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\(x^6-y^6\)
\(=\left(x^3-y^3\right)\left(x^3+y^3\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)\)
hk
tốt
a. Ta có : (x + y)[(x - y)2 + xy]
= (x + y)(x2 - 2xy + y2 + xy)
= (x + y)(x2 - xy + y2)
= x3 + y3
b. Ta có : x3 + y3 - xy(x + y)
= x3 + y3 - x2y - xy2
=x2(x - y) + y2(y - x)
= (x - y)(x2 - y2)
= (x - y)2.(x + y) đpcm
c) Ta có (x + y)3 - 3xy(x + y)
= (x + y)[(x + y)2 - 3xy)
= (x + y)(x2 + 2xy + y2 - 3xy)
= (x + y)(x2 - xy + y2) (đpcm)
a) VP = ( x + y )( x2 - 2xy + y2 + xy ) = ( x + y )( x2 - xy + y2 ) = x3 + y3 = VT ( đpcm )
b) VP = ( x + y )( x - y )2 = ( x + y )( x2 - 2xy + y2 ) = x3 - 2x2y + xy2 + x2y - 2xy2 + y3 = x3 + y3 - x2y - xy2 = x3 + y3 - xy( x + y ) = VT ( đpcm )
c) VP = x3 + 3x2y + 3xy2 + y3 - 3x2y - 3xy2 = x3 + y3 = ( x + y )( x2 - xy + y2 ) = VT ( đpcm )
(3x-1)^2 - 16 = (3x-1)^2 - 4^2
= (3x-1-4)(3x-1+4)
= (3x-5)(3x+3)
\(\left(3x-1\right)^2-16\)
\(=\left(3x-1\right)^2-4^2\)
\(=\left(3x-1-4\right)\left(3x-1+4\right)\)
\(=\left(3x-5\right)\left(3x+3\right)\)
\(=3\left(x+1\right)\left(3x-5\right)\)
Bài 2:
a: \(\left(2x-5\right)^2-4x\left(x-3\right)=0\)
=>\(4x^2-20x+25-4x^2+12x=0\)
=>-8x+25=0
=>-8x=-25
=>\(x=\frac{25}{8}\)
b: \(2\left(x+5\right)-x^2-5x=0\)
=>2(x+5)-x(x+5)=0
=>(x+5)(2-x)=0
=>\(\left[\begin{array}{l}x+5=0\\ 2-x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-5\\ x=2\end{array}\right.\)
c: \(6x^2-7x+2=0\)
=>\(6x^2-3x-4x+2=0\)
=>3x(2x-1)-2(2x-1)=0
=>(2x-1)(3x-2)=0
=>\(\left[\begin{array}{l}2x-1=0\\ 3x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac12\\ x=\frac23\end{array}\right.\)
Bài 1:
a: x(x-5)+(x+3)(x-3)
\(=x^2-5x+x^2-9\)
\(=2x^2-5x-9\)
b: \(\frac{x}{x-1}+\frac{2x-4}{x^2-1}-\frac{5}{x+1}\)
\(=\frac{x\left(x+1\right)+2x-4-5\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2+x+2x-4-5x+5}{\left(x-1\right)\left(x+1\right)}=\frac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}=\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x-1}{x+1}\)
c: \(\left(20x^2+7x-6\right):\left(5x-2\right)\)
\(=\left(20x^2-8x+15x-6\right):\left(5x-2\right)\)
\(=\frac{4x\left(5x-2\right)+3\left(5x-2\right)}{5x-2}=4x+3\)
\(299.301=\left(300-1\right)\left(300+1\right)=300^2-1^2=90000-1=89999\)
\(299\cdot301=300^2-1=89999\)
Cảm ơn bạn rất nhiều
Cảm ơn bạn rất nhiều