Em muốn hỏi bài nào vậy?

dạ bài 3 ạ

Lời giải:
$A=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$

$A=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$

$=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)=(2^4-1)(2^4+1)(2^8+1)(2^{16}+1)$

$=(2^8-1)(2^8+1)(2^{16}+1)=(2^{16}-1)(2^{16}+1)=2^{32}-1$

P/s: Lần sau bạn lưu ý ghi đầy đủ yêu cầu đề bài.

Lời giải:
a. ĐKXĐ: $x\neq \pm 1$

b.

\(A=\left[\frac{x+1}{(x-1)(x+1)}+\frac{x^2}{(x-1)(x+1)}-\frac{x(x-1)}{(x+1)(x-1)}\right].\frac{x+1}{2x+1}\)

\(=\frac{x+1+x^2-(x^2-x)}{(x-1)(x+1)}.\frac{x+1}{2x+1}=\frac{2x+1}{(x-1)(x+1)}.\frac{x+1}{2x+1}=\frac{1}{x-1}\)

c.

Khi $x=2$ thì $A=\frac{1}{2-1}=1$

x²+3x+3y+xy

=x(x+3)+y(3+x)

=(x+3)(x+y)

     x² + 3x + 3y + xy

=   ( x² + xy) + ( 3x + 3y)

= x( x + y) + 3 ( x + y)

= ( x + y) ( x + 3) 

Theo Bezout  đa thức  F(n) = 2n² + n - 18 chia hết cho đa thức n - 3 

⇔  F(3) ⋮ n- 3 ⇔ 2.3² + 3 - 18  ⋮ n - 3 ⇔ 3 ⋮ n - 3

n - 3 ⋮ Ư(3) = {  -3; -1; 1; 3} ⇔ n ϵ { 0; 2; 4; 6}

bạn có chép sai đề bài không ạ?

\(\dfrac{x^3\left(x-1\right)^3}{\left(x-1\right)^3}+\dfrac{x^3}{\left(x-1\right)^3}+\dfrac{3x^2\left(x-1\right)^2}{\left(x-1\right)^3}=28\)

ĐK: \(x\ne1\)

\(x^3+\dfrac{x^3}{\left(x-1\right)^3}+\dfrac{3x^2}{x-1}-28=0\)

\(x^3\left(x-1\right)^3+x^3+3x^2\left(x-1\right)^2-28\left(x-1\right)^3=0\)

\(\left(x^2-x\right)^3+3\left(x^2-x\right)^2+x^3-28\left(x^3-3x^2+3x-1\right)=0\)

\(\left(x^2-x\right)^3+3\left(x^2-x\right)^2+3\left(x^2-x\right)+1-\left(27x^3-81x^2+81x-27\right)=0\)

\(\left(x^2-x+1\right)^3-\left(3x-3\right)^3=0\)

\(\left(x^2-x+1-3x+3\right)\left[\left(x^2-x+1\right)^2+\left(x^2-x+1\right)\left(3x-3\right)+\left(3x-3\right)^2\right]=0\)

\(x^2-4x+4=0\)

\(x=2\) (TMĐK)

Ta có:

+ ⇒ MTC = 2x( x + 3 )

Khi đó ta có:

`-x^2+6x-11`

`=-x^2+6x-9-2`

`=-(x-3)^{2}-2`

  Vì \(-(x-3)^{2} \le 0\)

\(<=>-(x-3)^{2}-2 \le -2\)

  Hay \(-x^2+6x-11 \le -2\)

Dấu "`=`" xảy ra `<=>x-3=0<=>x=3`

-x²+6x-11

=>-(x²-6x+11)

=>-(x-3)²-2

vì-(x-3)²\(\le0\)

=>-(x-3)²-2\(\le\)-2

dấu bằng xảy ra khi x-3=0=>x=3

vậy gtln của -x²+6x-11 là -2 khi x=3

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