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\(x^2+11x+24=x^2+3x+8x+24\)
\(=x\left(x+3\right)+8\left(x+3\right)\)
\(=\left(x+3\right)\left(x+8\right)\)
\(x^2+11x+24=\left(x^2+8x\right)+\left(3x+24\right)\)
\(=x\left(x+8\right)+3\left(x+8\right)\)
\(=\left(x+3\right)\left(x+8\right)\)
\(x^4+x^3+2x^2+x+1\)
\(=\left(x^4+2x^2+1\right)+\left(x^3+x\right)\)
\(=\left(x^2+1\right)^2+x\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(x^2+1+x\right)\)
x^4+x^3+2x^2+x+1
=(x^4+2x^2+1)+(x^3+x)
=(x^2+1)^2+x(x^2+1)
=(x^2+1)(x^2+x+1)
\(-3xy^2+x^2y^2-5x^2y\)
\(=-xy\left(3y+xy-5x\right)\)
\(x\left(y-1\right)+3\left(y^3+2y+1\right)\)
\(=3y^3+6y+3+xy-x\)
Xem lại nhé ko phân tích được
\(12xy^2-12xy+3x\)
\(=3x\left(4y^2-4y+1\right)\)
\(=3x\left(2y-1\right)^2\)
\(10x^2\left(x+y\right)-5\left(2x+2y\right)y^2\)
\(=10x^2\left(x+y\right)-10\left(x+y\right)y^2\)
\(=10\left(x+y\right)\left(x-y\right)\left(x+y\right)\)
\(=10\left(x+y\right)^2\left(x-y\right)\)
a) \(x^2y-xy^2-3x+3y\)
\(=xy\left(x-y\right)-3\left(x-y\right)\)
\(=\left(x-y\right)\left(xy-3\right)\)
b) \(5x^4-20=5\left(x^4-4\right)\)
\(=5\left(x^2-2\right)\left(x^2+2\right)\)
\(=5\left(x^2+2\right)\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\)
c) \(x^2-9x+20\)
\(=x^2-4x-5x+20\)
\(=x\left(x-4\right)-5\left(x-4\right)\)
\(=\left(x-4\right)\left(x-5\right)\)
\(x^2y-xy^2-3x+3y=xy.\left(x-y\right)-3.\left(x-y\right)=\left(x-y\right).\left(xy-3\right)\)
\(5x^4-20=5.\left[\left(x^2\right)^2-2^2\right]=5.\left(x^2-2\right).\left(x^2+2\right)\)
\(x^2-9x+20=\left(x^2-2.4,5x+4,5^2\right)-0,25=\left(x-20,25\right)^2-0,5^2\)\(=\left(x-20,25+0,5\right).\left(x-20,25-0,5\right)=\left(x-19,75\right).\left(x-20,75\right)\)
Tham khảo nhé~
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)
a, x2y - xy2 - 3x + 3y
= ( x2y - xy2 ) + ( 3x - 3y )
= xy( x - y ) + 3( x - y )
= ( x - y ) + ( xy + 3 )
b, 5x4-20
=5x4 - 5.4
=5( x4 - 4 )
\(A=\left(x+1\right)\left(x-4\right)\left(x+2\right)\left(x-8\right)+4x^2\)
\(A=\left[\left(x+1\right)\left(x-8\right)\right]\left[\left(x-4\right)\left(x+2\right)\right]+4x^2\)
\(A=\left(x^2-7x-8\right)\left(x^2-2x-8\right)+4x^2\)
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 ~~
\(x^4+x^2y^2+y^4\)
\(=x^4+2x^2y^2+y^4-x^2y^2\)
\(=\left(x^2+y^2\right)^2-\left(xy\right)^2\)
\(=\left(x^2+y^2-xy\right)\left(x^2+y^2+xy\right)\)
(2x2 - 4)2 + 9 = 4x4 - 16x2 + 25
= (4x4 + 20x2 + 25) - 36x2 = (2x2 + 5)2 - 36x2
= (2x2 - 6x + 5)(2x2 + 6x + 5)
(2x2-4)2+9
=(2x2-4)2+32
=(2x2 - 4+3)(2x2-4-3)
=(2x2 - 7)(2x2 - 1)
Tích cho mk nha!
\(x^3+2x^2+x\)
\(=x\left(x^2+2x+1\right)\)
\(=x\left(x+1\right)^2\)