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\(a,a^3-7a-6\)
\(\Leftrightarrow a^3+a^2-a^2-a-6a-6\)
\(\Leftrightarrow a^2\left(a+1\right)-a\left(a+1\right)-6\left(a+1\right)\)
\(\Leftrightarrow\left(a+1\right)\left(a^2-a-6\right)\)
\(\left(x+1\right)\left(x+2\right)\left(x-3\right)\)
\(b,a^3+4a^2-7a-10\)
\(\Leftrightarrow a^3+5a^2-a^2-5a-2a-10\)
\(\Leftrightarrow a^2\left(a+5\right)-a\left(a+5\right)-2\left(a+5\right)\)
\(\Leftrightarrow\left(a+5\right)\left(a+1\right)\left(a-2\right)\)
\(d,\left(a^2+a\right)^2+4\left(a^2+a\right)-12\)
Đặt a^2+a=y ta có
y^2+4y-12=(y+6)(y-2)
<=> (a^2+a+6)(a^2+a-2)
<=> (a^2+a+6)(x-1)(x+2)
Phân tích đa thức thành nhân tử
a. 3ab ( x+ y) - 6ab ( y+ x)
=( x + y) ( 3ab - 6ab )
= ( x +y ) ( - 3ab)
b.7a (x - 3)+a2(x2 - 9)
=7a( x- 3) + a2 ( x2 - 32)
=7a ( x - 3 ) + a2 ( x- 3 ) ( x+3 )
= ( x- 3) . 7a + a2 ( x + 3)
= ( x- 3) ( 7a +a2x + 3a2)
c. 34 (x + y) -x -y
= 34 ( x+ y) - ( x+y)
=(x +y ) ( 34 - 1) = 33 ( x+ y)
d. 25 x4 - 942
=( 5x2 )2 - 942
=( 5x2 - 94 ) ( 5x2+94)
e.( 5a - b )2 - ( 2a +3b)2
=( 5a -b -2a - 3b) (5a -b + 2a + 3b)
=(3a - 4b) (7a+ 2b)
k. 22 -3a - b2 +3b
=( 22 - b2 ) + ( -3a +3b)
=( 2-b) (2+b) + 3( -a +b)
a) \(a^6-b^6=\left(a^2\right)^3-\left(b^2\right)^3=\left(a^2-b^2\right)\left(a^4+a^2b^2+b^{\text{4}}\right)\)
\(=\left(a-b\right)\left(a+b\right)\left(a^{\text{4}}+a^2b^2+b^{\text{4}}\right)\)
c) \(\left(x^2+x\right)^2+2\left(x^2+x\right)+1=\left(x^2+x+1\right)^2\)
e) \(\left(x^2-10x+25\right)-4y^2=\left(x-5\right)^2-\left(2y\right)^2\)
\(=\left(x-5-2y\right)\left(x-5+2y\right)\)
g) \(x^6+27=\left(x^2\right)^3+3^3=\left(x^2+3\right)\left(x^4-3x^2+9\right)\)
Còn lại tớ làm sau nhé, bây h muộn rùi
4. Đặt t= a^2 +a
Suy ra t^2 +4t - 12 = (t-2)(t+6) = (a^2+a-2) (a^2+a +6) = (a-1)(a+2)(a^2+a+6)
5. Đặt t = x^2 +x+1
Ta có: t(t+1) -12
= t^2 +t-12
= (t-3)(t+4)
= ( x^2 +x -2 ) (x^2+x+5)
= (x-1) ( x+2) (x^2+x+5)
6. x^8 + x^7 + x^6 - x^7- x^6 - x^5 + x^5+ x^4 + x^3- x^4- x^3- x^2 + x^2 + x +1
= (x^2 +x+1) ( x^6 - x^5 +x^3 -x^2 +1)
7. x^10 + x^9 +x^8 - x^9- x^8- x^7 +x^7+x^6+x^5 - x^6-x^5 - x^4 + x^5+ x^4 + x^3 - x^3 - x^2 - x + x^2 + x +1
= (x^2 + x + 1) ( x^8 -x^7 + x^5 - x^4 + x^3 -x + 1)
a3 - 7a - 6
= a3 - a - 6a - 6
= a ( a2 - 1 ) - 6 ( a + 1 )
= a ( a - 1 ) ( a + 1 ) - 6 ( a + 1 )
= ( a + 1 ) [ ( a ( a - 1 ) - 6 ]
= ( a + 1 ) ( a2 - a - 6 )
= ( a + 1 ) ( a2 + 2a - 3a - 6 )
= ( a + 1 ) ( a + 2 ) ( a - 3 )
\(3y^2\left(a-3x\right)-a\left(a-3x\right)=\left(3y^2-a\right)\left(a-3x\right)\)
Bài 2:
a) \(A=ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)\)
\(=\left(a-b\right)\left(c-a\right)\left(c-b\right)\)
b) \(B=a\left(b^2-c^2\right)+b^2\left(c^2-a^2\right)+c\left(a^2-b^2\right)\)
\(=\left(b-a\right)\left(c-a\right)\left(c-b\right)\)
c) \(C=\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
p/s: từ sau bn đăng 1-2 bài thôi nhé, nhiều thế này người lm bài cx hơi bất tiện để đọc đề
còn mấy câu nữa bn đăng lại nhé
a) Ta có: \(x^2-x-6\)
\(=x^2-x-9+3\)
\(=\left(x^2-9\right)-\left(x-3\right)\)
\(=\left(x-3\right)\left(x+3\right)-\left(x-3\right)\)
\(=\left(x-3\right)\left(x+2\right)\)
b) Sử dụng phương pháp Hệ số bất định
a) \(x^2\)\(-5x+6\)
=\(x^2\)\(-3x-2x+6\)
=\(x\left(x-3\right)-2\left(x-3\right)\)
=\(\left(x-2\right)\left(x-3\right)\)
b) \(3x^2\)\(+9x-30\)
=\(3x^2\)\(-6x+15x-30\)
=\(3x\left(x-2\right)+15\left(x-2\right)\)
=\(\left(x-2\right)\left(3x+15\right)\)
c)\(x^2\)\(-3x+2\)
=\(x^2\)\(-2x-x+2\)
=\(x\left(x-2\right)-\left(x-2\right)\)
=\(\left(x-2\right)\left(x-1\right)\)
d) \(12x^2\)\(+7x-12\)
=\(12x^2\)\(-9x+16x-12\)
=\(3x\left(4x-3\right)+4\left(4x-3\right)\)
=\(\left(3x+4\right)\left(4x-3\right)\)
e) \(15x^2\)\(+7x-2\)
=\(15x^2\)\(-3x+10x-2\)
=\(3x\left(5x-1\right)+2\left(5x-1\right)\)
=\(\left(3x+2\right)\left(5x-1\right)\)
f) \(a^2\)\(-5a-14\)
=\(a^2\)\(-7a+2a-14\)
=\(a\left(a-7\right)+2\left(a-7\right)\)
=\(\left(a+2\right)\left(a-7\right)\)
g) \(x^2\)\(-\left(a+b\right)x+ab\)
=\(x^2\)\(-ax-bx+ab\)
=\(x\left(x-a\right)-b\left(x-a\right)\)
=\(\left(x-a\right)\left(x-b\right)\)
\(a,Sửa:a^2-b^2=\left(a-b\right)\left(a+b\right)\\ b,=a^4+2a^2b^2+b^4-2a^2b^2\\ =\left(a^2+b^2\right)^2-2a^2b^2=\left(a^2+b^2-ab\sqrt{2}\right)\left(a^2+b^2+ab\sqrt{2}\right)\\ c,=a\left(a-1\right)\\ d,=a^2-a-2a+2=\left(a-1\right)\left(a-2\right)\\ e,=a^2-2a-3a+6=\left(a-2\right)\left(a-3\right)\\ g,=a^2-3a-4a+12=\left(a-3\right)\left(a-4\right)\)