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4x^4+y^4= 4x^4 +4x^2y^2+y^4-4x^2y^2= ( 2x^2 + y^2 ) ^2 - ( 2xy ) ^2
= (2x^2 + 2xy +y^2)( 2x^2 - 2xy + y^2)
\(\Rightarrow x^3+y^3-\left(x+y\right)\)
\(\Rightarrow\left(x+y\right)\left(x^2-xy+y^2\right)-\left(x+y\right)\)
\(\Rightarrow\left(x+y\right)\left(x^2-xy+y^2-1\right)\)
Chúc bạn học tốt
T I C K cho mình nhé
\(x^2+5x-6\)
\(\Leftrightarrow x^2-x+6x-6\)
\(\Leftrightarrow x\left(x-1\right)+6\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x+6\right)\)
Chúc bạn học tốt
\(x^3-x+3x^2y+xy^2+y^3-y\)
\(=\left(x^3+3x^2y+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)
\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
Ta có :
x3-x+y3-y
=(x3+y3)-(x+y)
=(x+y)(x2-xy+y2)-(x+y)
=(x+y)(x2-xy+y2-1)
a)\(x^2-y^2-x+3y-2=\left(x^2+xy-2x\right)-\left(xy+y^2-2y\right)+\left(x+y-2\right)\)
\(=x\left(x+y-2\right)-y\left(x+y-2\right)+\left(x+y-2\right)\)
\(=\left(x+y-2\right)\left(x-y+1\right)\)
b)\(x^3+y^3+6xy+x+y-10\)
\(=\left(x^3+xy^2-x^2y+2x^2+2xy+5x\right)+\left(y^3+x^2y+xy^2+2y^2+2xy+5y\right)-\left(2x^2+2y^2-2xy+4x+4y+10\right)\)
\(=x\left(x^2+y^2-xy+2x+2y+5\right)+y\left(y^2+x^2-xy+2y+2x+5\right)-2\left(x^2+y^2-xy+2x+2y+5\right)\)\(=\left(x+y-2\right)\left(x^2+y^2-xy+2x+2y+5\right)\)
a) 4x2 - y2 + 4x + 1 = 4x2 + 4x + 1 - y2
= ( 2x + 1 ) 2 - y2
= ( 2x + 1 - y ) ( 2x + 1 + y )
b) x3 - x + y3 - y = x3 + y3 - x - y
= ( x + y ) ( x2 - xy + y2 ) - ( x + y )
= ( x + y ) ( x2 - xy + y2 - 1 )
a. 4x2−y2+4x+14x2−y2+4x+1 =(4x2+4x+1)−y2=(2x+1)2−y2=(4x2+4x+1)−y2=(2x+1)2−y2
=(2x+1+y)(2x+1−y)=(2x+1+y)(2x+1−y)
b. x3−x+y3–yx3−x+y3–y =(x3+y3)−(x+y)=(x+y)(x2−xy+y2)−(x+y)=(x3+y3)−(x+y)=(x+y)(x2−xy+y2)−(x+y)
=(x+y)(x2−xy+y2−1)
Áp dụng hằng đẳng thức : \(\left(x+y\right)^3=x^3+3x^2y+3xy^2+y^3=x^3+y^3+3xy\left(x+y\right)\)
Ta có: \(\left(x+y+z\right)^3-x^3-y^3-z^3=\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+z^3+3\left(x+y\right).z.\left(x+y+z\right)-x^3-y^3-z^3\)
\(=x^3+y^3+3xy\left(x+y\right)+z^3+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3-z^3\)
\(=3\left(x+y\right)\left(xy+xz+yz+z^2\right)=3\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\)
\(=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
Dùng hằng đẳng thức là xong
a, \(\left(x+y\right)^3-x^3-y^3=x^3+3x^2y+3xy^2+y^3-x^3-y^3\)
\(=3x^2y+3xy^2=3xy\left(x+y\right)\)
b, \(x^2+6xy+9y^2=\left(x+3y\right)^2\)
x 3 – x + y 3 – y
= ( x 3 + y 3 ) − (x + y)
= (x + y)( x 2 – xy + y 2 ) − (x + y)
= (x + y)( x 2 – xy + y 2 − 1)