Giúp mìnhh với nha
a, (x+3)^3
b, (1/2-x)^3
c, (2x-y^2)^3
d, (2x-y^2/x)^3
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\(\)a: \(\left(x-2y\right)^3\)
\(=x^3-3\cdot x^2\cdot2y+3\cdot x\cdot\left(2y\right)^2-\left(2y\right)^3\)
\(=x^3-6x^2y+12xy^2-8y^3\)
b: \(\left(2x+y\right)^3=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot y+3\cdot2x\cdot y^2+y^3\)
\(=8x^3+12x^2y+6xy^2+y^3\)
c: \(\left(\dfrac{1}{3}x-1\right)^3=\left(\dfrac{1}{3}x\right)^3-3\cdot\left(\dfrac{1}{3}x\right)^2\cdot1+3\cdot\dfrac{1}{3}x\cdot1^2-1^3\)
\(=\dfrac{1}{27}x^3-\dfrac{1}{3}x^2+x-1\)
d: \(\left(x+\dfrac{1}{3}y\right)^3\)
\(=x^3+3\cdot x^2\cdot\dfrac{1}{3}y+3\cdot x\cdot\left(\dfrac{1}{3}y\right)^2+\left(\dfrac{1}{3}y\right)^3\)
\(=x^3+x^2y+\dfrac{1}{3}xy^2+\dfrac{1}{27}y^3\)
e: (2x-3y)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: \(\left(x^2-2y\right)^3\)
\(=\left(x^2\right)^3-3\cdot\left(x^2\right)^2\cdot2y+3\cdot x^2\cdot\left(2y\right)^2-\left(2y\right)^3\)
\(=x^6-6x^4y+12x^2y^2-8y^3\)
g: \(\left(\dfrac{1}{2}x-y\right)^3=\left(\dfrac{1}{2}x\right)^3-3\cdot\left(\dfrac{1}{2}x\right)^2\cdot y+3\cdot\dfrac{1}{2}x\cdot y^2-y^3\)
\(=\dfrac{1}{8}x^3-\dfrac{3}{4}x^2y+\dfrac{3}{2}xy^2-y^3\)
\(a,=27x^3+27x^2+9x+1\)
\(b,=\dfrac{x^3}{27}-\dfrac{x^2}{3}+x-1\)
\(c,=-\left(27x^3-27x^2y^2+9xy^4-y^6\right)\)
\(=-27x^3+27x^2y^2-9xy^4+y^6\)
\(d,=\dfrac{x^3}{y^3}-\dfrac{6x}{y}+\dfrac{12y}{x}-\dfrac{8y^3}{x^3}\)
a) \(\left(3x+1\right)^3=27x^3+27x^2+9x+1\)
b) \(\left(\dfrac{x}{3}-1\right)^3=\dfrac{x^3}{27}-\dfrac{x^2}{3}\)
c) \(\left(-y^2+3x\right)^3=27x^3-27x^2y^2+9xy^4-y^6\)
d) \(\left(\dfrac{x}{y}-\dfrac{2y}{x}\right)^3=\dfrac{x^3}{y^3}-\dfrac{6x}{y}+\dfrac{12y}{x}-\dfrac{8y^3}{x^3}\)
a: =-3x^2y*x^2y+3x^2y*2xy
=-3x^4y^2+6x^3y^2
b: =x^3-x^2y+x^2y+y^2=x^3+y^2
c: =x*4x^3-x*5xy+2x*x
=4x^4-5x^2y+2x^2
d: =x^3+x^2y+2x^3+2xy
=3x^3+x^2y+2xy
Bài này đề bài phải là khai triển biểu thức, chứ không phải là tính em nhé.
Lời giải:
Ta áp dụng hằng đẳng thức đáng nhớ thôi.
a. $(3+2x)^3=3^3+3.3^2.2x+3.3.(2x)^2+(2x)^3$
$=8x^3+36x^2+54x+27$
b.
$(\frac{1}{2}-y)^3=(\frac{1}{2})^3-3.(\frac{1}{2})^2.y+3.\frac{1}{2}y^2-y^3$
$=-y^3+\frac{3}{2}y^2-\frac{3}{4}y+\frac{1}{8}$
c.
$(x-5)(x^2+5x+25)=(x-5)^2(x^2+5x+5^2)$
$=x^3-5^3=x^3-125$
d.
$(3x+\frac{1}{2})(9x^2-\frac{3}{2}x+\frac{1}{4})$
$=(3x+\frac{1}{2})[(3x)^2-3x.\frac{1}{2}+(\frac{1}{2})^2]$
$=(3x)^3+(\frac{1}{2})^3=27x^3+\frac{1}{8}$
a) \(\left(2x+y\right)^2-\left(2x+y\right)\left(2x-y\right)+y\left(x-y\right)\)
\(=\left(2x+y\right)^2-\left(2x\right)^2+y^2+xy-y^2\)
\(=\left(2x+y+2x\right)\left(2x+y-2x\right)+xy\)
\(=\left(4x+y\right)y+xy\)
\(=\left[4\left(-2\right)+3\right].3+\left(-2\right).3\)
\(=\left(-8+3\right).3+1\)
\(=-15+1\)
\(=-14\)
\(a,\left(x+3\right)^3=x^3+9x^2+27x+27\\ b,\left(\dfrac{1}{2}-x\right)^3=\dfrac{1}{8}-\dfrac{3}{4}x+\dfrac{3}{2}x^2-x^3\\ c,\left(2x-y^2\right)^3=8x^3-12x^2y^2+6xy^4-y^6\\ d,\left(2x-\dfrac{y^2}{x}\right)^3=8x^3-\dfrac{12x^2y^2}{x}+\dfrac{6xy^4}{x^2}-\dfrac{y^6}{x^2}\\ =8x^3-12xy^2+6y^4-\dfrac{y^6}{x^2}\)
Cảm ơn bạn nhé!:333