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a: \(\left(2x^2y-3xy+4xy^2\right):2xy\)
\(=\frac{2x^2y}{2xy}-\frac{3xy}{2xy}+\frac{4xy^2}{2xy}\)
\(=x-\frac32+2y\)
b: \(\frac{1}{xy}-\frac{x^2-1}{y^2-xy}\)
\(=\frac{1}{xy}-\frac{x^2-1}{y\left(y-x\right)}\)
\(=\frac{y-x}{xy\left(y-x\right)}-\frac{x\left(x^2-1\right)}{xy\left(y-x\right)}=\frac{y-x-x^3+x}{xy\left(y-x\right)}=\frac{-x^3+y}{xy\left(y-x\right)}\)
c: \(\left\lbrack\frac{x}{xy-y^2}-\frac{2x-y}{x^2-xy}\right\rbrack:\left(\frac{1}{x}-\frac{1}{y}\right)\)
\(=\left\lbrack\frac{x}{y\left(x-y\right)}-\frac{2x-y}{x\left(x-y\right)}\right\rbrack:\frac{y-x}{xy}\)
\(=\frac{x^2-y\left(2x-y\right)}{xy\cdot\left(x-y\right)}\cdot\frac{xy}{-\left(x-y\right)}=\frac{x^2-2xy+y^2}{-\left(x-y\right)^2}=\frac{\left(x-y\right)^2}{-\left(x-y\right)^2}\)
=-1
Bài 3:
3: \(6x\left(x-y\right)-9y^2+9xy\)
\(=6x\left(x-y\right)+9xy-9y^2\)
\(=6x\left(x-y\right)+9y\left(x-y\right)\)
\(=\left(x-y\right)\left(6x+9y\right)\)
\(=3\left(2x+3y\right)\left(x-y\right)\)
Bài 4:
$(x-1)(x^2+x+1)-x^3-6x=11$
Dùng $(x-1)(x^2+x+1)=x^3-1$:
$x^3-1-x^3-6x=11$
$-6x-1=11$
$-6x=12$
$x=-2$
Vậy $x=-2$.
2.$16x^2-(3x-4)^2=0$
$(4x)^2-(3x-4)^2=0$
$(4x-3x+4)(4x+3x-4)=0$
$(x+4)(7x-4)=0$
$x=-4$ hoặc $x=\dfrac47$
Vậy $x=-4,\dfrac47$.
3.$x^3-x^2+3-3x=0$
$=x^2(x-1)-3(x-1)=0$
$=(x-1)(x^2-3)=0$
$x-1=0$ hoặc $x^2-3=0$
$x=1$ hoặc $x=\pm\sqrt3$
Vậy $x=1,\sqrt3,-\sqrt3$.
4.$\dfrac{x-1}{x+2}=\dfrac{x+2}{x+1}$
Điều kiện: $x\ne-2,-1$.
$(x-1)(x+1)=(x+2)^2$
$x^2-1=x^2+4x+4$
$-4x=5$
$x=-\dfrac54$
Vậy $x=-\dfrac54$.
5.$\dfrac1{x+2}+\dfrac2{x+1}=0$
Điều kiện: $x\ne-2,-1$.
$\dfrac{x+1+2(x+2)}{(x+2)(x+1)}=0$
$x+1+2x+4=0$
$3x+5=0$
$x=-\dfrac53$
Vậy $x=-\dfrac53$.
6.$\dfrac{9-x^2}{x}:(x-3)=1$
Điều kiện: $x\ne0,3$.
$\dfrac{9-x^2}{x(x-3)}=1$
$9-x^2=x(x-3)$
$9-x^2=x^2-3x$
$2x^2-3x-9=0$
$(2x+3)(x-3)=0$
$x=-\dfrac32$ hoặc $x=3$
Nhưng $x=3$ không thỏa điều kiện.
Vậy $x=-\dfrac32$.
\(2x\left(x^2-7x-3\right)=2x^3-14x-6x\)
\(4xy^2\left(-2x^3+y^2-7xy\right)=-8x^4y^2+4xy^5-28x^2y^3\)
Bài 1:
\(3a.\left(2a^2-ab\right)=6a^3-3a^2b\)
\(\left(4-7b^2\right).\left(2a+5b\right)=8a+20b-14ab^2-35b^3\)
Bài 2:
\(2x^2-6x+xy-3y=2x.\left(x-3\right)+y.\left(x-3\right)=\left(x-3\right).\left(2x+y\right)\)
Bài 3: Tại x = 3/2, y =1/3 thì Q = 67/9
Bài 4:
\(\left(\frac{1}{x+1}+\frac{2x}{1-x^2}\right).\left(\frac{1}{x-1}\right)\) \(\frac{1}{\left(x+1\right).\left(x-1\right)}+\frac{2x}{\left(1-x^2\right).\left(x-1\right)}=\frac{x-1}{\left(x+1\right).\left(x-1\right)^2}+\frac{-2x}{\left(x-1\right)^2.\left(x+1\right)}\)
= \(\frac{x-1-2x}{\left(x+1\right).\left(x-1\right)^2}=\frac{-\left(x+1\right)}{\left(x+1\right).\left(x-1\right)^2}=\frac{-1}{\left(x-1\right)^2}\)
Ta có: \(\left(x+y+1\right)\left(x+y-1\right)-\left(x-y\right)^2-4xy\)
\(=\left(x+y\right)^2-\left(x-y\right)^2-1-4xy\)
\(=x^2+2xy+y^2-x^2+2xy-y^2-1-4xy\)
=-1
`(x+y+1)(x+y-1)-(x-y)^{2}-4xy`
`=(x+y)^{2}-1-(x-y)^{2}-4xy`
`=(x+y+x-y)(x+y-x+y)-1-4xy`
`=2x.2y-4xy-1`
`=4xy-4xy-1`
`=-1`