Question

Đã trả lời bởi chuyên gia

Answer 1

Mỗi lần hỏi 1 bài em nhé

\[\begin{array}{l}
a.{\rm{ }}\left( {2x + 3} \right)\left( {5x - 3} \right) - 2x\left( {5x - 1} \right)\\
= 10{{\rm{x}}^2} - 6{\rm{x}} + 15{\rm{x}} - 9 - 10{{\rm{x}}^2} + 2{\rm{x}}\\
= 11{\rm{x}} - 9\\
b.{\rm{ }}{\left( {x - 2} \right)^2} + \left( {2x - 8} \right)\left( {x + 6} \right)\\
= {x^2} - 4{\rm{x}} + 4 + 2{{\rm{x}}^2} + 12{\rm{x}} - 8{\rm{x}} - 48\\
= 3{{\rm{x}}^2} - 44\\
c.{\rm{ }}\left( {x - 5} \right)\left( {x + 5} \right) - {\left( {x - 3} \right)^2}\\
= {x^2} - {5^2} - \left( {{x^2} - 6{\rm{x}} + 9} \right)\\
= {x^2} - 25 - {x^2} + 6{\rm{x}} - 9\\
= 6{\rm{x}} - 34\\
d.{\rm{ }}{\left( {2x + 1} \right)^2} + 2\left( {4{x^2} - 1} \right) + {\left( {2x + 1} \right)^2}\\
= 4{{\rm{x}}^2} + 4{\rm{x}} + 1 + 8{{\rm{x}}^2} - 2 + 4{{\rm{x}}^2} + 4{\rm{x}} + 1\\
= 16{{\rm{x}}^2} + 8{\rm{x}}
\end{array}\]

Answer 2

`1,` Rút gọn:

`a, (2x+3)(5x-3)-2x(5x-1) = 10x^2 + 15x - 6x - 9 - 10x^2 + 2x = 17x - 9`

`b, (x-2)² + (2x-8)(x+6) = (x^2 - 4x + 4) + (2x^2 + 12x - 16x - 48) = x^2 - 4x + 4 + 2x^2 - 4x - 48 = 3x^2 - 8x - 44`

`c, (x-5)(x+5) - (x-3)² = (x^2 - 25) - (x^2 - 6x + 9) = x^2 - 25 - x^2 + 6x - 9 = 6x - 34`

`d, (2x+1)² + 2(4x²-1) + (2x+1)² = 4x^2 + 4x + 1 + 8x^2 - 2 + 4x^2 + 4x + 1 = 16x^2 + 12x`

`2,` Tìm x:

`a, (x+7)(x-7) - x(x+2) = 1`

`<=>x^2 - 49 - x^2 - 2x = 1`

`<=>-2x - 49 = 1 -2x = 50`

`=>x = -25`

`b, (2x-1)² - (2x+5)(2x-5) = 18`

`<=>4x^2 - 4x + 1 - (4x^2 - 25)`

`<=>4x^2 - 4x + 1 - 4x^2 + 25 = 18`

`<=>-4x + 26 = 18`

`<=>-4x = -8`

`=>x = 2`

`c, (2x-1)² - 36 = 0`

<=> 4x^2 - 4x + 1 - 36 = 0`

<=>4x^2 - 4x - 35 = 0 `

`=>x=-5/2,x=7/2`

`d, x³ - 4x = 0 x(x² - 4) = 0`

`<=>x(x - 2)(x + 2) = 0`

`*` x = 0`

`* x - 2=0=>x = 2`

`* x + 2=0 => x = -2`

`@CREAM`

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Answer 3