\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

tớ thử nha

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

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

           = (x2-x+1)(x3+x2-1)

Ta có:\(x^5+x+1\)

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

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

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

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

\(x^5+x+1\)

\(=x^5-x^2+x^2+x+1\)

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

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

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

\(x^5+x^4+1\)

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

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

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

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

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

\(x^5+x^4+1=x^3\left(x^2+x+1\right)-x\left(x^2+x+1\right)+x^2+x+1=\left(x^2+x+1\right)\left(x^3-x+1\right)\)

DỂ QUÁ!!!!!!!!!!!!!!!!!!!!!!!!

tui hk biết làm

\(=\left(x^2+8x+15\right)\left(x^2+8x+7\right)+15\)

đặt:\(^{x^2+8x+11=t}\)

ta co \(\left(t+4\right)\left(t-4\right)+15=t^2-16+15=t^2-1\)

\(=\left(t-1\right)\left(t+1\right)\Rightarrow\left(x^2+8x+11-1\right)\left(x^2+8x+11+1\right)\)

\(\Rightarrow\left(x^2+8x+12\right)\left(x^2+8x+10\right)\)

Phân tích đa thức sau thành nhân tử:( x-5 )\(^2\)- 5 + x

Bài làm:

( x-5 )\(^2\)- 5 + x

= ( x - 5 )\(^2\)+ ( x - 5 )                                         Chỗ này có thể phân tích thành ( x - 5 ) ( x - 5 ) + ( x - 5 ) nha

= ( x - 5 ) \([\)( x - 5 ) + 1 \(]\)

= ( x- 5 ) ( x - 5 + 1 )

* Chúc bạn học tốt ^^

# Linh

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

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

= ( x³ - x - 1 ) ( x² - x + 1 )

x⁵+x⁴+1

= x⁵+x²-x²+x⁴-x+x+1

=x²(x³-1) + (x³+1) +x²+x+1

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

=( x² + x+1)( x³-x²+x²-x+1)

=(x² + x+1)( x³-x+1)

x^4+x^2+1 = (x^4+2x^2+1)-x^2 = (x^2+1)^2-x^2 = (x^2-x+1).(x^2+x+1)

k mk nha

bạn ơi bạn chưa bớt 2x^2 kìa

ngu dốt

x⁵-x⁴-1=x⁵-x³-x²-x⁴+x²+x+x³-x-1

=x².(x³-x-1)-x.(x³-x-1)+(x³-x-1)

=(x³-x-1)(x²-x+1)

x^4+x^2+1 = (x^4+2x^2+1)-x^2 = (x^2+1)^2-x^2 = (x^2-x+1).(x^2+x+1)

k mk nha

mik chịu