\(x^6+1=\left(x^2\right)^3+1^3=\left(x^2+1\right)\left(x^4-x^2+1\right)⋮\left(x^2+1\right)\)

\(x^6+1=x^6-x^2+x^2+1=x^2\left(x^4-1\right)+\left(x^2+1\right)\)

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

\(=\left(x^2+1\right)\left[x^2\left(x^2-1\right)+1\right]⋮x^2+1\) (đpcm)

1h45p+105p

=1,75h+1,75h

=3,5h

1,75hx8

=14h

#H

bi nhầm  ở dấu = thành dấu +

sai đè

2 giờ 15 phút = 2,25 giờ

2 ^m3 3 ^dm3 = 2,003 ^m3

2 giờ 15 phút = 2,25 giờ

2 ^m3 3 ^dm3 = 2,003 ^m3

225 h

2003 m3

\(999\times2786=2783214\)

2783214!

2783214 k mik nha

\(\frac{2^{11}+3.\left(-2^{10}\right)}{10.\left(-4\right)^5}=\frac{2048+3.1024}{10.\left(-1024\right)}=\frac{2048+3072}{-10240}=\frac{5120}{-10240}=\left(-0.5\right)\)

TH1  \(x\ge0;\left|2x-3\right|=2x-3\)

\(2x-3-5=7x+1\)

\(\Leftrightarrow2x-7x=3+5+1=9\)

\(\Leftrightarrow-5x=9\Rightarrow x=-\frac{9}{5}\left(ktm\right)\)

TH2:\(x< 0;\left|2x-3\right|=-\left(2x-3\right)\)

\(-\left(2x-3\right)-5=7x+1\)

\(\Leftrightarrow-2x+3-5=7x+1\)

\(\Leftrightarrow-2x-7x=-3+5+1=3\)

\(\Leftrightarrow-9x=3\Rightarrow x=-\frac{3}{9}=-\frac{1}{3}\left(tm\right)\)

Vậy \(x=-\frac{1}{3}\)

     x⁷ = 3²⁰ : 3⁶
=> x⁷ = 3^{20 - 6}
=> x⁷ = 3¹⁴
=> x⁷ = \(\left(3^2\right)^7\)
=> x  = 3²
=> x  = 9

x=9 đó

x⁷=3²⁰:3⁶

(x²)⁷=3¹⁴

x¹⁴=3¹⁴

=>x=3

Vậy x=3

\(\frac{x^4-5x+4}{x^2-2}=5\left(x-1\right)\)

\(\Leftrightarrow\frac{x^4-5x+4}{x^2-2}\left(x^2-2\right)=5\left(x-1\right)\left(x^2-2\right)\)

\(\Leftrightarrow x^4-5x+4=5\left(x-1\right)\left(x^2-2\right)\)

\(\Rightarrow\hept{\begin{cases}x=\pm1\\x=2\\x=3\end{cases}}\)

P/s: ko chắc

ĐKXĐ : X² \(\ne\)2

Ta có: ^{\(\frac{x^4-5x+4}{x^2-2}\)= \(5\left(x-1\right)\)\(\Leftrightarrow\frac{\left(x-1\right)\left(x^3+x^2+x-4\right)}{x^2-2}=5\left(x-1\right)\)}

\(\Leftrightarrow\left(x-1\right)\left(\frac{x^3+x^2+x-4}{x^2-2}-5\right)\)\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\\frac{x^3+x^2+x-4}{x^2-2}-5=0\end{cases}}\)

\(+x-1=0\Rightarrow x=1\)

+)\(\frac{x^3+x^2+x-4}{x^2-2}-5=0\Leftrightarrow x^3+x^2+x-4-5x^2+10=0\)

\(\Leftrightarrow x^3-4x^2+x+6=0\Leftrightarrow\left(x-2\right)\left(x+1\right)\left(x-3\right)=0\)\(\Leftrightarrow x=2\)hoặc \(x=3\)

hoặc x=-1

Bạn tự kết luận nhé..

lam sao

ta có : x^4+x^3-x^2+1= (x^4+x^3)-(x^2-1^2)=x^3*(x+1)-(x-1)*(x+1)=(x^3-x+1)(x-1)

\(x^4+x^3-x^2+1\)

\(=\left(x^4+x^3\right)-\left(x^2+1\right)\)

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

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