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3\(x^2\).(5\(x\) + 1) + 6\(x^3\).(5\(x\) + 2) = 9\(x^3\) .(5\(x\) + 3)
15\(x^3\) + 3\(x^2\) + 30\(x^4\) + 12\(x^3\) = 45\(x^4\) + 27\(x^3\)
(15\(x^3\) + 12\(x^3\)) + 3\(x^2\) + 30\(x^4\) - 45\(x^4\) - 27\(x^3\) = 0
27\(x^3\) + 3\(x^2\) - 15\(x^4\) - 27\(x^3\) = 0
3\(x^2\) - 15\(x^4\) = 0
3\(x^2\).(1 - 5\(x^2\)) = 0
\(\left[{}\begin{matrix}x^2=0\\1-5x^2=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\5x^2=1\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=\mp\dfrac{\sqrt{5}}{5}\end{matrix}\right.\)
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ta co:2x^2-2xy=5x-y-19 | |||||||||||||||||||||||||||||||||||
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a, 5x - 16 = 40 + x
5x - x = 40 + 16
4x = 56
x = 14
b, 4x - 10 = 5x - x
4x - 5x + x = 10
0x = 10
Ko có giá trị của x
HOK TỐT
a) \(5x-16=40+x\)
\(\Leftrightarrow5x-x=40+16\)
\(\Leftrightarrow4x=56\)
\(\Leftrightarrow x=\frac{56}{4}\)
\(\Leftrightarrow x=14\)
b) \(4x-10=5x-x\)
\(\Leftrightarrow4x-5x+x=10\)
\(\Leftrightarrow0=10\left(vl\right)\)
Ta có : 5x.5x.5x=5.5.5.x.x.x=5^3.x^3=125.x^3
\(=125x^3\)