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a: =>7-x=0
hay x=7
b: \(\Leftrightarrow\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\left(x+5\right)\left(3x-8\right)=0\)
hay \(x\in\left\{\sqrt{2};-\sqrt{2};-5;\dfrac{8}{3}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>-x+7=0
hay x=7
b: \(\Leftrightarrow\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\left(x+5\right)\left(3x-8\right)=0\)
hay \(x\in\left\{\sqrt{2};-\sqrt{2};-5;\dfrac{8}{3}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
-2x^2 + 3x -1 = -2x^2 +2x + x -1
=-2x(x-1) + (x - 1)
=(-2x+1)(x-1)
![](https://rs.olm.vn/images/avt/0.png?1311)
b) \(16x-5x^2-3=5x\left(3-x\right)-\left(3-x\right)=\left(3-x\right)\left(5x-1\right)\)
c) \(2x^2+3x-5=2x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(2x+5\right)\)
d) \(2x^2+3x-5=2x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(2x+5\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(5x^2\left(3x^3-2x^2+x+2\right)\)
\(=15x^5-10x^4+5x^3+10x^2\)
b: \(3x^4\left(-2x^3+5x^2-\dfrac{2}{3}x+\dfrac{1}{3}\right)\)
\(=-6x^7+15x^6-2x^5+x^4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(2x2 – 3x)(5x2 – 2x + 1)
= 2x2(5x2 – 2x + 1) + (-3x)(5x2 – 2x + 1)
= 2x2.5x2 + 2x2.(-2x) + 2x2.1 + (–3x).5x2 + (-3x).(-2x) + (-3x).1
= (2.5)(x2.x2) + (2. (-2)).(x2.x) + 2x2 + [(-3).5].(x.x2) + [(-3).(-2).(x.x) + (-3x)
= 10x4 – 4x3 + 2x2 – 15x3 + 6x2 – 3x
= 10x4 – (4x3 + 15x3) + (2x2 + 6x2) – 3x
= 10x4 – 19x3 + 8x2 – 3x
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(5x^2-2x+1\right)\left(2x^2-3x\right)=10x^4-15x^3-4x^3+6x^2+2x^2-3x\)
\(=10x^4-19x^3+8x^2-3x\)
2x2 hay 2x3 bn?
ta có : \(2x^2-5x^2+3x=0\)
\(\Leftrightarrow-3x^2+3x=0\)
\(\Leftrightarrow-3x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-3x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy tập nghiệm \(S=\left\{0;1\right\}\)