\(\frac{3x^4-8x^3-10x^2+8x-5}{3x^2-2x+1}\)

\(=\frac{x^2\left(3x^2-2x+1\right)-2x\left(3x^2-2x+1\right)-5\left(3x^2-2x+1\right)}{3x^2-2x+1}\)

\(=\frac{\left(3x^2-2x+1\right)\cdot\left(x^2-2x-5\right)}{3x^2-2x+1}\)

\(=x^2-2x-5\)

\(\frac{2x^3-9x^2+19x-15}{x^2-3x+5}\)

\(=\frac{2x\left(x^2-3x+5\right)-3\left(x^2-3x+5\right)}{x^2-3x+5}\)

\(=\frac{\left(x^2-3x+5\right)\left(2x-3\right)}{x^2-3x+5}\)

\(=2x-3\)

a: \(\left(4x^5-8x^3\right):\left(-2x^3\right)\)

\(=-4x^5:2x^3+8x^3:2x^3\)

\(=-2x^2+4\)

b: \(\left(9x^3-12x^2+3x\right):\left(-3x\right)\)

\(=-\frac{9x^3}{3x}+\frac{12x^2}{3x}-\frac{3x}{3x}\)

\(=-3x^2+4x-1\)

c: \(\left(xy^2+4x^2y^3-3x^2y^4\right):\left(-\frac12x^2y^3\right)\)

\(=-xy^2:\frac12x^2y^3-4x^2y^3:\frac12x^2y^3+3x^2y^4:\frac12x^2y^3\)

\(=-\frac{2}{xy}-8+6y\)

d: \(\left\lbrack2\left(x-y\right)^3-7\left(y-x\right)^2-\left(y-x\right)\right\rbrack:\left(x-y\right)\)

\(=\left\lbrack2\left(x-y\right)^3-7\left(x-y\right)^2+\left(x-y\right)\right\rbrack:\left(x-y\right)\)

\(=\frac{2\left(x-y\right)^3}{x-y}-\frac{7\left(x-y\right)^2}{x-y}+\frac{\left(x-y\right)}{x-y}\)

\(=2\left(x-y\right)^2-7\left(x-y\right)+1\)

$\dfrac{5x+10}{4x-8}\cdot\dfrac{4-2x}{x+2}$

$=\dfrac{5(x+2)}{4(x-2)}\cdot\dfrac{-2(x-2)}{x+2}$

$=-\dfrac{10}{4}$

$=-\dfrac{5}{2}$

$\dfrac{6x^2y^3}{8x^3y^2}$

$=\dfrac{3y}{4x}$

$\dfrac{x^3-x}{3x+3}$

$=\dfrac{x(x^2-1)}{3(x+1)}$

$=\dfrac{x(x-1)(x+1)}{3(x+1)}$

$=\dfrac{x(x-1)}{3}$

$\dfrac{x^2+3xy}{x^2-9y^2}$

$=\dfrac{x(x+3y)}{(x-3y)(x+3y)}$

$=\dfrac{x}{x-3y}$

$\dfrac{x^2+4x+4}{3x+6}$

$=\dfrac{(x+2)^2}{3(x+2)}$

$=\dfrac{x+2}{3}$

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 $x=-2$.

$16x^2-(3x-4)^2=0$

$(4x)^2-(3x-4)^2=0$

$(4x-3x+4)(4x+3x-4)=0$

$(x+4)(7x-4)=0$

$x=-4$ hoặc $x=\dfrac47$

Vậy $x=-4,\dfrac47$.

$x^3-x^2+3-3x=0$

$=x^2(x-1)-3(x-1)=0$

$=(x-1)(x^2-3)=0$

$x-1=0$ hoặc $x^2-3=0$

$x=1$ hoặc $x=\pm\sqrt3$

Vậy $x=1,\sqrt3,-\sqrt3$.

$\dfrac{x-1}{x+2}=\dfrac{x+2}{x+1}$

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 $x=-\dfrac54$.

$\dfrac1{x+2}+\dfrac2{x+1}=0$

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 $x=-\dfrac53$.

$\dfrac{9-x^2}{x}:(x-3)=1$

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 $x=-\dfrac32$.

$2xy(3xy+2xy^2)$

$=2xy\cdot3xy+2xy\cdot2xy^2$

$=6x^2y^2+4x^2y^3$

$(2x-1)(x^2+2x+4)-(x^2-3x)\cdot2x$

$=2x^3+4x^2+8x-x^2-2x-2x^3+6x^2$

$=9x^2+6x$

$4x^3y-8x^2y^2+4xy^3$

$=4xy(x^2-2xy+y^2)$

$=4xy(x-y)^2$

$2xy+3xz+6y^2+xz$

$=2xy+4xz+6y^2$

$=2(xy+2xz+3y^2)$

Biểu thức trong ngoặc không phân tích tiếp được bằng các phương pháp thông thường.

$y^2-4x-4xy+4x^2+2y$

$=y^2+2y+1+4x^2-4xy-4x-1$

$=(y+1)^2+(2x-y)^2-4x-1$

Biểu thức này không có dạng nhân tử đơn giản với hệ số nguyên. Có khả năng đề câu c bị chép sai.

a , x(3-x)-y(y-2x)+y(y-2)-x(2y-x)

=3x-x²-y²+2xy+y²-2y-2xy+x²

=3x-(-x²+x²)+(2xy-2xy)-2y

=3x-2y

b,(x²-7)(x+2)-(2x-1)(x-14)+x(x²-2x-22)+35

=x³+2x²-7x-14-2x²+29x-14+x³-2x²-22x+35

=(x³+x³)+(2x²-2x²-2x²)-(-7x+29x-22x)-14-14+35

=2x³-2x²+7

c,(2x+y)(4x²-2xy+y²)-8x³-y³

=2xy²-4x²y+8x³+y³-2xy²+4x²y-8x³-y³

=(2xy²-2xy²)-(-4x²y+4x²y)+(8x³-8x³)+(y³-y³)

=0