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bài này 1h rùi,chắc chờ tui ngủ dậy làm;
= (x+y)3 - (x+y) + xy(x+y) =
= (x+y)((x+y)2 -1 +xy)) = (x+y)(x2 +3xy +y2 -1)
\(x^2-y^2-ax+ay\)
\(=\left(x-y\right)\left(x+y\right)-a\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-a\right)\)
\(2xy-x^2-y^2+16\)
\(=4^2-\left(x^2-2xy+y^2\right)\)
\(=4^2-\left(x-y\right)^2\)
\(=\left(4-x+y\right)\left(4+x-y\right)\)
\(x^2+5x+4\)
\(=\left(x^2+x\right)+\left(4x+4\right)\)
\(=x\left(x+1\right)+4\left(x+1\right)\)
\(=\left(x+1\right)\left(x+4\right)\)
\(x^4+x^2+1=\left(x^4+2x^2+1\right)-x^2=\left(x^2+1\right)-x^2=\left(x^2-x+1\right)\left(x^2+x+1\right)\)
Hướng dẫn thôi :
a) x ( x + 2 ) ( x^2 - 6x + 4 )
b) ( x + 1 ) ( x + 2 ) ( x - 2 )
f)\(\left(x-y\right)^2-4=\left(x-y-4\right)\left(x-y+4\right)\)
h) \(x^3+27=\left(x+3\right)\left(x^2-3x+9\right)\)
i)\(10x-x^2-25=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\)
k)\(4x^2-12xy+9y^2=\left(2x\right)^2-2.2x.3y+\left(3y\right)^2=\left(2x-3y\right)^2\)
mấy bài này cơ bản mà, mở sgk toán 8 ra có các dạng đấy, đăng cũng đăng ít chứ, đăng nhiều quá
a)\(6x^3-9y^2=3\left(2x^3-3y^2\right)\)
b)\(4x^2y-8xy^2+18x^2y^2=2xy\left(2x-4y+9xy\right)\)
c)\(18x^2y-12x^3=6x^2\left(3y-2x\right)\)
d) \(5x\left(x-1\right)-3y\left(x-1\right)=\left(x-1\right)\left(5x-3y\right)\)
e)\(\left(x+1\right)^2-3\left(x+1\right)=\left(x+1\right)\left(x+1-3\right)=\left(x+1\right)\left(x-2\right)\)
g)\(\left(4x^2-4x+4\right)-\left(x+1\right)^2=\left(4x^2-4x+4\right)-\left(x^2+2x+1\right)\)
\(=4x^2-4x+4-x^2-2x-1\)\(=3x^2-6x+3\)\(=3\left(x^2-2x+1\right)\)
\(=3\left(x-1\right)^2\)
\(x^4+x^2y^2+y^4\)
\(=x^4+2x^2y^2+y^4-x^2y^2\)
\(=\left(x^2+y^2\right)^2-\left(xy\right)^2\)
\(=\left(x^2+y^2-xy\right)\left(x^2+y^2+xy\right)\)
a,Ta có:
x³ + y³ + z³ - 3xyz
= (x+y)³ - 3xy(x-y) + z³ - 3xyz
= [(x+y)³ + z³] - 3xy(x+y+z)
= (x+y+z)³ - 3z(x+y)(x+y+z) - 3xy(x-y-z)
= (x+y+z)[(x+y+z)² - 3z(x+y) - 3xy]
= (x+y+z)(x² + y² + z² + 2xy + 2xz + 2yz - 3xz - 3yz - 3xy)
= (x+y+z)(x² + y² + z² - xy - xz - yz)
b, Từ:
x + y + z = 0
=> x + y = -z
<=> (x + y)^3 = (-z)^3
<=> x^3 + 3x^2y + 3xy^2 + y^3 = -z^3
<=> x^3 + y^3 + z^3 = -3x^2y - 3xy^2
<=> x^3 + y^3 + z^3 = -3xy(x+y)
<=> x^3 + y^3 + z^3 = -3xy(-z)
<=> x^3 + y^3 + z^3 = 3xyz
\(A=\left(x+1\right)\left(x-4\right)\left(x+2\right)\left(x-8\right)+4x^2\)
\(A=\left[\left(x+1\right)\left(x-8\right)\right]\left[\left(x-4\right)\left(x+2\right)\right]+4x^2\)
\(A=\left(x^2-7x-8\right)\left(x^2-2x-8\right)+4x^2\)
Đặt \(p=x^2-4,5x-8\)ta có :
\(A=\left(p-2,5x\right)\left(p+2,5x\right)+4x^2\)
\(A=p^2-\left(2,5x\right)^2+4x^2\)
\(A=p^2-6,25x^2+4x^2\)
\(A=p^2-2,25x^2\)
\(A=p^2-\left(1,5x\right)^2\)
\(A=\left(p-1,5x\right)\left(p+1,5x\right)\)
Thay \(p=x^2-4,5x-8\)vào A ta có :
\(A=\left(x^2-4,5x-8-1,5x\right)\left(x^2-4,5x-8+1,5x\right)\)
\(A=\left(x^2-6x-8\right)\left(x^2-3x-8\right)\)
\(\left(x+1\right)\left(x-4\right)\left(x+2\right)\left(x-8\right)+4x^2\)
\(=\left(x+1\right)\left(x-8\right)\left(x-4\right)\left(x+2\right)+4x^2\)
\(=\left(x^2-7x-8\right)\left(x^2-2x-8\right)+4x^2\)
Đặt \(x^2-2x-8=t\)
Ta có : \(\left(t-5x\right)t+4x^2\)
\(=t^2-5xt+4x^2\)
\(=t^2-2.\frac{5}{2}xt+\frac{25}{4}x^2-\frac{9}{4}x^2\)
\(=\left(t-\frac{5}{2}\right)^2-\frac{9}{4}x^2\)
\(=\left(t-\frac{5}{2}-\frac{3}{2}x\right)\left(t-\frac{5}{2}+\frac{3}{2}x\right)\)
Học tốt ~~