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a) Đặt y=x2+x+1
Thay y vào biểu thức ta được
y(y+1)-12
=y2 + y - 12
= y2 - 3y + 4y -12
= y(y-3) + 4(y-3)
= (y-3)(y+4)
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
\(A=\left(x^2+x\right)^2-14\left(x^2+x\right)+24\)
Đặt \(x^2+x=t\), ta có:
\(A=t^2-14t+24\)
\(=t^2-2t-12t+24\)
\(=t\left(t-2\right)-12\left(t-2\right)\)
\(=\left(t-2\right)\left(t-12\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x-12\right)\)
\(B=\left(x^2+x\right)^2+4x^2+4x-12\)
\(=\left(x^2+x\right)^2+4\left(x^2+x\right)-12\)
Đặt \(x^2+x=t\), ta có:
\(B=t^2+4t-12\)
\(=t^2+6t-2t-12\)
\(=t\left(t+6\right)-2\left(t+6\right)\)
\(=\left(t+6\right)\left(t-2\right)\)
\(=\left(x^2+x+6\right)\left(x^2+x-2\right)\)
\(C=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)
Đặt \(x^2+5x+4=t\), ta có:
\(C=t\left(t+2\right)+1\)
\(=t^2+2t+1\)
\(=\left(t+1\right)^2\)
\(=\left(x^2+5x+4+1\right)^2\)
\(=\left(x^2+5x+5\right)^2\)
\(D=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)
\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
Đặt \(x^2+8x+7=t\), ta có:
\(D=t\left(t+8\right)+15\)
\(=t^2+8t+15\)
\(=t^2+3t+5t+15\)
\(=t\left(t+3\right)+5\left(t+3\right)\)
\(=\left(t+3\right)\left(t+5\right)\)
\(=\left(x^2+8x+7+3\right)\left(x^2+8x+7+5\right)\)
\(=\left(x^2+8x+10\right)\left(x^2+8x+12\right)\)
\(F=\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
Đặt \(x^2+x+1=t\), ta có:
\(F=t\left(t+1\right)-12\)
\(=t^2+t-12\)
\(=t^2+4t-3t-12\)
\(=t\left(t+4\right)-3\left(t+4\right)\)
\(=\left(t+4\right)\left(t-3\right)\)
\(=\left(x^2+x+1+4\right)\left(x^2+x+1-3\right)\)
\(=\left(x^2+x+5\right)\left(x^2+x-2\right)\)
\(E=x^4+2x^3+5x^2+4x-12\)
\(=x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12\)
\(=x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)\)
\(=\left(x-1\right)\left(x^3+3x^2+8x+12\right)\)
\(=\left(x-1\right)\left(x^3+2x^2+x^2+2x+6x+12\right)\)
\(=\left(x-1\right)\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]\)
\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)
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: \(\left[\left(x+2\right)\left(x+5\right)\right]\cdot\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
\(=\left(t-1\right)\left(t+1\right)-24\)
\(=t^2-25\)
\(=\left(t-5\right)\left(t+5\right)\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
a)9(2x+1)2 - 4(x-1) 2
<=>33(2x+1)2-22(x+1)2
<=>(3(2x+1)) 2-(2(x+1))2
<=>(6x+3)2-(2x+1)2
<=>((6x+3)-(2x+1)) ((6x+3)+(2x+1))
<=>(6x+3-2x-1)(6x+3+2x+1)
<=.>(4x+2)(8x+4)
b) x3 - 19x- 30
<=>x3-25x+6x-30
<=.>x(x2-52)+6(x-5)
<=>x(x+5)(x-5)+6(x-5)
<=>(x-5) (x2+5x+6)
<=>(x-5) (x2+2x+3x+6)
<=>(x-5) ( x(x+2)+3(x+2))
<=>(x-5) (x+2)(x+3)
c) x4+ x2 +1
<=>x4+x2+1
<=>x4−x+x2+x+1
<=>x(x3−1)+(x2+x+1)
<=>x(x−1)(x2+x+1)+(x2+x+1)
<=>(x2+x+1)[x(x−1)+1]
<=>(x2+x+1)(x2−x+1)
câu d mình chịu :(((
a) x2 - 4y2
= x2 - ( 2y )2
= ( x - 2y )( x + 2y )
b) x2 + x - 12
= x2 - 3x + 4x - 12
= x( x - 3 ) + 4( x - 3 )
= ( x - 3 )( x + 4 )
c) x2 + 2xy + y2 - 11
= ( x2 + 2xy + y2 ) - 11
= ( x + y )2 - ( √11 )2
= ( x + y - √11 )( x + y + √11 )
d) x4 + 1
= ( x4 + 2x2 + 1 ) - 2x2
= ( x2 + 1 )2 - ( √2x )2
= ( x2 - √2x + 1 )( x2 + √2x + 1 )
a) \(x^2-4y^2\)
\(=x^2-\left(2y\right)^2\)
\(=\left(x-2y\right).\left(x+2y\right)\)
b) \(x^2+x-12\)
\(=x^2+4x-3x-12\)
\(=\left(x^2+4x\right)-\left(3x+12\right)\)
\(=x.\left(x+4\right)-3.\left(x+4\right)\)
\(=\left(x+4\right).\left(x-3\right)\)
c) \(x^2+2xy+y^2-11\)
\(=\left(x^2+2xy+y^2\right)-11\)
\(=\left(x+y\right)^2-11\)
\(=\left(x+y\right)^2-\left(\sqrt{11}\right)^2\)
\(=\left(x+y-\sqrt{11}\right).\left(x+y+\sqrt{11}\right)\)
minh moi bn vao link nay dang ky roi tra loi minigame nha : https://alfazi.edu.vn/question/5b7768199c9d707fe5722878
a, x4 - 3x3 - x + 3
= (x4 - x) - (3x3 - 3)
= x(x3 - 1) - 3(x3 - 1)
= (x - 3)(x3 - 1)
b, x2 - x - 12
= x2 - x - 16 + 4
= (x2 - 16) - (x - 4)
= (x2 - 42) - (x - 4)
= (x + 4)(x - 4) - (x - 4)
= (x + 4 - 1)(x - 4)
= (x + 3)(x - 4)
c, x2 - 7x + 12
= x2 - 3x - 4x + 12
= (x2 - 3x) - (4x - 12)
= x(x - 3) - 4(x - 3)
= (x - 4)(x - 3)
d, x2 - 2x - 8
= x2 - 4x + 2x - 8
= (x2 - 4x) + (2x - 8)
= x(x - 4) + 2(x - 4)
= (x + 2)(x - 4)
5, x2 - 10x + 21
= x2 - 3x - 7x + 21
= (x2 - 3x) - (7x - 21)
= x(x - 3) - 7(x - 3)
= (x - 7)(x - 3)
f, x7 - x2 - 1
= t không bt
\(a,\left(x^2+x+1\right)\left(x^2+x+2\right)-12.\)
Đặt \(x^2+x+1=a\)
\(\Rightarrow a\left(a+1\right)-12\)\(=a^2+a-12\)
\(=a^2-3a+4a-12\)
\(=a\left(a-3\right)+4\left(a-3\right)\)
\(=\left(a-3\right)\left(a+4\right)\)
\(=\left(x^2+x+1-3\right)\left(x^2+x+1+4\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x+5\right)\)
\(b,\left(x^2+x\right)^2+4\left(x^2+x\right)-12\)
Đặt \(x^2+x=a\)
\(\Rightarrow a^2+4a-12\)
\(=a^2-2a+6a-12\)
\(=a\left(a-2\right)+6\left(a-2\right)\)
\(=\left(a-2\right)\left(a+6\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x+6\right)\)
Trả lời:
Đặt x^2+x+1=t
<=> t ( t + 1 ) - 12 = t^2 + t - 12 = t^2 + 4t - 3t - 12 = ( t + 4 ) ( t - 3 )
thay vào cách đặt
=> ( t + 4 )( t - 3 )=(x^2+x+5)(x^2+x-2)