Phân tích đa thức thành nhân tử :
a
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\(=\left(3x+1\right)^3+\dfrac{1}{3}\left(3x+1\right)=\left(3x+1\right)\left(9x^2+6x+1+\dfrac{1}{3}\right)\\ =\left(3x+1\right)\left(9x^2+6x+\dfrac{4}{3}\right)\)
12) \(\sqrt{11+2\sqrt{30}}=\sqrt{\left(\sqrt{6}\right)^2+2.\sqrt{6}.\sqrt{5}+\left(\sqrt{5}\right)^2}\)
\(=\sqrt{\left(\sqrt{6}+\sqrt{5}\right)^2}=\sqrt{6}+\sqrt{5}\)
\(=\sqrt{6+2\cdot\sqrt{6}\cdot\sqrt{5}+5}\)
\(=\sqrt{\left(\sqrt{6}+\sqrt{5}\right)^2}=\sqrt{6}+\sqrt{5}\)
\(\Delta=27^2-4.168=57>0\)
pt có 2 nghiệm pb
\(x=\dfrac{27\pm\sqrt{57}}{2}\)
x2+4x+3=x2+x+3x+3=(x2+x)+(3x+3)=x(x+1)+3(x+1)=(x+3)(x+1)
\(4x^2+4x-3\)
\(4x^2+4x+1-4\)
\(\left(2x+1\right)^2-2^2\)
\(\left(2x+1-2\right)\left(2x+1+2\right)\)
\(\left(2x-1\right)\left(2x+3\right)\)
\(x^3-64x=x\left(x^2-64\right)=x\left(x-8\right)\left(x+8\right)\)
\(64x^4+1\)
\(=64x^4+16x^2+1-16x^2\)
\(=\left(8x^2-4x+1\right)\left(8x^2+4x+1\right)\)
Đề sai nhé .Sửu lại
\(x^2-4x^2y^2+4+4x\)
\(=\left(x^2+4x+4\right)-4x^2y^2\)
\(=\left(x+2\right)^2-\left(2xy\right)^2\)
\(=\left(x+2+2xy\right)\left(x+2-2xy\right)\)
7a/
$x^3y+x-y-1=(x^3y-y)+(x-1)=y(x^3-1)+(x-1)$
$=y(x-1)(x^2+x+1)+(x-1)=(x-1)[y(x^2+x+1)+1]$
$=(x-1)(x^2y+xy+y+1)$
7b/
$x^2(x-2)+4(2-x)=x^2(x-2)-4(x-2)=(x-2)(x^2-4)$
$=(x-2)(x-2)(x+2)=(x-2)^2(x+2)$
7c/
$x^3-x^2-20x=x(x^2-x-20)=x[(x^2+4x)-(5x+20)]$
$x[x(x+4)-5(x+4)]=x(x+4)(x-5)$
7d/
$(x^2+1)^2-(x+1)^2=[(x^2+1)-(x+1)][(x^2+1)+(x+1)]$
$=(x^2-x)(x^2+x+2)$
$=x(x-1)(x^2+x+2)$
7e/
$6x^2-7x+2=(6x^2-3x)-(4x-2)=3x(2x-1)-2(2x-1)=(2x-1)(3x-2)$
7f/
$x^4+8x^2+12=(x^4+6x^2)+(2x^2+12)=x^2(x^2+6)+2(x^2+6)$
$=(x^2+6)(x^2+2)$
7g/
$(x^3+x+1)(x^3+x)-2=(t+1)t-2$ (đặt $x^3+x=t$)
$=t^2+t-2=(t^2+2t)-(t+2)=t(t+2)-(t+2)$
$=(t+2)(t-1)=(x^3+x+2)(x^3+x-1)$
$=[(x^3+x^2)-(x^2+x)+(2x+2)](x^3+x-1)$
$=[x^2(x+1)-x(x+1)+2(x+1)](x^3+x-1)$
$=(x+1)(x^2-x+2)(x^3+x-1)$