Question

1.Tìm x a) (x+2)*(3x+2)-(3x-1)*(x-5)=11 b) 2*(2x+1)*(8x-3)+(3-4x)*(8x-7)=6x+73 2. Rút gọn A=(x^3-x+1)*(2x+1)+(x+1)*(x-2) B=(2x+5)*(2x-3)-5*(x+3) (* là giấu nhân) Giúp e với ạ . E cảm ơn

Answer 1

\(a,\left(x+2\right)\left(3x+2\right)-\left(3x-1\right)\left(x-5\right)=11\\ \Leftrightarrow\left(3x^2+6x+2x+4\right)-\left(3x^2-15x-x+5\right)=11\\ \Leftrightarrow3x^2+8x+4-3x^2+16x-5=11\\ \Leftrightarrow24x-1=11\\ \Leftrightarrow24x=11+1=12\\ \Leftrightarrow x=\dfrac{12}{24}=\dfrac{1}{2}\\ b,2\left(2x+1\right)\left(8x-3\right)+\left(3-4x\right)\left(8x-7\right)=6x+73\\ \Leftrightarrow\left(4x+2\right)\left(8x-3\right)-\left(4x-3\right)\left(8x-7\right)=6x+73\\ \Leftrightarrow\left(32x^2-12x+16x-6\right)-\left(32x^2-28x-24x+21\right)=6x+73\\ \Leftrightarrow32x^2+4x-6-32x^2+52x-21=6x+73\\ \Leftrightarrow56x-27=6x+73\\ \Leftrightarrow56x-6x=73+27\\ \Leftrightarrow50x=100\\ \Leftrightarrow x=\dfrac{100}{50}\\ \Leftrightarrow x=2\)

Answer 2

Câu `1`
`a,(x+2)*(3x+2)-(3x-1)*(x-5)=11`
`=> 3x^2 + 8x + 4 - (3x^2 - 16x + 5 )=11`
`=> 3x^2 + 8x + 4 - 3x^2 + 16x - 5 = 11`
`=> 24x - 1  =11`
`=> 24x = 12`
`=> x=1/2`
Vậy: `x=1/2`
`b, 2(2x + 1)(8x - 3) + (3 - 4x)(8x - 7) = 6x + 73`
`=> 32x^2 + 4x - 6 - 32x^2 + 52x - 21 = 6x + 73`
`=> 56x - 27 = 6x + 73`
`=> 56x - 6x = 73 + 27`
`=> 50x = 100`
`=> x=2`
Vậy: `x=2`
Câu `2`
`(x^3-x+1)(2x+1)+(x+1)(x-2)`
`= 2x^4 + x^3 - 2x^2 + 1 + x^2- x -2`
`= 2x^4 + x^3 + (-2x^2 + x^2) -x + (1 - 2)`
`= 2x^4 + x^3 - x^2 - x - 1`
`b, (2x + 5)(2x - 3)-5(x+3)`
`= 4x^2 + 4x - 15 - 5x - 15`
`= 4x^2 + (4x - 5x) + (-15 - 15)`
`= 4x^2 - x - 30`

Answer 3

Bài 1:

a)
\[
(x+2)(3x+2) - (3x-1)(x-5) = 11
\]
  \[
  (x+2)(3x+2) = 3x^2 + 2x + 6x + 4 = 3x^2 + 8x + 4
  \]
  \[
  (3x-1)(x-5) = 3x^2 - 15x - x + 5 = 3x^2 - 16x + 5
  \]
  \[
  3x^2 + 8x + 4 - (3x^2 - 16x + 5) = 11
  \]
  \[
  3x^2 + 8x + 4 - 3x^2 + 16x - 5 = 11
  \]
  \[
  24x - 1 = 11
  \]
  \[
  24x = 12 \quad \Rightarrow \quad x = \frac{12}{24} = \frac{1}{2}
  \]

b)
\[
2(2x+1)(8x-3) + (3-4x)(8x-7) = 6x + 73
\]
  \[
  2(2x+1)(8x-3) = 2(16x^2 - 6x + 8x - 3) = 32x^2 + 4x - 6
  \]
  \[
  (3-4x)(8x-7) = 24x - 32x^2 - 21 + 28x = -32x^2 + 52x - 21
  \]
  \[
  32x^2 + 4x - 6 - 32x^2 + 52x - 21 = 6x + 73
  \]

  \[
  56x - 27 = 6x + 73
  \]
  \[
  50x = 100 \quad \Rightarrow \quad x = \frac{100}{50} = 2
  \]

 Bài 2:

a)
  \[
  (x^3 - x + 1)(2x + 1) = 2x^4 + x^3 - 2x^2 - x + 2x + 1 = 2x^4 + x^3 - 2x^2 + 1
  \]
  \[
  (x + 1)(x - 2) = x^2 - 2x + x - 2 = x^2 - x - 2
  \]
  \[
  A = 2x^4 + x^3 - 2x^2 + 1 + x^2 - x - 2 = 2x^4 + x^3 - x^2 - x - 1
  \]

b)
  \[
  (2x + 5)(2x - 3) = 4x^2 - 6x + 10x - 15 = 4x^2 + 4x - 15
  \]
  \[
  5(x + 3) = 5x + 15
  \]
  \[
  B = 4x^2 + 4x - 15 - 5x - 15 = 4x^2 - x - 30
  \]