\(5x^2+xy+5y^2\)
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\(=\left(5x-5y\right)+\left(x^{^2}y+xy^{^2}\right)=5\left(x-y\right)-xy\left(x-y\right)=\left(5-xy\right)\left(x-y\right)\)
xy+x+y=2
xy+x+y+1=2+1
(xy+x)+(y+1)=3
x(y+1)+(y+1)=3
(x+1)(y+1)=3=1.3=3.1=-1.-3=-3.-1
\(\Rightarrow\left[{}\begin{matrix}x+1=1;y+1=3\\x+1=3;y+1=1\\x+1=-1;y+1=-3\\x+1=-3;y+1=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0;y=2\\x=2;y=0\\x=-2;y=-4\\x=-4;y=-2\end{matrix}\right.\)
Vậy:.................
xy+14+2y+7x= -10
\(\Leftrightarrow\)y(x+2)+7(x+2)=-10
\(\Leftrightarrow\)(y+7)(x+2)=-10=1.(-10)=2.(-5)=5.(-2)=10.(-1)
y+7 | 1 | 2 | 5 | 10 |
x+2 | -10 | -5 | -2 | -1 |
y | -6 | -5 | -2 | 3 |
x | -12 | -7 | -4 | -3 |
=\(^{\dfrac{-x^2-xy}{5\left(x^2-y^2\right)}}\).\(\dfrac{3\left(x^3-y^3\right)}{x^2-xy}\)
=\(\dfrac{-3\left(x-y\right)}{5}\)
\(\dfrac{5x+y^2}{x^2y}-\dfrac{5y+x^2}{xy^2}\)
\(=\dfrac{y\left(5x+y^2\right)}{x^2y^2}-\dfrac{x\left(5y-x^2\right)}{x^2y^2}\)
\(=\dfrac{5xy+y^3}{x^2y^2}-\dfrac{5xy+x^3}{x^2y^2}\)
\(=\dfrac{5xy+y^3-5xy+x^3}{x^2y^2}\)
\(=\dfrac{\left(5xy-5xy\right)+x^3+y^3}{x^2y^2}\)
\(=\dfrac{x^3+y^3}{x^2y^2}\)
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.
$C=(6x^2y^2-6xy^3)-(8x^3-8x^2y^2)-(5x^2y^2-5xy^3)$
$=-8x^3+9x^2y^2-xy^3$
$=(-2x)^3+(3xy)^2-xy^3$
$=(-2.\frac{1}{2})^3+(3.\frac{1}{2}.2)^2-\frac{1}{2}.2^3$
$=(-1)^3+3^2-4=4$
\(5x\left(4x^2+2x+1\right)-2x\left(10x^2-5x-2\right)\)
\(=20x^3+10x^2+5x-20x^3+10x^2+4x\)
\(=20x^2+9x\)
thay x = 15 ta được
\(20.15^2+9.15=4635\)
câu b tương tự
5x - xy + y2 - 5y
= ( 5x - 5y ) - ( xy - y2 )
= 5( x - y ) - y( x - y )
= ( 5 - y )( x - y )
Ta có M= \(x.y.x-x.y.y-5x+5y-12\)
=> \(M=5x-5y-5x+5y-12\)=-12
Vậy M=-12