a)x⁷+x⁵+1=x⁷+x⁶-x⁶+2x⁵-x⁵+x⁴-x⁴+x³-x³+x²-x²+1

=x⁷-x⁶+x⁵-x³+x²+x⁶-x⁵+x⁴-x²+x+x⁵-x⁴+x³-x+1

=x²(x⁵-x⁴+x³-x+1)+x(x⁵-x⁴+x³-x+1)+1(x⁵-x⁴+x³-x+1)

=(x²+x+1)(x⁵-x⁴+x³-x+1)

b)4x⁴-32x²+1=4x⁴+12x³+2x²-12x³-36x²-6x+2x²+6x+1

=2x²(2x²+6x+1)-6x(2x²+6x+1)+1(2x²+6x+1)

=(2x²-6x+1)(2x²+6x+1)

c)x⁶+27=(x²+3)(x²-3x+3)(x²+3x+3)

d)3(x⁴+x²+1)-(x²+x+1)

=3x⁴-3x³+2x²+3x³-3x²+2x+3x²-3x+2

=x²(3x²-3x+2)+x(3x²-3x+2)+1(3x²-3x+2)

=(x²+x+1)(3x²-3x+2)

e)bạn tự làm nhé

Ta có \(\frac{32x-8x^2+2x^3}{x^3+64}=\frac{x\left(32-8x+2x^2\right)}{\left(x+4\right)\left(x^2-4x+16\right)}=\frac{2x\left(x^2-4x+16\right)}{\left(x+4\right)\left(x^2-4x+16\right)}=\frac{2x}{x+4}\)

<=> 32( \(x^2-\frac{3}{16}-\frac{13}{16}\)) = 0

<=> 32\(\left(x^2-x+\frac{13}{16}x-\frac{13}{16}\right)\)= 0

<=> 32[ x(x-1)+ 13/16( x- 1)] = 0

<=> 32( x-1) (x+13/16)= 0

<=> x-1=0  hoặc  x+13/16= 0

<=> x= 1 hoặc  x= -13/16

32x^2 - 6x - 26 = 0

<=> (x- 1) (x+\(\frac{13}{16}\)) = 0

<=> x= 1 hoặc  x= \(\frac{-13}{16}\)

bài nay bn p dùng máy tính, ko thì ko lm đc đâu

sao lại ra 13/13 z ạ

3^2x-3 =3².3³ = 3⁵

2x - 3 = 5

2x = 8

x=4

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

Thay \(m=3\) vào \(\left(1\right)\)

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

\(\Rightarrow x^2+3x+2=0\)

\(\Rightarrow x^2+x+2x+2=0\)

\(\Rightarrow x\left(x+1\right)+2\left(x+1\right)=0\)

\(\Rightarrow\left(x+2\right)\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+2=0\\x+1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-2\\x=-1\end{matrix}\right.\)

Vậy \(S=\left\{-2;-1\right\}\) khi \(m=3\)

2x³ - 32x = 0

=> 2x.(x² - 16) = 0

=> 2x = 0 hoặc x² - 16 = 0

=> x = 0 hoặc x² = 16

=> x = 0 hoặc x thuộc {4 ; -4}

Vậy x thuộc {0 ; 4 ; -4}

2x³-32x=0

2x(x²-16)=0

2x=0 hoặc x²-16=0

x=0 hoặc x²=16

x=0 hoặc x=-4 hoặc x=4

\(2x^3-32x=0\)

=>\(2x\left(x^2-16\right)=0\)

=>\(2x\left(x-4\right)\left(x+4\right)=0\)

=>x=0 hoặc x-4=0 hoặc x+4=0

=>x=0 hoặc x=4 hoặc x=-4

Mẫu thức chung: 4 x 2 - 2 x = 2 x 2 x - 1

\(\sqrt{2x}-\sqrt{32x}+\sqrt{8x}\)

\(=\sqrt{2x}-\sqrt{4.8x}+\sqrt{8x}\)

\(=\sqrt{2x}-2\sqrt{8x}+\sqrt{8x}\)

\(=\sqrt{2x}-\sqrt{8x}\)

\(=\sqrt{2x}-\sqrt{2.4x}\)

\(=\sqrt{2x}-2\sqrt{2x}\)

\(=-\sqrt{2x}\)