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B1:
a) \(5\left(x^2+y^2\right)-20x^2y^2\)
\(=5\left(x^2-4x^2y^2+y^2\right)\)
b) \(=2\left(x^8-16\right)=2\left(x^4-4\right)\left(x^4+4\right)=2\left(x^2-2\right)\left(x^2+2\right)\left(x^4+4\right)\)
B2:
a) Đặt \(x^2-3x+1=y\)
=> \(y^2-12y+27\)
\(=\left(y^2-12y+36\right)-9\)
\(=\left(y-6\right)^2-3^2\)
\(=\left(y-9\right)\left(y-3\right)\)
\(=\left(x^2-3x-10\right)\left(x^2-3x-4\right)\)
\(=\left(x+1\right)\left(x-4\right)\left(x^2-3x-10\right)\)
b) Đặt \(x^2+7x+11=t\)
Ta có: \(\left[\left(x+2\right)\left(x+5\right)\right]\cdot\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
\(=\left(t-1\right)\left(t+1\right)-24\)
\(=t^2-25\)
\(=\left(t-5\right)\left(t+5\right)\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
\(3x\left(x-5\right)-x\left(4+3x\right)=43\)
\(\Leftrightarrow3x^2-15x-4x-3x^2=43\)
\(\Leftrightarrow-19x=43\)
\(\Leftrightarrow x=\frac{-43}{19}\)
a,\(xy+3x-7y-21\)
\(=x\left(y+3\right)-7\left(y+3\right)\)
\(=\left(y+3\right)\left(x-7\right)\)
\(b,2xy-15-6x+5y\)
\(=\left(2xy-6x\right)+\left(-15+5y\right)\)
\(=2x\left(y-3\right)-5\left(3-y\right)\)
\(=2x\left(y-3\right)+5\left(y-3\right)\)
\(=\left(y-3\right)\left(2x+5\right)\)
bÀI LÀM
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
F=x2+2xy+y2-x-y-12
= (x + y)^2 - (x + y) - 12
= (x + y)(x + y - 1) - 12
đặt x + y = t
F = t(t - 1) - 12
= t^2 - t - 12
= (t - 4)(t + 3)
G=(x2-3x-1)2-12(x2-3x-1)+27
đăth x^2 - 3x - 1 = t
G = t^2 - 12t + 27
= (t - 3)(t - 9)
có t = x^2 - 3x - 1
thay vào
Câu F ( kiểm tra lại đề )
Câu G . Đặt x^2 -3x-1=t
t^2 -12t+27 ( thực hiện pp tách)
A = xy + y - 2x - 2
= y( x + 1 ) - 2( x + 1 )
= ( x + 1 )( y - 2 )
B = x2 - 3x + xy - 3y
= x( x - 3 ) + y( x - 3 )
= ( x - 3 )( x + y )
C = 3x2 - 3xy - 5x + 5y
= 3x( x - y ) - 5( x - y )
= ( x - y )( 3x - 5 )
D = xy + 1 + x + y
= y( x + 1 ) + ( x + 1 )
= ( x + 1 )( y + 1 )
E = ax - bx + ab - x2
= ( ax - x2 ) + ( ab - bx )
= x( a - x ) + b( a - x )
= ( a - x )( x + b )
F = x2 + ab + ax + bx
= ( ax + x2 ) + ( ab + bx )
= x( a + x ) + b( a + x )
= ( a + x )( x + b )
G = a3 - a2x - ay + xy
= a2( a - x ) - y( a - x )
= ( a - x )( a2 - y )
Bonus : = ( a - x )[ a2 - ( √y )2 ]
= ( a - x )( a - √y )( a + √y )
H = 2xy + 3z + 6y + xz
= ( 6y + 2xy ) + ( 3z + xz )
= 2y( 3 + x ) + z( 3 + x )
= ( 3 + x )( 2y + z )
A = xy + y - 2x - 2 = y(x + 1) - 2(x + 1) = (y - 2)(x + !1
B = x2 - 3x + xy - 3y = x(x - 3) + y(x - 3) = (x + y)(x - 3)
C = 3x2 - 3xy - 5x + 5y = 3x(x - y) - 5(x - y) = (3x - 5)(x - y)
D = xy + 1 + x + y = xy + x + y + 1 = x(y + 1) + (y + 1) = (x + 1)(y + 1)
E = ax - bx + ab - x2 = ax - x2 + ab - bx = a(a - x) - b(a - x) = (a - b)(a - x)
F = x2 + ab + ax + bx = ab + ax + bx + x2 = a(b + x) + x(b + x) = (a + x)(b + x)
G = a3 - a2x - ay + xy = a2(a - x) - y(a - x) = (a2 - y)(a - x)
H = 2xy + 3z + 6y + xz = 2xy + 6y + 3z + xz = 2y(x + 3) + z(x + 3) = (2y + z)(x + 3)
b: \(=x^4+10x^2-9x^2-90\)
\(=\left(x^2+10\right)\left(x^2-9\right)\)
\(=\left(x^2+10\right)\left(x-3\right)\left(x+3\right)\)
c: \(=\left(5x^2-2x\right)^2-\left(5x^2-2x\right)-42\)
\(=\left(5x^2-2x-7\right)\left(5x^2-2x+6\right)\)
\(=\left(x-1\right)\left(5x+7\right)\left(5x^2-2x+6\right)\)
d: \(=\left(x+4y\right)^2+2\left(x+4y\right)-3\)
\(=\left(x+4y+3\right)\left(x+4y-1\right)\)
e: \(=\left(x^2+3x\right)^2+3\left(x^2+3x\right)-4\)
\(=\left(x^2+3x+4\right)\left(x^2+3x-1\right)\)