phân tích đa thức \(\dfrac{1}{3}x^5+\dfrac{7}{2}x^2+2x+1\) thành nhân tử
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2 tháng 1 2022
\(\dfrac{1}{x-y}-\dfrac{1}{x+y}+\dfrac{2x}{\left(x-y\right)\left(x+y\right)}\\ \dfrac{x+y}{\left(x-y\right)\left(x+y\right)}-\dfrac{x-y}{\left(x-y\right)\left(x+y\right)}+\dfrac{2x}{\left(x-y\right)\left(x+y\right)}\\ \dfrac{x+y-x+y+2x}{\left(x-y\right)\left(x+y\right)}\\ \dfrac{2x+2y}{\left(x-y\right)\left(x+y\right)}\\ \dfrac{2}{x-y}\)
2 tháng 1 2022
\(\dfrac{1}{x-y}+\dfrac{1}{x+y}+\dfrac{2x}{x^2-y^2}=\dfrac{x+y}{\left(x-y\right)\left(x+y\right)}+\dfrac{x-y}{\left(x-y\right)\left(x+y\right)}+\dfrac{2x}{\left(x-y\right)\left(x+y\right)}=\dfrac{4x}{\left(x-y\right)\left(x+y\right)}\)
18 tháng 7 2021
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, \(4x^2-1=\left(2x-1\right)\left(2x+1\right)\)
18 tháng 7 2021
`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)`
20 tháng 11 2021
\(a,=\dfrac{1}{2}\left[\left(x^2+y^2\right)^2-4x^2y^2\right]\\ =\dfrac{1}{2}\left(x^2-2xy+y^2\right)\left(x^2+2xy+y^2\right)\\ =\dfrac{1}{2}\left(x-y\right)^2\left(x+y\right)^2\\ b,=\left(3x-\dfrac{1}{2}y\right)\left(9x^2+\dfrac{3}{2}xy+\dfrac{1}{4}y^2\right)\\ c,=\dfrac{1}{2}\left(x^2+\dfrac{1}{2}x+\dfrac{1}{16}\right)=\dfrac{1}{2}\left(x+\dfrac{1}{4}\right)^2\)
19 tháng 11 2021
\(a,=2\left(\dfrac{1}{4}x^2-y^2\right)=2\left(\dfrac{1}{2}x-y\right)\left(\dfrac{1}{2}x+y\right)\\ b,=\dfrac{1}{3}x\left(y+3xz+3z\right)\\ c,=2x\left(9x^2-\dfrac{4}{25}\right)=2x\left(3x-\dfrac{2}{5}\right)\left(3x+\dfrac{2}{5}\right)\)
\(d,=x^2\left(\dfrac{2}{5}+5x+y\right)\\ e,=\dfrac{1}{2}\left[\left(x^2+y^2\right)^2-4x^2y^2\right]\\ =\dfrac{1}{2}\left(x^2-2xy+y^2\right)\left(x^2+2xy+y^2\right)\\ =\dfrac{1}{2}\left(x-y\right)^2\left(x+y\right)^2\\ f,=\left(3x-\dfrac{1}{2}y\right)\left(9x^2+\dfrac{3}{2}xy+\dfrac{1}{4}y^2\right)\\ g,=\dfrac{1}{2}\left(x^2+\dfrac{1}{2}x+\dfrac{1}{16}\right)=\dfrac{1}{2}\left(x+\dfrac{1}{4}\right)^2\)
20 tháng 11 2021
câu b sai r
\(\dfrac{1}{3}xy+x^2z+xz=3x\left(\dfrac{1}{9}y+\dfrac{1}{3}xz+\dfrac{1}{3}z\right)\)
AH
Akai Haruma
Giáo viên
20 tháng 11 2021
Lời giải:
a.
$=\frac{1}{2}(x^2-4y^2)=\frac{1}{2}[x^2-(2y)^2]=\frac{1}{2}(x-2y)(x+2y)$
b.
$=\frac{1}{3}x(y+3xz+3z)$
c.
$=\frac{2}{25}x(225x^2-4)=\frac{2}{25}(15x-2)(15x+2)$
d.
$=\frac{1}{5}x^2(2+25x+5y)$