> k3 = (-8)
=> k = -2
Vì \(\dfrac{x}{4}=-2=>x=-8\)
\(\dfrac{y}{6}=-2=>y=-12\)
\(\dfrac{z}{-3}=-2=>z=6\)
Vậy x = -8 ; y = -12 ; z=6
