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![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=3\left(x-1\right)^2-\left(x+1\right)^2+2\left(x+3\right)\left(x-3\right)-\left(2x-3\right)^2-\left(5-20x\right)\)
\(=3\left(x^2-2x+1\right)-\left(x^2+2x+1\right)+2\left(x^2-9\right)-\left(4x^2-12x+9\right)-5+20x\)
\(=3x^2-6x+3-x^2-2x-1+2x^2-18-4x^2+12x-9-5+20x=24x-30\)
Vậy biểu thức phụ thuộc giá trị biến x
\(B=-x\left(x+2\right)^2+\left(2x+1\right)^2+\left(x+3\right)\left(x^2-3x+9\right)-1\)
\(=-x\left(x^2+4x+4\right)+4x^2+4x+1+x^3+27-1\)
\(=-x^3-4x^2-4x+4x^2+4x+1+x^3+27-1=27\)
Vậy biểu thức ko phụ thuộc giá trị biến x
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,-x^3+\left(x-3\right)\left[\left(2x+1\right)^2-2\left(\dfrac{3}{2}x^2+\dfrac{1}{2}x-4\right)\right]\\ =-x^3+\left(x-3\right)\left(4x^2+4x+1-3x^2-x+8\right)\\ =-x^3+\left(x-3\right)\left(x^2+3x+9\right)\\ =-x^3+\left(x^3-27\right)=-27\)
\(b,\left(x+2y\right)^3-\left(x-3y\right)\left(x^2+3xy+9y^2\right)-6y\left(x^2+2xy-\dfrac{35}{6}y^2\right)\\ =x^3+6x^2y+12xy^2+8y^3-x^3+27y^3-6x^2y-12xy^2+35y^3\\ =0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\frac{1}{3}-2x\right)\left(4x^2+\frac{2}{3}x+\frac{1}{9}\right)-\left(\frac{1}{27}-8x^3\right)\)
\(=\frac{1}{3}\left(4x^2+\frac{2}{3}x+\frac{1}{9}\right)-2x\left(4x^2+\frac{2}{3}x+\frac{1}{9}\right)-\frac{1}{27}+8x^3\)
\(=\frac{4}{3}x^2+\frac{2}{9}x+\frac{1}{27}-8x^3-\frac{4}{3}x^2-\frac{2}{9}x-\frac{1}{27}+8x^3\)
\(=\left(\frac{4}{3}x^2-\frac{4}{3}x^2\right)+\left(\frac{2}{9}x-\frac{2}{9}x\right)+\left(\frac{1}{27}-\frac{1}{27}\right)+\left(-8x^3+8x^3\right)\)
= 0 =>không phụ thuộc vào biến x
Ta có: \(\left(\frac{1}{3}-2x\right)\left(4x^2+\frac{2}{3}x+\frac{1}{9}\right)-\left(\frac{1}{27}-8x^3\right)\)
\(=\left(\frac{1}{3}-2x\right)\left[\left(\frac{1}{3}\right)^2+\frac{1}{3}\cdot2x+\left(2x\right)^2\right]-\left(\frac{1}{27}-8x^3\right)\)
\(=\left(\frac{1}{27}-8x^3\right)-\left(\frac{1}{27}-8x^3\right)\)
\(=0\)
=> đpcm
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(\left(x-1\right)\left(x-2\right)\left(x^2+x+1\right)\left(x^2+2x+4\right)-x^6+9x^3\)
\(=\left(x-1\right)\left(x^2+x+1\right)\left(x-2\right)\left(x^2+2x+4\right)-x^6+9x^3\)
\(=\left(x^3-1\right)\left(x^3-8\right)-x^6+9x^3\)
\(=x^6-9x^3+8-x^6+9x^3=8\)
b) Ta có: \(\left(\dfrac{1}{3}+2x\right)\left(\dfrac{1}{9}-\dfrac{2}{3}x+4x^2\right)-\left(2x-\dfrac{1}{3}\right)\left(4x^2+\dfrac{2}{3}x+\dfrac{1}{4}\right)\)
\(=\dfrac{1}{27}+8x^3-8x^3+\dfrac{1}{27}\)
\(=\dfrac{2}{27}\)
c) Ta có: \(\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3x\left(1-x\right)\)
\(=x^3-3x^2+3x-1-x^3+1-3x+3x^2\)
=0
d) Ta có: \(\left(x^2-y^2\right)\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)-x^6+y^6\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)-x^6+y^6\)
\(=\left(x^3-y^3\right)\left(x^3+y^3\right)-x^6+y^6\)
\(=x^6-y^6-x^6+y^6=0\)