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a) ( 4n2 - 6mn + 9m2 ) . ( 2n + 3m ) =
= ( 2n + 3m ) . ( 4n2 - 6mn + 9m2 )
= 8n3 + 27m3
b) ( 7 + 2b ) . ( 4b2 - 14b + 49 ) =
= ( 2b + 7 ) . ( 4b2 - 14b + 49 )
= 8b3 + 343
MÌnh nghĩ là chỗ - 4b phải viết là -14b
c) ( 25a2 + 10ab + 4b2 ) . ( 5a - 2b ) =
= ( 5a - 2b ) . ( 25a2 + 10ab + 4b2 )
= 125a3 - 8b3
HOk tốt!!!!!!!!!!!!
a ) \(\left(a^6-3a^3+9\right)\left(a^3+3\right)=a^9+27\)
b ) Đặt \(a-y=t\) , ta có :
\(\left(t-x\right)^3-\left(t+x\right)^3\)
\(=\left(t-x-t-x\right)\left[\left(t-x\right)^2+\left(t-x\right)\left(t+x\right)+\left(t+x\right)^2\right]\)
\(=-2x\left[t^2-2tx+x^2+t^2-x^2+t^2+2tx+x^2\right]\)
\(=-2x\left[\left(t^2+t^2+t^2\right)+\left(x^2-x^2+x^2\right)+\left(2tx-2tx\right)\right]\)
\(=-2x\left(3t^2+x^2\right)\)
\(=-2x\left[3\left(a-y\right)^2+x^2\right]\)
\(=-2x\left(3a^2-6ay+3y^2+x^2\right)\)
c ) \(\left(4n^2-6mn+9m^2\right)\left(2n+3m\right)=8n^3+27m^3\)
d ) \(\left(25a^2+10ab+4b^2\right)\left(5a-2b\right)=125a^3-8b^3\)
a, ( a6 - 3a3 + 9 )(a3+ 3) = (a3)3 - 33 = a9 - 27
b, ( a-x-y)3 - (a+x-y)3 = (a-x-y-a+x-y)(a-x-y+a+x-y)
= (-2y)(2a-2y) = -2y.2(a-y)
c, (4n2- 6mn + 9m2)(2n + 3m) = (2n)3 + (3m)3
= 8n3 + 27m3
d, (25a2 + 10ab +4b2)( 5a - 2b ) = 125a3 - 8b3
a) (x - 1)(x + 1)(x2 + 1)(x4 + 1)(x8 + 1)
= (x2 - 1)(x2 + 1)(x4 + 1)(x8 + 1)
= (x4 - 1)(x4 + 1)(x8 + 1)
= (x8 - 1)(x8 + 1)
= x16 - 1
b) (a2 - 2b)(a2 + 2b)(a4 + 4b2)(a8 + 16b4)
= (a4 - 4b2)(a4 + 4b2)(a8 + 16b4)
= (a8 - 16b4)(a8 + 16b4)
= a16 - 256b8
a) (a - 2b)2 = a2 - 2.a.2b + 4b2
= a2 - 4ab + 4b2
b) m2 - 4n2 = m2 - (2n)2 = (m - 2n)(m + 2n)
. Bài 1:
a; 9m^2 + n^2 - 6mn
= (3m)^2 - 2.3m.n + (n)^2
= ( 3m-n )^2
b; x^2-x+1/4
= x^2-2.(x).1/2+(1/2)^2
= (x-1/2)^2
a) \(\left(4n^2-6nm+9m^2\right)\left(2n+3m\right)\)
\(=\left(2n+3m\right)\left[\left(2n\right)^2-2n.3m+\left(3m\right)^2\right]\)
\(=\left(2n\right)^3+\left(3m\right)^3\)
\(=8n^3+27m^3\)
b) Sửa đề \(\left(7+2b\right)\left(4b^2-14b+49\right)\)
\(=\left(7+2b\right)\left[\left(2b\right)^2-2b.7+7^2\right]\)
\(=7^3+\left(2b\right)^3\)
\(=343+8b^3\)
c) \(\left(25a^2+10ab+4b^2\right)\left(5a-2b\right)\)
\(=\left(5a-2b\right)\left[\left(5a\right)^2+5a.2b+\left(2b\right)^2\right]\)
\(=\left(5a\right)^3-\left(2b\right)^3\)
\(=125a^3-8b^3\)
d) \(\left(x^2+x+2\right)\left(x^2-x-2\right)\)
\(=\left[x^2+\left(x+2\right)\right]\left[x^2-\left(x+2\right)\right]\)
\(=x^4-\left(x+2\right)^2\)
a)x2-4x+5+y2+2y=x2-4x+4+y2+2y+1=(x-2)2+(y+1)2
b)2x2+y2-2xy+10x+25=x2-2xy+y2+x2+10x+25=(X+Y)2+(X+5)2
c)a2+2ab+5b2+4b+1=a2+2ab+b2+4b2+4b+1=(a+b)2+(2b+1)2
d)2x2+2b2+4x+4b+4=2x2+4x+2+2b2+4b+2=(\(\sqrt{2}x+\sqrt{2}\))2+(\(\sqrt{2}b+\sqrt{2}\))2
e)X4+13-6x2+4y+y2=x4-6x2+9+y2+4y+4=(x2-3)2+(y+2)2
f)-6x+9x2-8y+4y+y2+5= 9x2-6x+1+4y2-8y+4= (3x-1)2+(2y-2)2
a3-4a2b-4b3+5ab2=0
==>(a-b)3 - b (a-b)2 =0
==>a-b = b ==> a=2b
thay a=2b vào biểu thức ta đc kết quả bằng 1
hình như mấy cái GP của Đinh Tuấn Việt là giả hay sao ấy nhỉ
a) \(=a^2-4b^2-2a-4b\)
\(=\left(a-2b\right)\left(a+2b\right)-2\left(a-2b\right)\)
\(=\left(a+2b\right)\left(a-2b-2\right)\)
a)5 a . ( a - 2 ) - a + 2
= 5a . ( a - 2 ) + (a - 2 )
= (5a + 1 ).(a - 2)
b)7. (a - 5) + 8a .(5 - a)
= 7. (a - 5) - 8a . (a - 5)
= (7 - 8a ).(a - 5)
c)25a2 - 4b2 + 4b - 1
= 25b2 - ( 4b2 + 4b - 1)
= 25b2 - ( 2b - 1)2
= ( 25b - 2b -1). ( 25b + 2b - 1)
= (23b - 1).( 27b - 1)
1/ Phân tích đa thức thành nhân tử:
a/ 5a.(a-2) - a+2 = 5a(a - 2) - (a - 2) = (a - 2)(5a - 1)
b/7.(a-5)+8a.(5-a) = 7(a - 5) - 8a(a - 5) = (a - 5) (7 - 8a)
c/ 25a2 -4b2 +4b -1 = 25a2 - (4b2 - 4b + 1) = 25a2 - (2b - 1)2
= (5a - 2b + 1)(5a +2b - 1)
d/6ax2 -36ax+54a = 6a( x2 - 6x + 9 ) = 6a(x - 3)2