( 0 + 10 ) - 10 : 0 x 0 - 10 = ?
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a, 7\(x\).(2\(x\) + 10) = 0
\(\left[{}\begin{matrix}x=0\\2x+10=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\2x=-10\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=-10:2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Vậy \(x\in\){-5; 0}
b, - 9\(x\) : (2\(x\) - 10) = 0
- 9\(x\) = 0
\(x\) = 0
c, (4 - \(x\)).(\(x\) + 3) = 0
\(\left[{}\begin{matrix}4-x=0\\x+3=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
Vậy \(x\in\) {-3; 4}
d, (\(x\) + 2023).(\(x\) - 2024) = 0
\(\left[{}\begin{matrix}x+2023=0\\x-2024=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-2023\\x=2024\end{matrix}\right.\)
Vậy \(x\) \(\in\) {-2023; 2024}
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a, 7\(x\).(2\(x\) + 10) =0
\(\left[{}\begin{matrix}x=0\\2x+10=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\2x=-10\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Vậy \(x\in\) {-5; 0}
b, -9\(x\) : (2\(x\) - 10) = 0
9\(x\) = 0
\(x\) = 0
c, (4 - \(x\)).(\(x\) + 3) = 0
\(\left[{}\begin{matrix}4-x=0\\x+3=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
Vậy \(x\in\) {-3; 4}
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![](https://rs.olm.vn/images/avt/0.png?1311)
nguyễn thị phương hoa ơi đáp án là 18505486,87
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Ta có:\(x^2+4y+4=0;y^2+4z+4=0;z^2+4x+4=0\)
\(\Leftrightarrow\left(x^2+4y+4\right)+\left(y^2+4z+4\right)+\left(z^2+4x+4\right)=0\)
\(\Leftrightarrow x^2+4x+4+y^2+4y+4+z^2+4z+4=0\)
\(\Leftrightarrow\left(x+2\right)^2+\left(y+2\right)^2+\left(z+2\right)^2=0\)
Mà\(\left(x+2\right)^2\ge0;\left(y+2\right)^2\ge0;\left(z+2\right)^2\ge0\)
\(\Leftrightarrow\left(x+2\right)^2+\left(y+2\right)^2+\left(z+2\right)^2\ge0\)
Dấu "=" xảy ra\(\Leftrightarrow\hept{\begin{cases}x+2=0\\y+2=0\\z+2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-2\\y=-2\\z=-2\end{cases}\Leftrightarrow}x=y=z=-2}\)
Vậy\(x^{10}+y^{10}+z^{10}=x^{10}+x^{10}+x^{10}\)
\(=3\cdot x^{10}=3\cdot\left(-2\right)^{10}=3\cdot1024=3072\)
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a) \(\left(x+1\right).10=0\)
\(x+1=0\)
\(x=0-1\)
\(x=-1\)
b) \(15\left(x+2\right)=0\)
\(x+2=0\)
\(x=0-2\)
\(x=-2\)
c) \(\left(x-1\right).10=10\)
\(x-1=10:10\)
\(x-1=1\)
\(x=1+1\)
x = 2
d) \(\left(x-2\right):5=0\)
\(x-2=0\)
\(x=2\)
{0+10} -10: 0 x 0 - 10
=10-10:0x0-10
=10-0x0-10
=10-0-10
=0
bằng 0