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a) = x^2 - y^2 - x - y
= ( x- y)(x + y) - ( x+ y)
= ( x+ y)( x- y - 1 )

1) (x+6)(3x-1)+x+6=0
⇔(x+6)(3x-1)+(x+6)=0
⇔(x+6)(3x-1+1)=0
⇔3x(x+6)=0
2) (x+4)(5x+9)-x-4=0
⇔(x+4)(5x+9)-(x+4)=0
⇔(x+4)(5x+9-1)=0
⇔(x+4)(5x+8)=0
3)(1-x)(5x+3)÷(3x-7)(x-1)
=\(\frac{\left(1-x\right)\left(5x+3\right)}{\left(3x-7\right)\left(x-1\right)}=\frac{\left(1-x\right)\left(5x+3\right)}{\left(7-3x\right)\left(1-x\right)}=\frac{\left(5x+3\right)}{\left(7-3x\right)}\)

1) =\(x^7-x+x^2+x\)+1
=\(x\left(x^6-1\right)+\left(x^2+x+1\right)\)
=\(x\left(x^3-1\right)\left(x^3+1\right)\)\(+\left(x^2+x+1\right)\)
=x(x^3+1)(x-1)(x^2+x+1)+(x^2+x+1)
=[(x^4+x)(x-1)+1](x^2+x+1)
=(x^5-x^4+x^2-x)(x^2+x+1)
Trả lời:
1, x7 + x2 + 1
= x7 + x2 + 1 + x6 - x6 + x5 - x5 + x4 - x4 + x3 - x3 + x2 - x2 + x - x
= ( x7 + x6 + x5 ) - ( x6 + x5 + x4 ) + ( x4 + x3 + x2 ) - ( x3 + x2 + x ) + ( x2 + x + 1 )
= x5 ( x2 + x + 1 ) - x4 ( x2 + x + 1 ) + x2 ( x2 + x + 1 ) - x ( x2 + x + 1 ) + ( x2 + x + 1 )
= ( x2 + x + 1 )( x5 - x4 + x2 - x + 1 )
b, x8 + x7 + 1
= x8 + x7 + 1 + x6 - x6 + x5 - x5 + x4 - x4 + x3 - x3 + x2 - x2 + x - x
= ( x8 + x7 + x6 ) - ( x6 + x5 + x4 ) + ( x5 + x4 + x3 ) - ( x3 + x2 + x ) + ( x2 + x + 1 )
= x6 ( x2 + x + 1 ) - x4 ( x2 + x + 1 ) + x3 ( x2 + x + 1 ) - x ( x2 + x + 1 ) + ( x2 + x + 1 )
= ( x2 + x + 1 )( x6 - x4 + x3 - x + 1 )

Bài 1
1.(x-3)(x+2)-x(x-7)=15
\(\Leftrightarrow x^2+2x-3x-6-x^2+7x=15\)
\(\Leftrightarrow-6+6x=15\)
\(\Leftrightarrow6x=15+6\) =21
\(\Rightarrow x=\dfrac{21}{6}=3,5\)
2.(x-5)(x+5)+x(3-x)=20
\(\Leftrightarrow x^2-25+3x-x^2=20\)
\(\Leftrightarrow-25+3x=20\)
\(\Leftrightarrow3x=20+25=45\)
\(\Rightarrow x=\dfrac{45}{3}=15\)
3.(x-7)2-x(2+x)=-7
\(\Leftrightarrow x^2-14x+49-2x-x^2=-7\)
\(\Leftrightarrow-16x+49=-7\)
\(\Leftrightarrow-16x=-7-49=-56\)
\(\Rightarrow x=\dfrac{-56}{-16}=\dfrac{7}{2}=3,5\)
Tiếp bài 1
4.(x-4)2-(x+4)(x-4)=-16
\(\Leftrightarrow x^2-8x+16-x^2-16=-16\)
\(\Leftrightarrow-8x=-16\)
\(\Rightarrow x=\dfrac{-16}{-8}=2\)
5.(x-5)(x+5)-x(2-3x)=4x2-7
\(\Leftrightarrow x^2-25-2x+3x^2=4x^2-7\)
\(\Leftrightarrow4x^2-25-2x+3x^2=4x^2-7\)
\(\Leftrightarrow4x^2-4x^2-2x=-7+25\)
\(\Leftrightarrow-2x=18\)
\(\Rightarrow x=\dfrac{18}{-2}=-9\)

x^2 -2x = 24
=> x^2 - 2x - 24=0
=>x^2 -8x+6x - 24 = 0
=> ( x^2- 8x)+( 6x-24) = 0
=> x(x-8) + 6(x-8) = 0
=> (x+6)(x-8)=0
=>\(\orbr{\begin{cases}x=-6\\x=8\end{cases}}\)

\(\left(4-3x\right)\left(10x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)
\(\left(7-2x\right)\left(4+8x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)
rồi thực hiện đến hết ...
Brainchild bé ngây thơ qus e , ko thực hiện đến hết như thế đc đâu :>
\(\left(x-3\right)\left(2x-1\right)=\left(2x-1\right)\left(2x+3\right)\)
\(2x^2-7x+3=4x^2+4x-3\)
\(2x^2-7x+3-4x^2-4x+3=0\)
\(-2x^2-11x+6=0\)
\(2x^2+11x-6=0\)
\(2x^2+12x-x-6=0\)
\(2x\left(x+6\right)-\left(x+6\right)=0\)
\(\left(x+6\right)\left(2x-1\right)=0\)
\(x+6=0\Leftrightarrow x=-6\)
\(2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)
\(3x-2x^2=0\)
\(x\left(2x-3\right)=0\)
\(x=0\)
\(2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)
Tự lm tiếp nha