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`1/2x^2(-2x^2y^2z).(-1/3)x^2y^3`
`=1/2 .(-2).(-1/3)x^{2+2+2}.y^{2+3}.z`
`=1/3x^6y^5z`
Ta có: \(\dfrac{1}{2}x^2\cdot\left(-2x^2y^2z\right)\cdot\left(-\dfrac{1}{3}\right)x^2y^3\)
\(=\left(\dfrac{1}{2}\cdot2\cdot\dfrac{1}{3}\right)\cdot\left(x^2\cdot x^2\cdot x^2\right)\cdot\left(y^2\cdot y^3\right)\cdot z\)
\(=\dfrac{1}{3}x^6y^5z\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)
\(\Leftrightarrow\left(x-1\right)^x.\left(x-1\right)^2=\left(x-1\right)^x.\left(x-1\right)^4\)
\(\Leftrightarrow\left(x-1\right)^2=\left(x-1\right)^4\)
\(\Rightarrow x-1=\left\{-1;1\right\}\)
Nếu x - 1 = - 1 thì x = 0
Nếu x - 1 = 1 thì x = 2
Vậy x mang 2 giá trị là: x =2 hoặc x = 0
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\(x^4\ge0;3x^2\ge0;1>0\Rightarrow x^4+3x^2+1>0\Rightarrowđpcm\)
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\(\frac{x-1}{5}=\frac{y+4}{-3}=\frac{z-2}{1}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x-1}{5}=\frac{y+4}{-3}=\frac{z-2}{1}=\frac{\left(x-1\right)-5\left(y+4\right)+\left(z-2\right)}{5-5.\left(-3\right)+1}=\frac{-5}{21}\)
\(\Leftrightarrow\hept{\begin{cases}x-1=-\frac{5}{21}.5\\y+4=\frac{-5}{21}.\left(-3\right)\\z-2=-\frac{5}{21}.1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{4}{21}\\y=\frac{-23}{7}\\z=\frac{37}{21}\end{cases}}\)
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Vô nghiệm
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