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`a,(25xy^3(2x-y)^2)/(75xy^2(y-2x))(x,y ne 0)(y ne 2x)`
`=(25xy^3(y-2x)^2)/(75xy^2(y-2x))`
`=(y(y-2x))/3`
`b,(x^2-y^2)/(x^2-y^2+xz-yz)`
`=((x-y)(x+y))/((x-y)(x+y)+z(x-y))`
`=(x+y)/(x+y+z)`
`c,((2x+3)-x^2)/(x^2-1)(x ne +-1)`
`=(-(x^2-3x+x-3))/((x-1)(x+1))`
`=(-x(x-3)+x-3)/((x-1)(x+1))`
`=((x-3)(1-x))/((x-1)(x+1))`
`=(3-x)/(1+x)`
`d,(3x^3-7x^2+5x-1)/(2x^3-x^2-4x+3)`
`=(3x^3-3x^2-4x^2+4x+x-1)/(2x^3-2x^2+x^2-x-3x+3)`
`=(3x^2(x-1)-4x(x-1)+x-1)/(2x^2(x-1)+x(x-1)-3(x-1))`
`=(3x^2-4x+1)/(2x^2+x-3)`
`=(3x^2-3x-x+1)/(2x^2-2x+3x-3)`
`=(3x(x-1)-(x-1))/(2x(x-1)+3(x-1))`
`=(3x-1)/(2x+3)`
a) Ta có: \(\dfrac{25xy^3\cdot\left(2x-y\right)^2}{75xy^2\cdot\left(y-2x\right)}\)
\(=\dfrac{25xy^2\cdot y\cdot\left(y-2x\right)^2}{25xy\cdot y\cdot\left(y-2x\right)\cdot3}\)
\(=\dfrac{y\left(y-2x\right)}{3}\)
a: \(P=\left(\dfrac{3x+6}{2\left(x^2+4\right)}-\dfrac{2x^2-x-10}{\left(x+1\right)\left(x^2+1\right)}\right):\left(\dfrac{10\left(x^2-1\right)+3\left(x^2+1\right)\left(x-1\right)-6\left(x+1\right)\left(x^2+1\right)}{\left(x^2+1\right)\left(x+1\right)\left(x-1\right)\cdot2}\right)\cdot\dfrac{2}{x-1}\)
\(=\left(\dfrac{\left(3x+6\right)\left(x^3+x^2+x+1\right)-\left(2x^2+8\right)\left(2x^2-x-10\right)}{2\left(x^2+4\right)\left(x+1\right)\left(x^2+1\right)}\right)\cdot\dfrac{\left(x^2+1\right)\left(x-1\right)\left(x+1\right)\cdot2}{-3x^3+x^2-3x-13}\cdot\dfrac{2}{x-1}\)
\(=\dfrac{-x^4+11x^3+13x^2+17x+16}{\left(x^2+4\right)}\cdot\dfrac{2}{-3x^3+x^2-3x-13}\)
` @ \color{Red}{m}`
` \color{lightblue}{Answer}`
\(\dfrac{x^2}{x^2-1}+\dfrac{x}{\left(1-x\right)\left(x+1\right)}\\ =\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}+\dfrac{x}{\left(1-x\right)\left(x+1\right)}\\ =\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^2-x}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x}{x+1}\)
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\(\dfrac{3}{2x+6}-\dfrac{x-3}{x^2+3x}\\ =\dfrac{3}{2\left(x+3\right)}-\dfrac{x-3}{x\left(x+3\right)}\\ =\dfrac{3x}{2x\left(x+3\right)}-\dfrac{2\left(x-3\right)}{2x\left(x+3\right)}\\ =\dfrac{3x}{2x\left(x+3\right)}-\dfrac{2x-6}{2x\left(x+3\right)}\\ =\dfrac{3x-\left(2x-6\right)}{2x\left(x+3\right)}\\ =\dfrac{3x-2x+6}{2x\left(x+3\right)}\\ =\dfrac{x+6}{2x\left(x+3\right)}\)
__
\(\dfrac{1}{1-x}+\dfrac{2x}{x^2-1}\\ =\dfrac{1}{1-x}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{1}{1-x}-\dfrac{2x}{\left(1-x\right)\left(1+x\right)}\\ =\dfrac{1+x}{\left(1-x\right)\left(1+x\right)}-\dfrac{2x}{\left(1-x\right)\left(1+x\right)}\\ =\dfrac{1+x-2x}{\left(1-x\right)\left(1+x\right)}\\ =\dfrac{1-x}{\left(1-x\right)\left(1+x\right)}\\ =\dfrac{1}{1+x}\)
\(\dfrac{x^2}{x^2-1}+\dfrac{x}{\left(1-x\right)\left(x+1\right)}\left(dkxd:x\ne\pm1\right)\)
\(=\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2-x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x}{x+1}\)
========================
\(\dfrac{3}{2x+6}-\dfrac{x-3}{x^2+3x}\left(dkxd:x\ne\pm3;x\ne0\right)\)
\(=\dfrac{3}{2\left(x+3\right)}-\dfrac{x-3}{x\left(x+3\right)}\)
\(=\dfrac{3x-2\left(x-3\right)}{2x\left(x+3\right)}\)
\(=\dfrac{3x-2x+6}{2x\left(x+3\right)}\)
\(=\dfrac{x+6}{2x^2+6x}\)
==========================
\(\dfrac{1}{1-x}+\dfrac{2x}{x^2-1}\left(dkxd:x\ne\pm1\right)\)
\(=-\dfrac{1}{x-1}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-\left(x+1\right)+2x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x-1+2x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{1}{x+1}\)
a: \(A=\left(\dfrac{2x^2+2}{x^3-1}+\dfrac{x^2-x+1}{x^4+x^2+1}-\dfrac{x^2+3}{x^3-x^2+3x-3}\right):\dfrac{1}{x-1}\)
\(=\left(\dfrac{2x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{x^2-x+1}{x^4+2x^2+1-x^2}-\dfrac{x^2+3}{x^2\left(x-1\right)+3\left(x-1\right)}\right)\cdot\dfrac{x-1}{1}\)
\(=\left(\dfrac{2x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{\left(x^2-x+1\right)}{\left(x^2+1\right)^2-x^2}-\dfrac{x^2+3}{\left(x-1\right)\left(x^2+3\right)}\right)\cdot\dfrac{x-1}{1}\)
\(=\left(\dfrac{2x^2+3}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{x^2-x+1}{\left(x^2+1+x\right)\left(x^2+1-x\right)}-\dfrac{1}{x-1}\right)\cdot\dfrac{x-1}{1}\)
\(=\left(\dfrac{2x^2+3}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{1}{x^2+x+1}-\dfrac{1}{x-1}\right)\cdot\dfrac{x-1}{1}\)
\(=\dfrac{2x^2+3+x-1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x-1}{1}\)
\(=\dfrac{x^2+1}{x^2+x+1}\)
b: Để A là số nguyên thì \(x^2+1⋮x^2+x+1\)
=>\(x^2+x+1-x⋮x^2+x+1\)
=>\(x⋮x^2+x+1\)
=>\(x^2+x⋮x^2+x+1\)
=>\(x^2+x+1-1⋮x^2+x+1\)
=>\(-1⋮x^2+x+1\)
=>\(x^2+x+1\in\left\{1;-1\right\}\)
=>\(x^2+x+1=1\)
=>x2+x=0
=>x(x+1)=0
=>\(x\in\left\{0;-1\right\}\)
câu d
\(D=\dfrac{\left(1-x^2\right)}{x}\left(\dfrac{x^2}{x+3}-1\right)+\dfrac{3x^2-14x+3}{x^2+3x}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{\left(1-x^2\right)\left(x^2-x-3\right)+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{x^2-x-3-x^4+x^3-3x^2+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x^4+x^3+x^2-15x}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x\left(x^3-x^2-x+15\right)}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-\left(x^3-x^2-x+15\right)}{\left(x+3\right)}\end{matrix}\right.\)
Bài 1:
a. $2x^3+3x^2-2x=2x(x^2+3x-2)=2x[(x^2-2x)+(x-2)]$
$=2x[x(x-2)+(x-2)]=2x(x-2)(x+1)$
b.
$(x+1)(x+2)(x+3)(x+4)-24$
$=[(x+1)(x+4)][(x+2)(x+3)]-24$
$=(x^2+5x+4)(x^2+5x+6)-24$
$=a(a+2)-24$ (đặt $x^2+5x+4=a$)
$=a^2+2a-24=(a^2-4a)+(6a-24)$
$=a(a-4)+6(a-4)=(a-4)(a+6)=(x^2+5x)(x^2+5x+10)$
$=x(x+5)(x^2+5x+10)$
Bài 2:
a. ĐKXĐ: $x\neq 3; 4$
\(A=\frac{2x+1-(x+3)(x-3)+(2x-1)(x-4)}{(x-3)(x-4)}\\ =\frac{2x+1-(x^2-9)+(2x^2-9x+4)}{(x-3)(x-4)}\\ =\frac{x^2-7x+14}{(x-3)(x-4)}\)
b. $x^2+20=9x$
$\Leftrightarrow x^2-9x+20=0$
$\Leftrightarrow (x-4)(x-5)=0$
$\Rightarrow x=5$ (do $x\neq 4$)
Khi đó: $A=\frac{5^2-7.5+14}{(5-4)(5-3)}=2$
ĐKXĐ: \(x\notin\left\{-1;2;-2\right\}\)
a) Ta có: \(A=\left(\dfrac{\left(x+1\right)^2}{\left(x+1\right)^2-3x}-\dfrac{2x^2+4x-1}{x^3+1}-\dfrac{1}{x+1}\right):\dfrac{x^2-4}{3x^2+6x}\)
\(=\left(\dfrac{\left(x+1\right)^2}{x^2-x+1}-\dfrac{2x^2+4x-1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{1}{x+1}\right):\dfrac{x^2-4}{3x^2+6x}\)
\(=\left(\dfrac{x^3+3x^2+3x+1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{2x^2+4x-1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\right):\dfrac{\left(x-2\right)\left(x+2\right)}{3x\left(x+2\right)}\)
\(=\dfrac{x^3+3x^2+3x+1-2x^2-4x+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}:\dfrac{x-2}{3x}\)
\(=\dfrac{x^3+1}{\left(x+1\right)\left(x^2-x+1\right)}\cdot\dfrac{3x}{x-2}\)
\(=\dfrac{3x}{x-2}\)
b) Để A nguyên thì \(3x⋮x-2\)
\(\Leftrightarrow3x-6+6⋮x-2\)
mà \(3x-6⋮x-2\)
nên \(6⋮x-2\)
\(\Leftrightarrow x-2\inƯ\left(6\right)\)
\(\Leftrightarrow x-2\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
hay \(x\in\left\{3;1;4;0;5;-1;8;-4\right\}\)
Kết hợp ĐKXĐ, ta được:
\(x\in\left\{3;1;4;0;5;8;-4\right\}\)
Vậy: Để A nguyên thì \(x\in\left\{3;1;4;0;5;8;-4\right\}\)
\(=\dfrac{x^2-x+1-3+3x+3}{\left(x+1\right)\left(x^2-x+1\right)}\cdot\dfrac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\dfrac{2x-2}{x\left(x+2\right)}\)
\(=\dfrac{x^2+2x+1}{x+1}\cdot\dfrac{3}{\left(x+1\right)\left(x+2\right)}-\dfrac{2x-2}{x\left(x+2\right)}\)
\(=\dfrac{3}{x+2}-\dfrac{2x-2}{x\left(x+2\right)}=\dfrac{3x-2x+2}{x\left(x+2\right)}=\dfrac{1}{x}\)