(3/4-x)^2020+(3/7*y-5)^4 be hon hoac bang 0 17/07/2021 Bởi Katherine (3/4-x)^2020+(3/7*y-5)^4 be hon hoac bang 0
Giải thích các bước giải: `(3/4-x)^2020+(3/7.y-5)^4<=0` Có $\left\{\begin{matrix}(\dfrac34-x)^{2020}\ge0\\(\dfrac37.y-5)^4\ge0\end{matrix}\right.$ `=>(3/4-x)^2020+(3/7.y-5)^4>=0` Mà `(3/4-x)^2020+(3/7.y-5)^4<=0` `=>(3/4-x)^2020+(3/7.y-5)^4=0` `=>`$\left\{\begin{matrix}\dfrac34-x=0\\\dfrac37.y-5=0\end{matrix}\right.$`=>`$\left\{\begin{matrix}x=\dfrac34\\\dfrac37.y=5\end{matrix}\right.$`=>`$\left\{\begin{matrix}x=\dfrac34\\y=\dfrac{35}3\end{matrix}\right.$ Vậy `x=3/4;y=35/3.` Bình luận
Giải thích các bước giải: Ta có:`(3/4-x)^2020ge0``(3/7.y-5)^4ge0`Mà `(3/4-x)^2020+(3/7.y-5)^4le0``=>(3/4-x)^2020+(3/7.y-5)=0``=>`$\left\{\begin{matrix} \dfrac{3}{4}-x=0& & \\ \dfrac{3}{7}.y-5=0& & \end{matrix}\right.$`=>`$\left\{\begin{matrix} x=\dfrac{3}{4}& & \\ \dfrac{3}{7}.y=5& & \end{matrix}\right.$`=>`$\left\{\begin{matrix} \dfrac{3}{4}-x=0& & \\ y=\dfrac{35}{3}& & \end{matrix}\right.$ Bình luận
Giải thích các bước giải:
`(3/4-x)^2020+(3/7.y-5)^4<=0`
Có $\left\{\begin{matrix}(\dfrac34-x)^{2020}\ge0\\(\dfrac37.y-5)^4\ge0\end{matrix}\right.$
`=>(3/4-x)^2020+(3/7.y-5)^4>=0`
Mà `(3/4-x)^2020+(3/7.y-5)^4<=0`
`=>(3/4-x)^2020+(3/7.y-5)^4=0`
`=>`$\left\{\begin{matrix}\dfrac34-x=0\\\dfrac37.y-5=0\end{matrix}\right.$`=>`$\left\{\begin{matrix}x=\dfrac34\\\dfrac37.y=5\end{matrix}\right.$`=>`$\left\{\begin{matrix}x=\dfrac34\\y=\dfrac{35}3\end{matrix}\right.$
Vậy `x=3/4;y=35/3.`
Giải thích các bước giải:
Ta có:
`(3/4-x)^2020ge0`
`(3/7.y-5)^4ge0`
Mà `(3/4-x)^2020+(3/7.y-5)^4le0`
`=>(3/4-x)^2020+(3/7.y-5)=0`
`=>`$\left\{\begin{matrix}
\dfrac{3}{4}-x=0& & \\
\dfrac{3}{7}.y-5=0& &
\end{matrix}\right.$
`=>`$\left\{\begin{matrix}
x=\dfrac{3}{4}& & \\
\dfrac{3}{7}.y=5& &
\end{matrix}\right.$
`=>`$\left\{\begin{matrix}
\dfrac{3}{4}-x=0& & \\
y=\dfrac{35}{3}& &
\end{matrix}\right.$