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b: x-y=2
=>x=y+2
\(A=y^2+4y+4+y^2-2y+4+4y=2y^2+6y+8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`x^2-2x-4y^2+4y`
`=(x^2-4y^2)-2x+4y`
`=(x-2y)(x+2y)-2(x-2y)`
`=(x-2y)(x+2y-2)`
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a.
$A=20x^3-10x^2+5x-(20x^3-10x^2-4x)$
$=9x=9.15=135$
b.
$B=(5x^2-20xy)-(4y^2-20xy)=5x^2-4y^2$
$=5(\frac{-1}{5})^2-4(\frac{-1}{2})^2=\frac{-4}{5}$
c.
$C=(6x^2y^2-6xy^3)-(8x^3-8x^2y^2)-(5x^2y^2-5xy^3)$
$=-8x^3+9x^2y^2-xy^3$
$=(-2x)^3+(3xy)^2-xy^3$
$=(-2.\frac{1}{2})^3+(3.\frac{1}{2}.2)^2-\frac{1}{2}.2^3$
$=(-1)^3+3^2-4=4$
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\(x^2+4y^2-2x-4xy+4y+2018=\left[x^2-2x\left(1+2y\right)+\left(1+2y\right)^2\right]+2017=\left(x-1-2y\right)^2+2017\ge2017>0\)
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Lời giải:
$2x^2-2xy-4y^2=2(x^2-xy-2y^2)$
$=2[(x^2-2xy)+(xy-2y^2)]$
$=2[x(x-2y)+y(x-2y)]$
$=2(x+y)(x-2y)$
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$x^2-2x-4y^2-4y=(x^2-2x+1)-(4y^2+4y+1)$
$=(x-1)^2-(2y+1)^2=(x-1-2y-1)(x-1+2y+1)$
$=(x-2y-2)(x+2y)$
-------------------
$x^2-4y^2-x-2y=(x^2-4y^2)-(x+2y)=(x-2y)(x+2y)-(x+2y)$
$=(x+2y)(x-2y-1)$
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ \(A=20x^3-10x^2+5x-20x^3+10x^2+4x=9x\)
Thay x = 15 vào bt A ta có
A = 9 . 15 = 135
b/ \(B=5x^2-20xy-4y^2+2xy=5x^2-4y^2\)
Thay x = -1/5 ; y = - 1/2 vào bt B ta có
\(B=5.\dfrac{1}{25}-4.\dfrac{1}{4}=\dfrac{1}{5}-1=-\dfrac{4}{5}\)
c/ \(C=6x^2y^2-6xy^3-8x^3+8x^2y^2-5x^2y^2+5xy^3\)
\(=9x^2y^2-xy^3-8x^3\)
Thay x = 1/2 ; y = 2 vào bt C ta có
\(C=9.4.\dfrac{1}{4}-\dfrac{1}{2}.8-8.\dfrac{1}{8}=9-4-1=4\)
d/ \(D=6x^2+10x-3x-5+6x^2-3x+8x-2\)
\(=12x^2+12x-3\)
\(\left|x\right|=2\Rightarrow x=\pm2\)
Thay x = 2 vào bt D có
\(D=12.4+12.2-3=69\)
Thay x = - 2 vào bt D ta có
\(D=12.4-12.2-3=21\)