a)=\(3x^3-15x^2+21x\)

b)\(=-2x^4y-10x^2y+2xy\)

c)\(=-x^3+6x^2+5x-4x^2+24x+20=-x^3+2x^2+29x+20\)

d)\(=2x^4-3x^3+4x^2-2x^2+3x-4=2x^4-3x^32x^2+3x-4\)

e)\(=x^2-4y^2\)

f)\(=-2x^2y^3+y-3\)

g)\(=3xy^4-\dfrac{1}{2}y^2+2x^2y\)

h)\(=9x^2-6x+1-7x^2-14=2x^2-6x-13\)

i)\(=x^2-x-3\)

j)\(=\left(x+2y\right)\left(x^2-2y+4y^2\right):\left(x+2y\right)=x^2-2y+4y^2\)

\(x^3-3x^2+3x-1=\left(x-1\right)^3\)

Tại \(x=101\)

\(\Rightarrow\left(x-1\right)^3=\left(101-1\right)^3=100^3=1000000\)

\(x^3-3x^2+3x-1=x^3-1-3x^2+3x\)

\(=\left(x-1\right)\left(x^2+x+1\right)-3x\left(x-1\right)=\left(x-1\right)\left(x^2+x+1-x+1\right)\)

\(=\left(x-1\right)\left(x^2+2\right)\)Thay x = 101 ta được 

\(=\left(101-1\right)\left(101^2+2\right)=100.10203=1020300\)

Ta có: \(x^3-3x^2+3x-1\)

\(=\left(x-1\right)^3\)

\(=\left(101-1\right)^3\)

\(=100^3=1000000\)

\(f\left(x\right)-g\left(x\right)=\left(x^5-3x^2+x^3-x^2-2x+5\right)-\left(x^2-3x+1+x^2-x^4+x^5\right)\)

\(f\left(x\right)-g\left(x\right)=x^5-3x^2+x^3-x^2-2x+5-x^2+3x-1-x^2+x^4-x^5\)

\(f\left(x\right)-g\left(x\right)=\left(x^5-x^5\right)+\left(-3x^2-x^2-x^2-x^2\right)+x^3+\left(-2x+3x\right)+\left(5-1\right)+x^4\)

\(f\left(x\right)-g\left(x\right)=-6x^2+x^3+x+4+x^4\)

\(f\left(x\right)-g\left(x\right)=x^4+x^3-6x^2+x+4\)

\(\Leftrightarrow\left(x+1\right)^3=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

<=>x³ +3x²1 +3x1²+1²=0

<=>(x+1)²=0

<=>x+1=0

<=>x=-1

đề là gì đấy nhỉ

\(\dfrac{3x^2-3xy}{3\left(y-x\right)^2}=\dfrac{3x\left(x-y\right)}{3\left(x-y\right)^2}=\dfrac{x}{x-y}\Rightarrow?=x\)

uu7

\(=\left(x-1\right)^3\)

(X-1)^3

\(a,=\left(x-2\right)^2\\ b,=\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\\ c,=\left(1-2x\right)\left(1+2x+4x^2\right)\\ d,=\left(x+1\right)^3\\ e,Sửa:\left(x+y\right)^2-9x^2=\left(x+y-3x\right)\left(x+y+3x\right)\\ =\left(y-2x\right)\left(4x+y\right)\\ f,=\left(x+3\right)^2\\ g,=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\\ h,=8\left(x^3-\dfrac{1}{64}\right)=8\left(x-\dfrac{1}{4}\right)\left(x^2+\dfrac{1}{4}x+\dfrac{1}{16}\right)\)

a) \(\left(x-2\right)^2\)

b) \(\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\)

c) \(\left(1-2x\right)\left(1+2x+4x^2\right)\)

d) \(\left(x+1\right)^3\)

e) \(\left(x+y-3\sqrt{x}\right)\left(x+y+3\sqrt{x}\right)\)

f) \(\left(x+3\right)^2\)

g) \(-\left(x-5\right)^2\)

h) \(\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)

\(x^3+3x^2+3x+1=\left(x+1\right)^3\)

\(x^3+3x^2+3x=0\\ \Leftrightarrow x\left(x^2+3x+3\right)=0\\ \Leftrightarrow x=0\left(x^2+3x+3=x^2+3x+\dfrac{9}{4}=\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}>0\right)\)

\(x^3+3x^2+3x=0\)

\(\Rightarrow x\left(x^2+3x+3\right)=0\)

Mà: \(x^2+3x+3>0\)

=> x = 0

\(x^3+3x^2+3x=0\)

\(\Leftrightarrow x\left(x^2+3x+3\right)=0\)

mà \(x^2+3x+3=\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\)

\(\Rightarrow x=0\)