Ta có: \(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=\dfrac{\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)

\(=\dfrac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)

\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)

\(=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)

\(=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)}{2}\)

\(=\dfrac{3^{32}-1}{2}\)

Rút gọn: (3 + 1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)(3³² + 1)
A=(3 + 1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(316 + 1)(3³² + 1)

A=(3-1)(3 + 1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)(3³² + 1)
A=(3²-1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)(3³² + 1)
A=(3⁴-1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)(3³² + 1)
A=(3⁸-1)(3⁸ + 1)(3¹⁶ + 1)(3³² + 1)
A=(3¹⁶-1)(3¹⁶ + 1)(3³² + 1)
A=(3³² - 1)(3³² + 1)
A=3⁶⁴-1
=>A=(3⁶⁴-1) /2

\(A=\frac{x^3-3x^2-7x-15}{x^5-x^4-10x^3-38x^2-51x-45}\)

\(=\frac{x^2\left(x-5\right)+2x\left(x-5\right)+3\left(x-5\right)}{x^4\left(x-5\right)+4x^3\left(x-5\right)+10x^2\left(x-5\right)+12x\left(x-5\right)+9\left(x-5\right)}\)

\(=\frac{\left(x-5\right)\left(x^2+2x+3\right)}{\left(x-5\right)\left(x^4+4x^3+10x^2+12x+9\right)}\)

\(=\frac{x^2+2x+3}{x^4+4x^3+10x^2+12x+9}\)

\(=\frac{x^2+2x+3}{\left(x^2\right)^2+2.x^2.2x+\left(2x\right)^2+6x^2+12x+9}\)

\(=\frac{x^2+2x+3}{\left(x^2+2x\right)^2+2.\left(x^2+2x\right).3+3^2}\)

\(=\frac{\left(x^2+2x+3\right)}{\left(x^2+2x+3\right)^2}=\frac{1}{x^2+2x+3}\)

b, \(A=\frac{1}{x^2+2x+3}=\frac{1}{\left(x+1\right)^2+2}\le\frac{1}{2}\forall x\)

Dấu "=" xảy ra khi: \(x+1=0\Rightarrow x=-1\)

Vậy GTLN của A là \(\frac{1}{2}\) khi x = -1

Lời giải:

Vì $x=9$ nên $x-9=0$
Ta có:

$F=(x^{2017}-9x^{2016})-(x^{2016}-9x^{2015})+(x^{2015}-9x^{2014})-....-(x^2-9x)+x-10$

$=x^{2016}(x-9)-x^{2015}(x-9)+x^{2014}(x-9)-....-x(x-9)+x-10$

$=x^{2016}.0-x^{2015}.0+x^{2014}.0-...-x.0+x-10$

$=x-10=9-10=-1$

D

D?

D

\(-\dfrac{2}{3}xy^2z.9x^2y^2=-6x^3y^4z\)

6xy4z

6xy4z

\(\left(3x+4\right)^2-10x-\left(x-4\right)\left(x+4\right)\)

\(=9x^2+24x+16-10x-x^2+16\)

\(=8x^2+14x\)

P = ( 3x + 4 )² - 10x - ( x - 4 )( x + 4 )

P = 9x² + 24x + 16 - 10x - ( x² - 16 )

P = 9x² + 24x + 16 - 10x - x² + 16

P = 8x² + 14x + 32

P = 2( 4x² + 7x + 16 )

a: \(=\left(4x-1\right)^3\)

b: \(=1000x^3-1-10x\left(100x^2-1\right)\)

\(=-1+10x\)

\(5xy.\sqrt{\frac{25x^2}{y^6}}=5xy.\sqrt{\frac{5^2x^2}{\left(y^3\right)^2}}=5xy.\sqrt{\frac{\left(5x\right)^2}{\left(y^3\right)^2}}5xy.\sqrt{\left(\frac{5x}{y^3}\right)^2}=5xy.\frac{5x}{y^3}=\frac{5^2x^2}{y^2}=\frac{\left(5x\right)^2}{y^2}=\left(\frac{5x}{y}\right)^2\)

Chúc bạn học tốt

5xy.\(\sqrt{\frac{25x^2}{y^6}}\)

=5xy.\(\frac{\left|5x\right|}{\left|y^3\right|}\){x<0 nên |5x|=-5x

=\(\orbr{\begin{cases}5xy.\frac{-5x}{y^3}\\5xy.\frac{-5x}{-y^3}\end{cases}}\)

=\(\orbr{\begin{cases}\frac{-25x^2}{y^3}\\\frac{25x^2}{y^3}\end{cases}}\)