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a) \(x^2+2x+1=\left(x+1\right)^2\)
b) \(9x^2+y^2+6xy=\left(3x+y\right)^2\)
c) \(25a^2+4b^2-20ab=\left(5a-2b\right)^2\)
d) \(x^2-x+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2\)
e) \(\left(2x+3y\right)^3+2\left(2x+3y\right)+1=\left(2x+3y+1\right)^2\)
f) mk chỉnh lại đề nha:
\(2xy^2+x^2y^4+1=\left(xy^2+1\right)^2\)
g) \(x^2+6xy+9y^2=\left(x+3y\right)^2\)
h) \(x^2-10xy+25y^2=\left(x-5y\right)^2\)
a) Ta có: \(\left(x-3\right)^3\)
\(=x^3-3\cdot x^2\cdot3+3\cdot x\cdot3^2-3^3\)
\(=x^3-9x^2+27x^2-27\)
b) Ta có: \(\left(2x-3\right)^3\)
\(=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot3+3\cdot2x\cdot3^2-3^3\)
\(=8x^3-36x^2+54x-27\)
c) Ta có: \(\left(x-\frac{1}{2}\right)^3\)
\(=x^3-3\cdot x^2\cdot\frac{1}{2}+3\cdot x\cdot\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^3\)
\(=x^3-\frac{3}{2}x^2+\frac{3}{4}x-\frac{1}{8}\)
d) Ta có: \(\left(x^2-2\right)^3\)
\(=\left(x^2\right)^3-3\cdot\left(x^2\right)^2\cdot2+3\cdot x^2\cdot2^2-2^3\)
\(=x^6-6x^4+12x^2-8\)
e) Ta có: \(\left(2x-3y\right)^3\)
\(=\left(2x\right)^3-2\cdot\left(2x\right)^2\cdot3y+2\cdot2x\cdot\left(3y\right)^2-\left(3y\right)^3\)
\(=8x^3-24x^2y+36xy^2-27y^3\)
f) Ta có: \(\left(\frac{1}{2}x-y^2\right)^3\)
\(=\left(\frac{1}{2}x\right)^3-3\cdot\left(\frac{1}{2}x\right)^2\cdot y^2+3\cdot\frac{1}{2}x\cdot\left(y^2\right)^2-\left(y^2\right)^3\)
\(=\frac{1}{8}x^3-\frac{3}{4}x^2y^2+\frac{3}{2}xy^4-y^6\)
\(a,\left(3x+5\right)^2=9x^2+30x+25\)
\(b,\left(2x-1\right)^3=8x^3-12x^2+6x-1\)
\(c,\left(3y+2x\right)\left(2x-3y\right)=4x^2-9y^2\)
\(27a^3-b^3+9ab^2-27a^2b\)
\(=\left(3a\right)^3-3\cdot\left(3a\right)^2b+3\cdot3a\cdot b^2-b^3\)
\(=\left(3a-b\right)^3\)
a) \(\left(x+\dfrac{1}{2}\right)^2-2x^2\)
\(=x^2+2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-2x^2\)
\(=x^2+x+\dfrac{1}{4}-2x^2\)
\(=-x^2+x+\dfrac{1}{4}\)
b) \(\left(x-2y\right)^2-4y^2\)
\(=x^2-2\cdot x\cdot2y+\left(2y\right)^2-4y^2\)
\(=x^2-4xy+4y^2-4y^2\)
\(=x^2-4xy\)
c) \(\left(x+\dfrac{1}{2}y\right)^3\)
\(=x^3+3\cdot x^2\cdot\dfrac{1}{2}y+3\cdot x+\left(\dfrac{1}{2}y\right)^2+\left(\dfrac{1}{2}y\right)^3\)
\(=x^3+\dfrac{3}{2}x^2y+\dfrac{3}{4}xy^2+\dfrac{1}{8}y^3\)
d) \(\left(2x^2-3y\right)^3\)
\(=\left(2x^2\right)^3-3\cdot\left(2x^2\right)^2\cdot3y+3\cdot2x^2\cdot\left(3y\right)^2-\left(3y\right)^3\)
\(=8x^6-36x^4y+54x^2y^2-27y^3\)
e) \(\left(x^2+y\right)^2-\left(x+y\right)^2\)
\(=\left[\left(x^2\right)^2+2\cdot x^2\cdot y+y^2\right]-\left(x^2+2\cdot x\cdot y+y^2\right)\)
\(=\left(x^4+2x^2y+y^2\right)-\left(x^2+2xy+y^2\right)\)
\(=x^4+2x^2y+y^2-x^2-2xy-y^2\)
\(=x^4+2x^2y-x^2-2xy\)
a) Ta có: \(\left(x+1\right)^3\)
\(=x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3\)
\(=x^3+3x^2+3x+1\)
b) Ta có: \(\left(2x+3\right)^3\)
\(=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot3+3\cdot2x\cdot3^2+3^3\)
\(=8x^3+3\cdot4x^2\cdot3+27\cdot2x+27\)
\(=8x^3+36x^2+54x+27\)
c) Ta có: \(\left(x+\frac{1}{2}\right)^3\)
\(=x^3+2\cdot x^2\cdot\frac{1}{2}+2\cdot x\cdot\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3\)
\(=x^3+x^2+\frac{1}{2}x+\frac{1}{8}\)
d) Ta có: \(\left(x^2+2\right)^3\)
\(=\left(x^2\right)^3+3\cdot\left(x^2\right)^2\cdot2+3\cdot x^2\cdot2^2+2^3\)
\(=x^6+6x^4+12x^2+8\)
e) Ta có: \(\left(2x+3y\right)^3\)
\(=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot3y+3\cdot2x\cdot\left(3y\right)^2+\left(3y\right)^3\)
\(=8x^3+36x^2y+54xy^2+27y^3\)
f) Ta có: \(\left(\frac{1}{2}x+y^2\right)^3\)
\(=\left(\frac{1}{2}x\right)^3+3\cdot\left(\frac{1}{2}x\right)^2\cdot y^2+3\cdot\frac{1}{2}x\cdot\left(y^2\right)^2+\left(y^2\right)^3\)
\(=\frac{1}{8}x^3+\frac{3}{4}x^2y^2+\frac{3}{2}xy^4+y^6\)
a) \(\left(2x^2-1\right)^2\)
\(=4x^4-4x^2+1\)
b)\(\left(\dfrac{1}{2}x+3y^2\right)^2\)
\(=\dfrac{1}{4}x^2+3xy^2+9y^4\)
a: \(\left(2x^2+3y\right)^3\)
\(=8x^6+3\cdot4x^4\cdot3y+3\cdot2x^2\cdot9y^2+27y^3\)
\(=8x^6+36x^4y+54x^2y^2+27y^3\)
b: \(\left(2a^2b+\dfrac{1}{3}ab^2\right)^2\)
\(=4a^4b^2+2\cdot2a^2b\cdot\dfrac{1}{3}ab^2+\dfrac{1}{9}a^2b^4\)
\(=4a^4b^2+\dfrac{4}{3}a^3b^3+\dfrac{1}{9}a^2b^4\)