\(2x^2-3x+4x\left(1-\dfrac{1}{2}x\right)\)

\(=2x^2-3x+4x-4x\cdot\dfrac{1}{2}x\)

\(=2x^2-3x+4x-\left(4\cdot\dfrac{1}{2}\right)\left(x\cdot x\right)\)

\(=2x^2-3x+4x-2x^2\)

\(=\left(2x^2-2x^2\right)+\left(-3x+4x\right)\)

\(=x\)

\(\left(-2x^3-\dfrac{1}{4}y-4yz\right)8xy^2\)

\(=-2x^3\cdot8xy^2-\dfrac{1}{4}y\cdot8xy^2-4yz\cdot8xy^2\)

\(=\left(-2\cdot8\right)\left(x^3\cdot x\right)y-\left(\dfrac{1}{4}\cdot8\right)\left(y\cdot y^2\right)x-\left(4\cdot8\right)\left(y\cdot y^2\right)zx\)

\(=-16x^4y-2y^3x-32y^3zx\)

\(-\dfrac{1}{3}xz\left(-9xy+15yz\right)+3x^2\left(2yz^2-yz\right)\)

\(=-\dfrac{1}{3}xz\cdot-9xy+-\dfrac{1}{3}xz\cdot15yz+3x^2\cdot2yz^2-3x^2\cdot yz\)

\(=\left(-\dfrac{1}{3}\cdot-9\right)\left(x\cdot x\right)yz+\left(-\dfrac{1}{3}\cdot15\right)xy\left(z\cdot z\right)+\left(3.2\right)\left(x^2\cdot yz^2\right)-3\left(x^2\cdot yz\right)\)

=\(3x^2yz+-5xyz^2+6x^2yz^2-3x^2yz\)

\(=\left(3x^2yx-3x^2yx\right)-5x^2yz^2+6x^2yz\)

\(=-5x^2yz^2+6x^2yz\)

a Để \(E\) nhận giá trị nguyên thì:

\(3-x\)\(⋮\) \(x-1\)

\(\Rightarrow\left(3-1\right)-\left(x-1\right)\)\(⋮\) \(x-1\)

\(\Rightarrow2-\left(x-1\right)\) \(⋮\) \(x-1\)

\(\Rightarrow2\) \(⋮\) \(x-1\)

\(\Rightarrow x-1\inƯ_{\left(2\right)}\)

\(\Rightarrow x-1\in\left\{\pm1;\pm2\right\}\)

Ta có bảng giá trị:

Vậy \(x=\left\{-1;0;2;3\right\}\)

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