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a) Cách 1:
\(6(y - x) - 2(x - y)\)
\( = 6y - 6x - 2x + 2y\)
\( = 8y - 8x\)
Cách 2:
\(6(y - x) - 2(x - y)\\= 6(y-x)+2(y-x)\\=(6+2).(y-x)\\=8.(y-x)\\=8y-8x\)
b) \(3{x^2} + x - 4x - 5{x^2}\)
\( = (3{x^2} - 5{x^2}) + (x - 4x)\)
\( = - 2{x^2} - 3x\)
Ta có: \(B=\left|x-\dfrac{1}{7}\right|-\left|x+\dfrac{3}{5}\right|+\dfrac{4}{5}\)
\(=-x+\dfrac{1}{7}-x-\dfrac{3}{5}+\dfrac{4}{5}\)
\(=-2x+\dfrac{12}{35}\)
a) \(|x|-x\)
\(\Rightarrow\orbr{\begin{cases}x< 0\rightarrow\left|x\right|-x=2\left|x\right|\\x>0\rightarrow\left|x\right|-x=0\end{cases}}\)
\(\Rightarrow x=0\rightarrow x=0\)
a) 4x2(5x2 + 3) – 6x(3x3 – 2x + 1) – 5x3 (2x – 1)
= 4x2 . 5x2 + 4x2 . 3 – [6x . 3x3 + 6x . (-2x) + 6x . 1] – [5x3 . 2x + 5x3 . (-1)]
= 20x4 + 12x2 – (18x4 – 12x2 + 6x) – (10x4 – 5x3)
= 20x4 + 12x2 - 18x4 + 12x2 - 6x - 10x4 + 5x3
= (20x4 – 18x4 - 10x4 ) + 5x3 + (12x2 + 12x2 ) – 6x
= -8x4 + 5x3 + 24x2 – 6x
\(\begin{array}{l}b)\dfrac{3}{2}x\left( {{x^2} - \dfrac{2}{3}x + 2} \right) - \dfrac{5}{3}{x^2}(x + \dfrac{6}{5})\\ = \dfrac{3}{2}x.{x^2} + \dfrac{3}{2}x.( - \dfrac{2}{3}x) + \dfrac{3}{2}x.2 - (\dfrac{5}{3}{x^2}.x + \dfrac{5}{3}{x^2}.\dfrac{6}{5})\\ = \dfrac{3}{2}{x^3} - {x^2} + 3x - (\dfrac{5}{3}{x^3} + 2{x^2})\\ = \dfrac{3}{2}{x^3} - {x^2} + 3x - \dfrac{5}{3}{x^3} - 2{x^2}\\ = (\dfrac{3}{2}{x^3} - \dfrac{5}{3}{x^3}) + ( - {x^2} - 2{x^2}) + 3x\\ = \dfrac{{ - 1}}{6}{x^3} - 3{x^2} + 3x\end{array}\)
Ta có : \(\hept{\begin{cases}\left|x+3\right|\ge0\forall x\\\left|4-x\right|\ge0\forall x\\\left|x\right|\ge0\forall x\end{cases}\Rightarrow}\hept{\begin{cases}\left|x+3\right|=x+3\\\left|4-x\right|=4-x\\\left|x\right|=x\end{cases}}\)
\(\Rightarrow3\left|x-3\right|+2\left|4-x\right|+\left|x\right|\)
\(=3.\left(x-3\right)+2.\left(4-x\right)+x\)
\(=3x-9+8-2x+x\)
\(=\left(3x-2x+x\right)-\left(9-8\right)\)
\(=2x+1\)
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