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\(ĐKXĐ:x\ne-3;2\)
\(\frac{x+2}{x+3}-\frac{5}{x^2+x-6}-\frac{1}{x-2}=\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x+2\right)}-\frac{1}{x-2}\)
\(=\frac{x^2+4x+4}{\left(x+3\right)\left(x+2\right)}-\frac{5}{\left(x+3\right)\left(x+2\right)}-\frac{x+3}{\left(x+2\right)\left(x+3\right)}\)
\(=\frac{x^2+4x+4-5-x-3}{\left(x+2\right)\left(x+3\right)}=\frac{x^2+3x-4}{\left(x+3\right)\left(x+2\right)}=\frac{\left(x+4\right)\left(x-1\right)}{\left(x+3\right)\left(x+2\right)}\)
\(x^2-9=0\Leftrightarrow x=3\left(vì:x\ne-3\right)\)
\(\Rightarrow P=\frac{7}{15}\)
\(P\inℤ\Leftrightarrow x^2+3x-4⋮x^2+5x+6\Leftrightarrow2x+10⋮x^2+5x+6\Leftrightarrow12⋮x^2+5xx+6\)
\(................\left(dễ\right)\)
P/s: shitbo sai rồi nha bạn!Nếu không tin thì thay x = 3 vào P ban đầu và giá trị P sau khi rút gọn sẽ thấy sự khác biệt =)
ĐK: \(x\ne-3;x\ne2\)
a) \(P=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}-\frac{1}{x-2}\)
\(=\frac{x^2-4}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{x+3}{\left(x-2\right)\left(x+3\right)}\)
\(=\frac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}=\frac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\frac{x-4}{x-2}\)
b) \(x^2-9=0\Leftrightarrow x^2=9\Leftrightarrow x=\pm3\)
Thay vào điều kiện,tìm loại x = -3 .Tìm được x =3
Ta có: \(P=\frac{x-4}{x-2}=\frac{3-4}{3-2}=-1\)
c)Ta có: \(P=\frac{x-4}{x-2}=\frac{x-2-2}{x-2}=1-\frac{2}{x-2}\)
Để P có giá trị nguyên thì \(\frac{2}{x-2}\) nguyên hay \(x-2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
Suy ra \(x=\left\{0;1;3;4\right\}\)
Câu 1 :
a) Rút gọn P :
\(P=\dfrac{x+1}{3x-x^2}:\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{12x^2}{x^2-9}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left[\dfrac{\left(3+x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{\left(3-x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{12x^2}{\left(3-x\right)\left(3+x\right)}\right]\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left(\dfrac{9+6x+x^2-9+6x-x^2-12x^2}{\left(3-x\right)\left(3+x\right)}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\dfrac{12x-12x^2}{\left(3-x\right)\left(x+3\right)}\)
\(P=\dfrac{x+1}{x\left(3-x\right)}.\dfrac{\left(3-x\right)\left(x+3\right)}{12x\left(1-x\right)}\)
\(P=\dfrac{\left(x+1\right)\left(x+3\right)}{12x^2\left(1-x\right)}\)
Bài 1:
a, Ta có:
\(\dfrac{x.\dfrac{xy}{x-y}}{x+\dfrac{xy}{x-y}}-\dfrac{y.\dfrac{xy}{x-y}}{y-\dfrac{xy}{x-y}}\)
\(=\dfrac{\dfrac{x^2y}{x-y}}{x+\dfrac{xy}{x-y}}-\dfrac{\dfrac{xy^2}{x-y}}{y-\dfrac{xy}{x-y}}\)
\(=\dfrac{\left(\dfrac{x^2y}{x-y}\right)\left(y-\dfrac{xy}{x-y}\right)-\left(\dfrac{xy^2}{x-y}\right)\left(x+\dfrac{xy}{x-y}\right)}{\left(x+\dfrac{xy}{x-y}\right)\left(y-\dfrac{xy}{x-y}\right)}\)
\(=\dfrac{\dfrac{x^2y^2}{x-y}-\dfrac{x^3y^2}{\left(x-y\right)^2}-\dfrac{x^2y^2}{x-y}-\dfrac{x^2y^3}{\left(x-y\right)^2}}{xy-\dfrac{x^2y}{x-y}+\dfrac{xy^2}{x-y}-\dfrac{x^2y^2}{\left(x-y\right)^2}}\)
\(=\dfrac{-\left(\dfrac{x^3y^2+x^2y^3}{\left(x-y\right)^2}\right)}{xy-\left(\dfrac{x^2y-xy^2}{x-y}\right)-\dfrac{x^2y^2}{\left(x-y\right)^2}}\)
\(=-\dfrac{\dfrac{x^2y^2\left(x+y\right)}{\left(x-y\right)^2}}{xy-\left(\dfrac{xy\left(x-y\right)}{\left(x-y\right)}\right)-\dfrac{x^2y^2}{\left(x-y\right)^2}}\)
\(=\dfrac{\dfrac{x^2y^2\left(x+y\right)}{\left(x-y\right)^2}}{\dfrac{x^2y^2}{\left(x-y\right)^2}}=x+y\)
Chúc bạn học tốt!! Làm một câu mà toát cả mồ hôi!
mk nghỉ bài này đề sai
a) điều kiện : \(x\ne0;x\ne-1;x\ne2\)
ta có : \(A=1+\left(\dfrac{x+1}{x^3+1}-\dfrac{1}{x-x^2-1}+\dfrac{2}{x+1}\right):\dfrac{x^3-2x^2}{x^3-x^2+x}\)
\(\Leftrightarrow A=1+\left(\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{1}{x^2-x+1}+\dfrac{2}{x+1}\right):\dfrac{x\left(x-2\right)}{x^2-x+1}\) \(\Leftrightarrow A=1+\left(\dfrac{x+1+x+1+2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\right):\dfrac{x\left(x-2\right)}{x^2-x+1}\) \(\Leftrightarrow A=1+\left(\dfrac{2x^2+4}{\left(x+1\right)\left(x^2-x+1\right)}\right):\dfrac{x^2-x+1}{x\left(x-2\right)}\) \(\Leftrightarrow A=1+\dfrac{2x^2+4}{x\left(x+1\right)\left(x-2\right)}=\dfrac{2x^2+4+x\left(x+1\right)\left(x-2\right)}{x\left(x+1\right)\left(x-2\right)}\)\(\Leftrightarrow A=\dfrac{x^3+x^2-2x+4}{x\left(x+1\right)\left(x-2\right)}\)
b) ta có : \(\left|x-\dfrac{3}{4}\right|=\dfrac{5}{4}\) \(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{5}{4}\\x-\dfrac{3}{4}=\dfrac{-5}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\left(L\right)\\x=\dfrac{-1}{2}\end{matrix}\right.\)
thế vào \(A\) ta có : \(A=\dfrac{41}{5}\)
vậy ...............................................................................................................
Với `x \ne +-2` có:
`M=[x^3]/[x^2-4]-x/[x-2]-2/[x+2]`
`M=[x^3-x(x+2)-2(x-2)]/[(x-2)(x+2)]`
`M=[x^3-x^2-2x-2x+4]/[(x-2)(x+2)]`
`M=[x^3-x^2-4x+4]/[(x-2)(x+2)]`
`M=[x^2(x-1)-4(x-1)]/[x^2-4]`
`M=[(x-1)(x^2-4)]/[x^2-4]`
`M=x-1`
\(M=\dfrac{x^3}{x^2-4}-\dfrac{x}{x-2}-\dfrac{2}{x+2}\)
\(=\dfrac{x^3-x\left(x+2\right)-2\left(x-2\right)}{x^2-4}\)
\(=\dfrac{x^3-x^2-2x-2x+4}{x^2+4}=\dfrac{x^3-4x-x^2+4}{x^2-4}=\dfrac{x\left(x^2-4\right)-\left(x^2-4\right)}{x^2-4}\)
\(=\dfrac{\left(x^2-4\right)\left(x-1\right)}{x^2-4}=x-1\)