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bài 11
a) \(x^2-xy+x\\ =x\left(x-y+1\right)\)
b)
\(x^2-2xy-4+y^2\\ =\left(x^2-2xy+y^2\right)-4\\ =\left(x-y\right)^2-4\\ =\left(x-y-2\right)\left(x-y+2\right)\)
c)
\(x^3-x^2-16x+16\\ =x^2\left(x-1\right)-16\left(x-1\right)\\ =\left(x-1\right)\left(x-4\right)\left(x+4\right)\)
bài 12
\(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(2x^2-10x-3x-2x^2=26\)
\(-13x=26\\ x=-2\)
b)
\(2\left(x+5\right)-x^2-5x=0\\ 2\left(x+5\right)-x\left(x+5\right)=0\\ \left(x+5\right)\left(2-x\right)=0\\ \left[{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
b)(x2+x+1)(x2+x+2)-12
Đặt t=x2+x+1
t(t+1)-12=t2+t-12
=(t-3)(t+4)=(x2+x+1-3)(x2+x+1+4)
=(x2+x-2)(x2+x+5)
=(x-1)(x+2)(x2+x+5)
c)(x2+8x+7)(x2+8x+15)+15
Đặt t=x2+8x+7
t(t+8)+15=t2+8t+15
=(t+3)(t+5)
=(x2+8x+7+3)(x2+8x+7+15)
=(x2+8x+10)(x2+8x+22)
d)(x+2)(x+3)(x+4)(x+5)-24
=(x2+7x+10)(x2+7x+12)-24
Đặt t=x2+7x+10
t(t+2)-24=(t-4)(t+6)
=(x2+7x+10-4)(x2+7x+10+6)
=(x2+7x+6)(x2+7x+16)
=(x+1)(x+6)(x2+7x+16)
a/ Đặt x2 + 4x + 8 = a
Thì đa thức ban đầu thành
a2 + 3ax + 2x2 = (a2 + 2ax + x2) + (ax + x2)
= (a + x)2 + x(a + x) = (a + x)(a + 2x)
\(a.10x\left(x-y\right)-6y\left(y-x\right)\\ =10x\left(x-y\right)+6y\left(x-y\right)\\ =\left(10x-6y\right)\left(x-y\right)\\ =2\left(5x-3y\right)\left(x-y\right)\)
\(b.14x^2y-21xy^2+28x^3y^2\\ =7xy\left(x-y+xy\right)\)
\(c.x^2-4+\left(x-2\right)^2\\ =\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2\\ =\left(x-2\right)\left(x+2+x-2\right)\\ =2x\left(x-2\right)\)
\(d.\left(x+1\right)^2-25\\ =\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right)\)
Bạn nên tách bài ra để đăng. Không nên đăng 1 loạt như thế này.
1: \(=\left(x^2+x\right)^2+3\left(x^2+x\right)+2-12\)
=(x^2+x)^2+3(x^2+x)-10
=(x^2+x+5)(x^2+x-2)
=(x^2+x+5)(x+2)(x-1)
2: \(=\left(x^2+5ax+4a^2\right)\left(x^2+5ax+6a^2\right)+a^4\)
\(=\left(x^2+5ax\right)^2+10a^2\left(x^2+5ax\right)+25a^2\)
\(=\left(x^2+5ax+5a^2\right)^2\)
3: \(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
5: \(M=\left(n+1\right)\left(n^2+2n\right)+360\)
=n(n+1)(n+2)+360 chia hết cho 6
6A
7D
a)x4+2x3+5x2+4x-12
=(x4+2x3+x2)+(4x2+4x)-12
=(x2+x)2+4(x2+x)-12
Đặt t=x2+x
=t2+4t-12=(t-2)(t+6)
=(x2+x-2)(x2+x+6)
=(x-1)(x+2)(x2+x+6)
b)(x+1)(x+2)(x+3)(x+4)+1
=(x2+5x+4)(x2+5x+6)+1
Đặt x2+5x+4=t
t(t+2)+1=t2+2t+1
=(t+1)2=(x2+5x+4+1)2
=(x2+5x+5)2
c)(x+1)(x+3)(x+5)(x+7)+15
=(x2+8x+7)(x2+8x+15)+15
Đặt t=x2+8x+7
t(t+8)+15=(t+3)(t+5)
=(x2+8x+7+3)(x2+8x+7+5)
=(x2+8x+10)(x+2)(x+6)
d)(x+1)(x+2)(x+3)(x+4)-24
=(x2+5x+4)(x2+5x+6)-24
Đặt t=x2+5x+4
t(t+2)-24=(t-4)(t+6)
=(x2+5x+4-4)(x2+5x+4+6)
=x(x+5)(x2+5x+10)
c) ( x2 + x + 1 )( x2 + x + 2 ) - 12
Đặt t = x2 + x + 1
<=> t( t + 1 ) - 12
= t2 + t - 12
= t2 - 3t + 4t - 12
= t( t - 3 ) + 4( t - 3 )
= ( t - 3 )( t + 4 )
= ( x2 + x + 1 - 3 )( x2 + x + 1 + 4 )
= ( x2 + x - 2 )( x2 + x + 5 )
= ( x2 - x + 2x - 2 )( x2 + x + 5 )
= [ x( x - 1 ) + 2( x - 1 ) ]( x2 + x + 5 )
= ( x - 1 )( x + 2 )( x2 + x + 5 )
d) ( x + 2 )( x + 3 )( x + 4 )( x + 5 ) - 24
= [ ( x + 2 )( x + 5 ) ][ ( x + 3 )( x + 4 ) ] - 24
= ( x2 + 7x + 10 )( x2 + 7x + 12 ) - 24
Đặt t = x2 + 7x + 10
<=> t( t + 2 ) - 24
= t2 + 2t - 24
= t2 - 4t + 6t - 24
= t( t - 4 ) + 6( t - 4 )
= ( t - 4 )( t + 6 )
= ( x2 + 7x + 10 - 4 )( x2 + 7x + 10 + 6 )
= ( x2 + 7x + 6 )( x2 + 7x + 16 )
= ( x2 + 6x + x + 6 )( x2 + 7x + 16 )
= [ x( x + 6 ) + ( x + 6 ) ]( x2 + 7x + 16 )
= ( x + 6 )( x + 1 )( x2 + 7x + 16 )
a, Sửa đề:\(\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)
Đặt \(t=x^2+x\)
\(\Rightarrow t^2-2t-15\)
\(=t^2-5t+3t-15\)
\(=t\left(t-5\right)+3\left(t-5\right)\)
\(=\left(t+3\right)\left(t-5\right)\)
\(=\left(x^2+x+3\right)\left(x^2+x-5\right)\)
b,\(x^2+2xy+y^2-x-y-12\)
\(=\left(x+y\right)^2-\left(x+y\right)-12\)
Đặt \(t=x+y\)
\(\Rightarrow t^2-t-12\)
\(=t^2-4t+3t-12\)
\(=t\left(t-4\right)+3\left(t-4\right)\)
\(=\left(t+3\right)\left(t-4\right)\)
\(=\left(x+y+3\right)\left(x+y-4\right)\)
c,\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
Đặt \(t=x^2+x+1\)
\(\Rightarrow t\left(t+1\right)-12\)
\(=t^2+t-12\)
\(=t^2-3t+4t-12\)
\(=t\left(t-3\right)+4\left(t-3\right)\)
\(=\left(t+4\right)\left(t-3\right)\)
\(=\left(x^2+x+5\right)\left(x^2+x-2\right)\)
\(=\left(x^2+x+5\right)\left(x^2-x+2x-2\right)\)
\(=\left(x^2+x+5\right)\left[x\left(x-1\right)+2\left(x-1\right)\right]\)
\(=\left(x^2+x+5\right)\left(x+2\right)\left(x-1\right)\)
d,\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt \(t=x^2+7x+10\)
\(\Rightarrow t\left(t+2\right)-24\)
\(=t^2+2t-24\)
\(=t^2-4t+6t-24\)
\(=t\left(t-4\right)+6\left(t-4\right)\)
\(=\left(t+6\right)\left(t-4\right)\)
\(=\left(x^2+7x+16\right)\left(x^2+7x+6\right)\)
\(=\left(x^2+7x+16\right)\left(x^2+x+6x+6\right)\)
\(=\left(x^2+7x+16\right)\left[x\left(x+1\right)+6\left(x+1\right)\right]\)
\(=\left(x^2+7x+16\right)\left(x+6\right)\left(x+1\right)\)