\(a,=\dfrac{1}{2}\left[\left(x^2+y^2\right)^2-4x^2y^2\right]\\ =\dfrac{1}{2}\left(x^2-2xy+y^2\right)\left(x^2+2xy+y^2\right)\\ =\dfrac{1}{2}\left(x-y\right)^2\left(x+y\right)^2\\ b,=\left(3x-\dfrac{1}{2}y\right)\left(9x^2+\dfrac{3}{2}xy+\dfrac{1}{4}y^2\right)\\ c,=\dfrac{1}{2}\left(x^2+\dfrac{1}{2}x+\dfrac{1}{16}\right)=\dfrac{1}{2}\left(x+\dfrac{1}{4}\right)^2\)

a, \(\dfrac{x^2}{4}-xy+y^2=\left(\dfrac{x}{2}\right)^2-xy+y^2=\left(\dfrac{x}{2}\right)^2-2.\dfrac{x}{2}.y+y^2\)

\(=\left(\dfrac{x^2}{2}-y\right)^2\)

b, \(x^2+x+\dfrac{1}{4}=x^2+\dfrac{1}{2}.2.x+\left(\dfrac{1}{2}\right)^2=\left(x+\dfrac{1}{2}\right)^2\)

c, \(x^2+2\sqrt{3}x+3=x^2+2\sqrt{3}x+\left(\sqrt{3}\right)^2=\left(x+\sqrt{3}\right)^2\)

d, \(4x^2-1=\left(2x-1\right)\left(2x+1\right)\)

`x^2/4-2*x/2*y+y^2`

`=(x/2-y)^2`

`x^2+x+1/4`

`=x^2+2*x*1/2+(1/2)^2`

`=(x+1/2)^2`

`x^2+2sqrt3x+3`

`=x+2xsqrt3+sqrt3^2`

`=(x+sqrt3)^2`

`4x^2-1`

`=(2x)^2-1`

`=(2x-1)(2x+1)`

a: \(\frac12x^2-2y^2=\frac12\left(x^2-4y^2\right)\)

\(=\frac12\left(x-2y\right)\left(x+2y\right)\)

b: \(\frac13xy+x^2z+xz\)

\(=x\cdot\frac13y+x\cdot xz+x\cdot z\)

\(=x\left(\frac13y+xz+z\right)\)

c: \(18x^3-\frac{8}{25}x=2x\left(9x^2-\frac{4}{25}\right)\)

\(=2x\left\lbrack\left(3x\right)^2-\left(\frac25\right)^2\right\rbrack\)

\(=2x\left(3x-\frac25\right)\left(3x+\frac25\right)\)

d: \(\frac25x^2+5x^3+x^2y=x^2\cdot\frac25+x^2\cdot5x+x^2\cdot y=x^2\left(\frac25+5x+y\right)\)

e: \(\frac12\left(x^2+y^2\right)^2-2x^2y^2\)

\(=\frac12\left\lbrack\left(x^2+y^2\right)^2-4x^2y^2\right\rbrack\)

\(=\frac12\left(x^2+y^2-2xy\right)\left(x^2+y^2+2xy\right)\)

\(=\frac12\left(x-y\right)^2\cdot\left(x+y\right)^2\)

f: \(27x^3-\frac18y^3=\left(3x\right)^3-\left(\frac12y\right)^3\)

\(=\left(3x-\frac12y\right)\left(9x^2+\frac32xy+\frac14y^2\right)\)

\(a,=2\left(\dfrac{1}{4}x^2-y^2\right)=2\left(\dfrac{1}{2}x-y\right)\left(\dfrac{1}{2}x+y\right)\\ b,=\dfrac{1}{3}x\left(y+3xz+3z\right)\\ c,=2x\left(9x^2-\dfrac{4}{25}\right)=2x\left(3x-\dfrac{2}{5}\right)\left(3x+\dfrac{2}{5}\right)\)

\(d,=x^2\left(\dfrac{2}{5}+5x+y\right)\\ e,=\dfrac{1}{2}\left[\left(x^2+y^2\right)^2-4x^2y^2\right]\\ =\dfrac{1}{2}\left(x^2-2xy+y^2\right)\left(x^2+2xy+y^2\right)\\ =\dfrac{1}{2}\left(x-y\right)^2\left(x+y\right)^2\\ f,=\left(3x-\dfrac{1}{2}y\right)\left(9x^2+\dfrac{3}{2}xy+\dfrac{1}{4}y^2\right)\\ g,=\dfrac{1}{2}\left(x^2+\dfrac{1}{2}x+\dfrac{1}{16}\right)=\dfrac{1}{2}\left(x+\dfrac{1}{4}\right)^2\)

đáp án: a là đúng

A

k làm đc k cần phải ghi zậy mô ha

1.

\(y^2+y\left(x^3+x^2+x\right)+x^5-x^4+2x^3-2x^2\)

\(\Delta=\left(x^3+x^2+x\right)^2-4\left(x^5-x^4+2x^3-2x^2\right)\)

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

\(\Rightarrow\left[{}\begin{matrix}y=\dfrac{-x^3-x^2-x+x^3-x^2+3x}{2}=-x^2+x\\y=\dfrac{-x^3-x^2-x-x^3+x^2-3x}{2}=-x^3-2x\end{matrix}\right.\)

Hay đa thức trên có thể phân tích thành:

\(\left(x^2-x+y\right)\left(x^3+2x+y\right)\)

Dựa vào đó em tự tách cho phù hợp

Sửa đề: \(\frac13x^3+\frac72x^2+2x+1\)

\(\frac13x^3+\frac72x^2+2x+1\)

\(=\frac16\cdot21x^3+\frac16\cdot21x^2+\frac16\cdot12x+\frac16\cdot6\)

\(=\frac16\left(2x^3+21x^2+12x+6\right)\)

Biểu thức này không phân tích thành nhân tử.