2x (x2 - 4)=0
giải hộ mik vs ạ
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c: \(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=-\sqrt{7}\\x=-5\\x=5\end{matrix}\right.\)
\(M=2x\left(-3x+2x^3\right)-x^2\left(3x^2-2\right)-x^2\left(x^2-4\right)\)
\(=-6x^2+4x^4-3x^4+2x^2-x^4+4x^2\)
\(=0\)
`a)(2x-1)^2-0,25=0`
`<=>(2x-1-0,5)(2x-1+0,5)=0`
`<=>(2x-1,5)(2x-0,5)=0`
`<=>[(x=0,75)(x=0,25):}`
`b)x^2+9=6x`
`<=>(x-3)^2=0`
`<=>x-3=0`
`<=>x=3`
`c)(x^2-4)-3x-6=0`
`<=>(x-2)(x+2)-3(x+2)=0`
`<=>(x+2)(x-2-3)=0`
`<=>(x+2)(x-5)=0`
`<=>[(x=-2),(x=5):}`
a: =>(2x-1-0,5)(2x-1+0,5)=0
=>(2x-1,5)(2x-0,5)=0
=>x=0,25 hoặc x=0,75
b: =>x^2-6x+9=0
=>(x-3)^2=0
=>x-3=0
=>x=3
c: =>(x-2)(x+2)-3(x+2)=0
=>(x+2)(x-5)=0
=>x=5 hoặc x=-2
\(\left(x^2+2x+3\right)^2-6\left(x^2+2x+3\right)=-9\\ \Rightarrow\left(x^2+2x+3\right)^2-6\left(x^2+2x+3\right)+9=0\\ \Rightarrow\left(x^2+2x+3-3\right)^2=0\\\Rightarrow \left(x^2+2x\right)^2=0\Rightarrow x^2+2x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
\(Q=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{x^2+y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+2x^2+2y^2}{2\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{2x^2+2y^2+4xy}{2\left(x-y\right)\left(x+y\right)}=\dfrac{2\left(x+y\right)^2}{2\left(x-y\right)\left(x+y\right)}=\dfrac{x+y}{x-y}\)
sửa đề :
\(x^3-3x^2+3x-1=0\)
\(\Leftrightarrow\left(x-1\right)^3=0\Leftrightarrow x=1\)
\(\left(x^2-9\right)^2-\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(x-3\right)^2\left(x+3\right)^2-\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(x-3\right)^2\left[\left(x+3\right)^2-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-3\right)^2=0\\\left(x+3\right)^2=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+3=1\\x+3=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\\x=-4\end{matrix}\right.\)
1:
2 ngày 8 giờ=56h
1,75 ngày=1 ngày 18h
3,4h=3h24p
309p=5h9p
\(\left\{{}\begin{matrix}2x=0\\x^2-4=0\end{matrix}\right.\) ==>\(\left\{{}\begin{matrix}x=0\\x=+,-2\end{matrix}\right.\)
sao ngắn vậy ạ ?