NT
H24
6 tháng 8 2023 lúc 11:39

`a,(2x-1)^3 -(2x+3)^3 + 48x(x+1)-2019`

`= (2x)^3 - 3.(2x)^2.1 + 3.2x.1^2 - 1^3 - [(2x)^3 + 3.(2x)^2.3+3.2x.3^2+3^3)]+48x.x+48x.1 -2019`

`=8x^3 - 12x^2 + 6x-1 - 8x^3 -36x^2- 54x+27 +48x^2+48x-2019`

`=-1993`

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`(7x+3)^2 -(2x+11)^2 - x(45x-2)-17`

\(=\left[\left(7x+3\right)-\left(2x+11\right)\right]\left[\left(7x+3\right)+\left(2x+11\right)\right]-\left(45x.x-x.2\right)-17\\ =\left(7x+3-2x-11\right)\left(7x+3+2x+11\right)-45x^2+2x-17\\ =\left(5x-8\right)\left(9x+14\right)-45x^2+2x-17\\ =45x^2+70x-72x-112-45x^2+2x-17\\ =-129\)

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NQ
6 tháng 8 2023 lúc 11:56

\(a,(2x−1)^3−(2x+3)^3+48x(x+1)−2019 \)

\(=8x^3-12x^2+6x-1-\left(8x^3+36x^2+54x+27\right)+48x^2+48x-2019\)

\(=8x^3-12x^2+6x-1-8x^3-36x^2-54x-27+48x^2+48x-2019\)

\(=-2047\)

\(b,\) \(( 7 x + 3 ) ^2 − ( 2 x + 11 ) ^2 − x ( 45 x − 2 ) − 17\)

\(=\left(7x+3-2x-11\right)\left(7x+3+2x+11\right)-45x^2+2x-17\)

\(=\left(5x-8\right)\left(9x+14\right)-45x^2+2x-17\)

\(=45x^2+70x-72x-112-45x^2+2x-17\)

\(=-129\)

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