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\(a,x^3+8y^3=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)
\(b,a^6-b^3=\left(a^2-b\right)\left(a^4+a^2b+b^2\right)\)
\(c,8y^3-125=\left(2y-5\right)\left(4y^2+10y+25\right)\)
\(d,8z^3+27=\left(2z+3\right)\left(4z^2-6z+9\right)\)
\(a)x^3+8y^3=x^3+\left(2y\right)^3=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)
\(b)a^6-b^3=\left(a^3-b^3\right)\left(a^3+b^3\right)\)
\(c)8y^3-125=\left(2y\right)^3-5^3=\left(2y-5\right)\left(4y^2+10y+25\right)\)
\(d)8z^3+27=\left(2z\right)^3+3^3=\left(2x+3\right)\left(4z^2-6z+9\right)\)
a ) \(x^3+8y^3=x^3+\left(2y\right)^3=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)
b ) \(a^6-b^3=\left(a^2\right)^3-b^3=\left(a^2-b\right)\left(a^4+a^2b+b^2\right)\)
c ) \(8y^3-125=\left(2y\right)^3-5^3=\left(2y-5\right)\left(4y^2+10y+25\right)\)
d ) \(8z^3+27=\left(2z\right)^3+3^3=\left(2z+3\right)\left(4z^2-6z+9\right)\)
a) x3 + 8y3 = x3 + (2y)3 = (x+2y)(x2+2xy+4y2)
b) a6 - b3 = (a2)3 - b3 = (a2-b)(a4 + a2b + b2)
c) 8y3 - 125 = (2y)3 - 53 = (2y - 5)(4y2 + 10y + 25)
d) 8x3 + 27 = (2z)3 + 33 = (2z + 3)(4z2 - 6x + 9)
\(8-27x^3\)
\(=2^3-\left(3x\right)^3\)
\(=\left(2-3x\right)\left(4+6x+9x^2\right)\)
a) \(8-27x^3=\left(2-x\right)\left(4+6x+9x^2\right)\)
b) \(27+27x+9x^2+x^3=\left(3+x\right)^3\)
c) \(x^3+8y^3=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)
\(a,y-x^2y+2xy^2-y^3=y(1-x^2+2xy-y^2) =y[1-(x^2-2xy+y^2)]=y[1-(x-y)^2] =y(1-x+y)(1+x-y) =y(x+y-1)(x-y+1) \)
BÀi 1 : xem lại đề
bài 2
a) 27 - x^3
= ( 3 -x )( 9 + 3x + x^2)
b) 8x^3 + 0,001
= (2x + 0,1) ( 4x^2 - 0,2x + 0,01)
\(\frac{x^3}{64}-\frac{y^3}{125}=\left(\frac{x}{4}-\frac{y}{5}\right)\left(\frac{x^2}{16}-\frac{xy}{20}+\frac{y^2}{25}\right)\)
a+b=7=>(a+b)2=49
=>a2+2ab+b2=49
Do ab=3
=>2ab=6
=>b2+a2=43
Ta có:a3+b3=(a+b)(a2-ab+b2)
Thay a2+b2=43 ab=3 a+b=7
=> a3+b3=7.(43-3)=7.40=280
a)27-x3=(3-x)(9+3x+x2)
b)8x3+0,001=(2x+0,1)(4x2-0,2x+0,01)
c)x3/64-y3/125=(x/4-y/5)(x2/16+xy/20+y2/25)
a) \(3xyz^2-5xzt=xz\left(3yz-5t\right)\)
b) \(49x^2-25=\left(7x-5\right)\left(7x+5\right)\)
c) \(x^3+8z^3=\left(x+2z\right)\left(x^2+4zx+4x^2\right)\)
a) \(2x^2-4xy+2y^2-8z^2=2\left(x^2-2xy+y^2-4z^2\right)=2\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
\(=2\left(x-y-2z\right)\left(x-y+2z\right)\)
b) \(x^3-3x^2-4x+12=x^2\left(x-3\right)-4\left(x-3\right)=\left(x^2-4\right)\left(x-3\right)=\left(x-2\right)\left(x+2\right)\left(x-3\right)\)
Bạn sai đề rồi :
a ) \(8y^3-125\)
\(=\left(2y\right)^3-5^3\)
\(=\left(2y-5\right)\left(4y^2+2y.5+5^2\right)\)
\(=\left(2y-5\right)\left(4y^2+10y+25\right)\)
b) Ta thấy \(8z^3=\left(2z\right)^3\)còn \(27=3^3\)
Trở thành : \(\left(2z\right)^3+3^3\)
Rồi bạn bạn tự làm nha
Thanks