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a) x2 + x - 12 = x2 - 3x + 4x - 12 = x( x - 3 ) + 4( x - 3 ) = ( x - 3 )( x + 4 )
b) x2 - 4x - 5 = x2 + x - 5x - 5 = x( x + 1 ) - 5( x + 1 ) = ( x + 1 )( x - 5 )
c) x2 - 2x - 3 = x2 + x - 3x - 3 = x( x + 1 ) - 3( x + 1 ) = ( x + 1 )( x - 3 )
d) x2 - 2x - 8 = x2 + 2x - 4x - 8 = x( x + 2 ) - 4( x + 2 ) = ( x + 2 )( x - 4 )
e) x2 - 5x - 6 = x2 + x - 6x - 6 = x( x + 1 ) - 6( x + 1 ) = ( x + 1 )( x - 6 )
f) x2 - 6x + 8 = x2 - 2x - 4x + 8 = x( x - 2 ) - 4( x - 2 ) = ( x - 2 )( x - 4 )
g) x2 + 4x + 3 = x2 + x + 3x + 3 = x( x + 1 ) + 3( x + 1 ) = ( x + 1 )( x + 3 )
h) x2 - 2x - 15 = x2 + 3x - 5x - 15 = x( x + 3 ) - 5( x + 3 ) = ( x + 3 )( x - 5 )
i) x2 + 7x + 12 = x2 + 3x + 4x + 12 = x( x + 3 ) + 4( x + 3 ) = ( x + 3 )( x + 4 )
j) x2 - 5x - 14 = x2 + 2x - 7x - 14 = x( x + 2 ) - 7( x + 2 ) = ( x + 2 )( x - 7 )
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)
a,x2 - 13x + 36
= x2 - 4x - 9x + 36
= x(x - 4) - 9(x - 4)
= (x - 4)(x - 9)
b,x2 + 3x - 18
= x2 - 9 + 3x - 9
= (x - 3)(x + 3) + 3(x - 3)
= (x - 3)(x + 3 + 3)
= (x - 3)(x + 6)
c,x2 - 5x - 24
= x2 + 3x - 8x - 24
= x(x + 3) - 8(x + 3)
= (x + 3)(x - 8)
d,3x2 - 16x + 5
= 3x2 - 15x - x + 5
= 3x(x - 5) - (x - 5)
= (x - 5)(3x - 1)
e, 8x2 + 30x + 7
= 8x2 + 2x + 28x + 7
= 2x(4x + 1) + 7(4x + 1)
= (4x + 1)(2x + 7)
g,2x2 - 5x - 12
= 2x2 - 8x + 3x - 12
= 2x(x - 4) + 3(x - 4)
= (x - 4)(2x + 3)
i,x4 + 4
= x4 + 4x2 + 4 - 4x2
= (x2 + 2)2 - (2x)2
= (x2 + 2 - 2x)(x2 + 2 + 2x)
Phân tích đa thức thành nhân tử bằng cách tách một số hạng tử thành nhiều số hạng khác
a) x^2+4x+3
b) 4x^2-4x-3
c)x^2-x-12
d) 4x^4-4x^2-8y^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)
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
\(A=\left(x^2+x\right)^2-14\left(x^2+x\right)+24\)
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:
\(F=t\left(t+1\right)-12\)
\(=t^2+t-12\)
\(=t^2+4t-3t-12\)
\(=t\left(t+4\right)-3\left(t+4\right)\)
\(=\left(t+4\right)\left(t-3\right)\)
\(=\left(x^2+x+1+4\right)\left(x^2+x+1-3\right)\)
\(=\left(x^2+x+5\right)\left(x^2+x-2\right)\)
\(E=x^4+2x^3+5x^2+4x-12\)
\(=x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12\)
\(=x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)\)
\(=\left(x-1\right)\left(x^3+3x^2+8x+12\right)\)
\(=\left(x-1\right)\left(x^3+2x^2+x^2+2x+6x+12\right)\)
\(=\left(x-1\right)\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]\)
\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)
Mấy bài kia phá tung tóe rồi rút gọn hết sức xong thay x vào, làm câu c thôi nhé:
c) \(C=x^{14}-10x^{13}+10x^{12}-10x^{11}+...+10x^2-10x+10\)
riêng câu này ta thay x = 9 vào luôn, vậy ta có:
\(C=9^{14}-10\cdot9^{13}+10\cdot9^{12}-10\cdot9^{11}+...+10\cdot9^2-10\cdot9+10\)
\(=9^{14}-\left(9+1\right)\cdot9^{13}+\left(9+1\right)\cdot9^{12}-\left(9+1\right)\cdot9^{11}+...+\left(9+1\right)\cdot9^2-\left(9+1\right)\cdot9+10\)
\(=9^{14}-9^{14}-9^{13}+9^{13}+9^{12}-9^{12}-9^{11}+...+9^3+9^2-9^2-9+10\)
\(=-9+10\)
\(=1\)
a) x2 + 7x + 12 = x2 + 3x + 4x + 12 = x( x + 3 ) + 4( x + 3 ) = ( x + 3 )( x + 4 )
b) x2 - 10x + 16 = x2 - 2x - 8x + 16 = x( x - 2 ) - 8( x - 2 ) = ( x - 2 )( x - 8 )
c) x2 + 6x + 8 = x2 + 2x + 4x + 8 = x( x + 2 ) + 4( x + 2 ) = ( x + 2 )( x + 4 )
d) x2 - 8x + 15 = x2 - 3x - 5x + 15 = x( x - 3 ) - 5( x - 3 ) = ( x - 3 )( x - 5 )
e) x2 - 8x - 9 = x2 + x - 9x - 9 = x( x + 1 ) - 9( x + 1 ) = ( x + 1 )( x - 9 )
f) x2 + 14x + 48 = x2 + 6x + 8x + 48 = x( x + 6 ) + 8( x + 6 ) = ( x + 6 )( x + 8 )
a) \(x^2-5x+6=x^2-2x-3x+6=\left(x-2\right)\left(x-3\right)\)
b)\(3x^2+9x-30=3x^2-6x+15x-30=3\left(x-2\right)\left(x+5\right)\)
c)\(x^2-7x+12=x^2-3x-4x+12=\left(x-3\right)\left(x-4\right)\)
d)\(x^2-7x+10=x^2-2x-5x+10=\left(x-2\right)\left(x-5\right)\)
a) \(x^2-5x+6=x^2-2x-3x+6=\left(x^2-2x\right)-\left(3x-6\right)\)
\(=x\left(x-2\right)-3\left(x-2\right)=\left(x-2\right)\left(x-3\right)\)
b) \(3x^2+9x-30=3\left(x^2+3x-10\right)=3\left(x^2-2x+5x-10\right)\)
\(=3\left[\left(x^2-2x\right)+\left(5x-10\right)\right]=3\left[x\left(x-2\right)+5\left(x-2\right)\right]\)
\(=3\left(x-2\right)\left(x+5\right)\)
c) \(x^2-7x+12=x^2-3x-4x+12=\left(x^2-3x\right)-\left(4x-12\right)\)
\(=x\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x-4\right)\)
d) \(x^2-7x+10=x^2-2x-5x+10=\left(x^2-2x\right)-\left(5x-10\right)\)
\(=x\left(x-2\right)-5\left(x-2\right)=\left(x-2\right)\left(x-5\right)\)
a) x2-5x+4
= x2-x-(4x-4)
= x(x-1)-4(x-1)
= (x-4)(x-1)
b) giống phần a
c) x2-7x+12
= x2-3x-(4x-12)
= x(x-3)-4(x-3)
= (x-4)(x-3)
d) x2+7x+12
= x2+3x+4x+12
= x(x+3)+4(x+3)
= (x+3)(x+4)
e) x2-8x+12
= x2-2x-(6x-12)
= x(x-2)-6(x-2)
=(x-6)(x-2)
f)x2+8x+12
= x2+2x+6x+12
=x(x+2)+6(x+2)
=(x+2)(x+6)
g) x2+x-12
=x2-3x+4x-12
=x(x-3)+4(x-3)
=(x-3)(x+4)
h)x2-3x-10
=x2+2x-(5x+10)
=x(x+2)-5(x+2)
=(x-5)(x+2)
i)x2+4x-21
=x2-3x+7x-21
=x(x-3)+7(x-3)
=(x+7)(x-3)