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a) \(\left(2x^3-y^2\right)^3\)
\(=\left(2x^3\right)^3-3\cdot\left(2x^3\right)^2\cdot y^2+3\cdot2x^3\cdot\left(y^2\right)^{^2}-\left(y^2\right)^3\)
\(=8x^9-3\cdot4x^6y^2+3\cdot2x^3y^4-y^6\)
\(=8x^9-12x^6y^2+6x^3y^4-y^6\)
b) \(\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)
\(=x^3-\left(3y\right)^3\)
\(=x^3-27y^3\)
c) \(\left(x+2y+z\right)\left(x+2y-z\right)\)
\(=\left(x+2y\right)^2-z^2\)
\(=x^2+4xy+4y^2-z^2\)
d) \(\left(2x^3y-0,5x^2\right)^3\)
\(=\left(2x^3y-\dfrac{1}{2}x^2\right)^3\)
\(=8x^9y^3-6x^8y^2+\dfrac{3}{2}x^7y-\dfrac{1}{8}x^6\)
e) \(\left(x^2-3\right)\left(x^4+3x^2+9\right)\)
\(=\left(x^2-3\right)\left(4x^2+9\right)\)
\(=4x^4+9x^2-12x^2-27\)
\(=4x^4-3x^2-27\)
f) \(\left(2x-1\right)\left(4x^2+2x+1\right)\)
\(=\left(2x\right)^3-1^3\)
\(=8x^3-1\)
\(a,\left(2x^3-y^2\right)^3=8x^9-12x^6y^2+6x^3y^4-y^6\)\(b,\left(x-3y\right)\left(x^2+3xy+9y^2\right)=x^3-27y^3\)
\(c,\left(x+2y+z\right)\left(x+2y-z\right)=\left(x+2y\right)^2-z^2=x^2+4xy+4y^2-z^2\)\(d,\left(2x^3y-0,5x^2\right)^3=8x^9y^3-6x^4y^2x^2+3x^3yx^4-0,125x^6=8x^9y^3-6x^6y^2+3x^7y-0,125x^6\)
\(a,\left(x-3y\right)\left(x^2+3xy+9y^2\right)=x^3-27y^3\)
\(b,\left(x^2-3\right)\left(x^4+3x^2+9\right)=x^6-27\)
\(c,\left(x+2y+z\right)\left(x+2y-z\right)=\left(x+2y\right)^2-z^2\)
\(=x^2+4xy+4y^2-z^2\)
\(d,\left(2x-1\right)\left(4x^2+2x+1\right)=8x^3-1\)
\(e,\left(5+3x\right)^3=125+225x+135x^2+27x^3\)
a: \(\left(2x+3\right)^3=8x^3+36x^2+54x+27\)
b: \(\left(x-3y\right)^3=x^3-9x^2y+27xy^2-27y^3\)
\(a,\left(x+4\right).\left(x^2-4x+16\right)=x^3-4x^2+16x+4x^2-16x+64\) \(64\)
\(=x^3+64\)
hoặc \(\left(x+4\right).\left(x^2-4x+16\right)=x^3+64\) ÁP Dụng hằng đẳng thức
\(b,\left(x-3y\right).\left(x^2+3xy+9y^2\right)=x^3-27\)
a) (x+4).(x^2-4x+16)
= (x+4).(x^2-x.4+4^2)
= x^3+4^3
= x^3+64
b) (x-3y).(x^2+3xy+9y^2)
= (x-3y).(x^2+x.3y+(3y)^2)
= x^3-(3y)^3
= x^3-27y^3
\(\left(\dfrac{1}{3}x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)\)
\(=\left(\dfrac{1}{3}x+2y\right)\left[\left(\dfrac{1}{3}xy\right)^2-\dfrac{1}{3}x.2y+\left(2y\right)^2\right]\)
\(=\left(\dfrac{1}{3}x\right)^3+\left(2y\right)^3=\dfrac{1}{27}x^3+8y^3\)
\(\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)\)
\(=\left(x^2-\dfrac{1}{3}\right)\left[\left(x^2\right)^2+\dfrac{1}{3}.x^2+\left(\dfrac{1}{3}\right)^2\right]\)
\(=\left(x^2\right)^3-\left(\dfrac{1}{3}\right)^3\)
\(=x^6-\dfrac{1}{27}\)
Giải:
+) \(\left(x-4\right)\left(x^2+4x+16\right)\)
\(=\left(x-4\right)\left(x^2+4.x+4^2\right)\)
\(=x^3-4^3\)
\(=x^3-64\)
+) \(\left(\dfrac{1}{3}x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)\)
\(=\left(\dfrac{1}{3}x+2y\right)\left[\left(\dfrac{1}{3}x\right)^2-\dfrac{1}{3}x.2y+\left(2y\right)^2\right]\)
\(=\left(\dfrac{1}{3}x\right)^3+\left(2y\right)^3\)
\(=\dfrac{1}{27}x^3+8y^3\)
+) \(\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)
\(=\left(x-3y\right)\left[x^2+x.3y+\left(3y\right)^2\right]\)
\(=x^3-\left(3y\right)^3\)
\(=x^3-27y^3\)
+) \(\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)\)
\(=\left(x^2-\dfrac{1}{3}\right)\left[\left(x^2\right)^2+\dfrac{1}{3}.x^2+\left(\dfrac{1}{3}\right)^2\right]\)
\(=\left(x^2\right)^3-\left(\dfrac{1}{3}\right)^3\)
\(=x^6-\dfrac{1}{27}\)
Vậy ...
a) \(\left(x^3-2y\right)^3\)
\(=\left(x^3-2y\right)\left[\left(x^3\right)^2+x^3.2y+\left(2y^2\right)\right]\)
\(=\left(x^3-2y\right)\left(x^6+2x^3y+4y^2\right)\)
b) \(\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)
\(=\left(x-3y\right)\left(x^2+x.3y+\left(3y\right)^2\right)\)
\(=\left(x-3y\right)^3\)