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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
\(x^2-x-6=x^2-3x+2x-6=x\left(x-3\right)+2\left(x-3\right)=\left(x-3\right)\left(x+2\right)\)
\(x^4+x^2+1=x^4+2x^2+1-x^2=\left(x^2+1\right)-x^2=\left(x^2+x+1\right)\left(x^2-x+1\right)\)\(x^3-19x-30=\left(x^3+8\right)-\left(19x-38\right)=\left(x+2\right)\left(x^2-2x+4\right)-19\left(x+2\right)=\left(x+2\right)\left(x^2-2x-15\right)=\left(x+2\right)\left(x^2-5x+3x-15\right)=\left(x+2\right)\left(x-5\right)\left(x+3\right)\)
\(x^4+4x^2-5=x^4+4x^2+4-9=\left(x^2+2\right)^2-9=\left(x^2+5\right)\left(x^2-1\right)=\left(x^2+5\right)\left(x-1\right)\left(x+1\right)\)
\(x^3-7x-6=0\Leftrightarrow\left(x^3+1\right)-\left(7x+7\right)=0\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)-7\left(x+1\right)=0\Leftrightarrow\left(x+1\right)\left(x^2-x-6\right)=0\Leftrightarrow\left(x+1\right)\left(x^2-3x+2x-6\right)=0\Leftrightarrow\left(x+1\right)\left(x-3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\\x=-1\end{matrix}\right.\)
\(x^3-3x^2-16x+48=x^2\left(x-3\right)-16\left(x-3\right)=\left(x^2-16\right)\left(x-3\right)=\left(x-4\right)\left(x+4\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-3=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\\x=-4\end{matrix}\right.\)
a) 4.(2a-b)2-16(a-b)2
= [2(2a-b)]2 - [4(a-b)]2
= [2(2a-b)-4(a-b)].[2(2a-b)+4(a-b)]
= [4a-2b-4a+4b].[4a-2b+4a-4b]
= 2b.(8a-6b)
b) 8x3-27y3
= (2x)3 - (3y)3
= (2x - 3y).[(2x)2+2x.3y+(3y)2]
= (2x-3y)(4x2+6xy+9y2)
c) 1/64x6-125y3
= (1/4x2)3 - (5y)3
= (1/4x2 - 5y)[(1/4x2)2 + 1/4x2.5y + (5y)2]
= (1/4x2 - 5y)(1/16x4 + 5/4x2y +25y2)
d) (x+3)3-8
= (x+3-2)[(x+3)2+(x+3).2+22]
= (x+1)(x2+6x+9+2x+6+4)
= (x+1)(x2+8x+19)
e) x6+1
= (x2)3 + 13
= (x2 + 1)[(x2)2 - x2 + 1]
= (x2 + 1)(x4-x2+1)
g) x9 + 1
= (x3)3 + 13
= (x3 + 1 )[(x3)2 - x6 + 1]
= (x+1)(x2+x+1)(x6-x6+1)
= (x+1)(x2+x+1)
Mình gõ hơi lâu mới làm được nhiêu đó thôi
\(h)x^3+12x^2+48x+64=\left(x+4\right)^3\)
\(i)27-27m+9m^2-m^3=\left(3-m\right)^3\)
a, \(4x^2+4xy+y^2=\left(4x\right)^2+2.2x.y+y^2\)
\(=\left(4x+y\right)^2\)
b, \(9m^2+n^2-6mn=\left(3m\right)^2-2.3m.n+n^2\)
\(=\left(3m-n\right)^2\)
c, \(16a^2+25b^2+40ab=\left(4a\right)^2+2.4a.5b+\left(5b\right)^2\)
\(=\left(4a+5b\right)^2\)
d, \(x^2-x+\dfrac{1}{4}=x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\)
\(=\left(x-\dfrac{1}{2}\right)^2\)
Chúc bạn học tốt!!!
Bài 1:
b:
x=9 nên x+1=10
\(M=x^{10}-x^9\left(x+1\right)+x^8\left(x+1\right)-x^7\left(x+1\right)+...-x\left(x+1\right)+x+1\)
\(=x^{10}-x^{10}-x^9+x^9+x^8-x^8-x^7+...-x^2-x+x+1\)
=1
c: \(N=\left(1+2+2^2+2^3+2^4\right)+2^5\left(1+2+2^2+2^3+2^4\right)+2^{10}\left(1+2+2^2+2^3+2^4\right)\)
\(=31\left(1+2^5+2^{10}\right)⋮31\)
Câu a : \(4x^2+4xy+y^2=\left(2x+y\right)^2\)
Câu b : \(9m^2+n^2-6mn=\left(3m-n\right)^2\)
Câu c : \(16a^2+25b^2+40ab=\left(4a+5b\right)^2\)
Câu d : \(x^2-x+\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2\)
\(a,4x^2+4xy+y^2=\left(2x\right)^2+4xy+y^2=\left(2x+y\right)^2\)
\(b,9m^2+n^2-6mn=\left(3m\right)^2-6mn+n^2=\left(3m-n\right)^2\)
\(c,16a^2+25b^2+40ab=\left(4a\right)^2+40ab+\left(5b\right)^2=\left(4a+5b\right)^2\)
@Yukru ơi! giúp câu D với!
Chúc bạn học tốt!
c: \(5\left(a+b\right)+x\left(a+b\right)\)
=(a+b)(x+5)
d: \(\left(a-b\right)^2-\left(b-a\right)\)
\(=\left(a-b\right)^2+\left(a-b\right)\)
=(a-b)(a-b+1)
e: \(=\left(12x^2+6x\right)\left(y+z+y-z\right)\)
\(=2y\cdot6x\cdot\left(2x+1\right)=12xy\left(2x+1\right)\)
Đề bài là j vậy bạn???