rút gọn: \(\dfrac{x^3+2x^2+x}{x^3-x}\)
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29 tháng 1
ĐKXĐ: \(x\notin\left\{1;\dfrac{1}{2}\right\}\)
\(\left(\dfrac{1}{x-1}+2+\dfrac{2x^3+x^2-x}{1-x^3}\right):\dfrac{1-2x}{x^3+x-2}\)
\(=\left(\dfrac{1}{x-1}+2-\dfrac{2x^3+x^2-x}{\left(x-1\right)\left(x^2+x+1\right)}\right)\cdot\dfrac{x^3+x-2}{1-2x}\)
\(=\dfrac{x^2+x+1+2\left(x^3-1\right)-2x^3-x^2+x}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^3-x^2+x^2-x+2x-2}{-\left(2x-1\right)}\)
\(=\dfrac{2x+1+2x^3-2-2x^3}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x^2+x+2\right)}{-\left(2x-1\right)}\)
\(=\dfrac{2x-1}{x^2+x+1}\cdot\dfrac{-\left(x^2+x+2\right)}{2x-1}=\dfrac{-x^2-x-2}{x^2+x+1}\)
5 tháng 5 2021
Câu 1:
Sửa đề: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
Ta có: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}+3}\right):\left(\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\sqrt{x}-6+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x+\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=1\)
5 tháng 5 2021
Câu 3:
Ta có: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)
\(=\left(\dfrac{a}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-2\right)^2}\)
\(=\dfrac{a+\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\cdot\dfrac{\sqrt{a}-2}{1}\)
\(=\sqrt{a}\left(\sqrt{a}-2\right)\)
\(=a-2\sqrt{a}\)
A
1
NH
5 tháng 2 2023
\(\left(\dfrac{x-3}{x}-\dfrac{x}{x-3}+\dfrac{9}{x^2-3x}\right):\dfrac{2x-2}{x}\)
\(=\left(\dfrac{x-3}{x}-\dfrac{x}{x-3}+\dfrac{9}{x\left(x-3\right)}\right):\dfrac{2x-2}{x}\)
\(=\left(\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}-\dfrac{x^2}{x\left(x-3\right)}+\dfrac{9}{x\left(x-3\right)}\right)\cdot\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{x^2-6x+9-x^2+9}{x\left(x-3\right)}\cdot\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{-6x+18}{x\left(x-3\right)}\cdot\dfrac{x}{2\left(x-1\right)}\\ =\dfrac{-6\left(x-3\right)}{x\left(x-3\right)}\cdot\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{-6}{x}\cdot\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{-3}{x-1}\)
BC
1
4 tháng 4 2021
\(\dfrac{x^3-3}{x^2-2x-3}\)\(-\dfrac{2\left(x-3\right)}{x+1}\)\(+\dfrac{x+3}{3-x}\)
=\(\dfrac{x^3-3}{\left(x+1\right)\left(x-3\right)}\)\(-\dfrac{2\left(x-3\right)}{x+1}\)\(-\dfrac{x+3}{x-3}\)
=\(\dfrac{x^3-3-2x^2+12x-18-x^2-4x-3}{\left(x+1\right)\left(x-3\right)}\)
=\(\dfrac{x^3-3x^2+8x-24}{\left(x+1\right)\left(x-3\right)}\)
=\(\dfrac{\left(x-3\right)\left(x^2+8\right)}{\left(x-3\right)\left(x+1\right)}\)=\(\dfrac{x^2+8}{x+1}\)
1 tháng 7 2021
a, đề này chắc sai ở đoạn \(\dfrac{2x}{x^2-3}\) sửa thành \(\dfrac{2x}{x-3}\)
\(=>đk:x\ne1,x\ne3\)
\(=>A=\dfrac{2x}{x-3}+\dfrac{2x}{x^2-4x+3}+\dfrac{x}{x-1}\)
\(=\dfrac{2x\left(x-1\right)+2x+x\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}=\dfrac{2x^2-2x+2x+x^2-3x}{\left(x-1\right)\left(x-3\right)}\)
\(=\dfrac{3x^2-3x}{\left(x-1\right)\left(x-3\right)}=\dfrac{3x\left(x-1\right)}{\left(x-1\right)\left(x-3\right)}=\dfrac{3x}{x-3}\)
b, \(A=\dfrac{3x}{x-3}=3+\dfrac{9}{x-3}\)
A nguyên <=>\(x-3\inƯ\left(9\right)=\left\{\pm1;\pm3;\pm9\right\}\)
\(=>x\in\left\{4;2;6;0;12;-6\right\}\left(TM\right)\)
H
7 tháng 6 2023
` @ \color{Red}{m}`
` \color{lightblue}{Answer}`
\(\dfrac{x^2}{x^2-1}+\dfrac{x}{\left(1-x\right)\left(x+1\right)}\\ =\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}+\dfrac{x}{\left(1-x\right)\left(x+1\right)}\\ =\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^2-x}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x}{x+1}\)
__
\(\dfrac{3}{2x+6}-\dfrac{x-3}{x^2+3x}\\ =\dfrac{3}{2\left(x+3\right)}-\dfrac{x-3}{x\left(x+3\right)}\\ =\dfrac{3x}{2x\left(x+3\right)}-\dfrac{2\left(x-3\right)}{2x\left(x+3\right)}\\ =\dfrac{3x}{2x\left(x+3\right)}-\dfrac{2x-6}{2x\left(x+3\right)}\\ =\dfrac{3x-\left(2x-6\right)}{2x\left(x+3\right)}\\ =\dfrac{3x-2x+6}{2x\left(x+3\right)}\\ =\dfrac{x+6}{2x\left(x+3\right)}\)
__
\(\dfrac{1}{1-x}+\dfrac{2x}{x^2-1}\\ =\dfrac{1}{1-x}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{1}{1-x}-\dfrac{2x}{\left(1-x\right)\left(1+x\right)}\\ =\dfrac{1+x}{\left(1-x\right)\left(1+x\right)}-\dfrac{2x}{\left(1-x\right)\left(1+x\right)}\\ =\dfrac{1+x-2x}{\left(1-x\right)\left(1+x\right)}\\ =\dfrac{1-x}{\left(1-x\right)\left(1+x\right)}\\ =\dfrac{1}{1+x}\)
Y
7 tháng 6 2023
\(\dfrac{x^2}{x^2-1}+\dfrac{x}{\left(1-x\right)\left(x+1\right)}\left(dkxd:x\ne\pm1\right)\)
\(=\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2-x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x}{x+1}\)
========================
\(\dfrac{3}{2x+6}-\dfrac{x-3}{x^2+3x}\left(dkxd:x\ne\pm3;x\ne0\right)\)
\(=\dfrac{3}{2\left(x+3\right)}-\dfrac{x-3}{x\left(x+3\right)}\)
\(=\dfrac{3x-2\left(x-3\right)}{2x\left(x+3\right)}\)
\(=\dfrac{3x-2x+6}{2x\left(x+3\right)}\)
\(=\dfrac{x+6}{2x^2+6x}\)
==========================
\(\dfrac{1}{1-x}+\dfrac{2x}{x^2-1}\left(dkxd:x\ne\pm1\right)\)
\(=-\dfrac{1}{x-1}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-\left(x+1\right)+2x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x-1+2x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{1}{x+1}\)
23 tháng 12 2021
\(A=\left(\dfrac{x-1}{x\left(x-2\right)}+\dfrac{x+1}{x\left(x+2\right)}-\dfrac{4}{x\left(x-2\right)\left(x+2\right)}\right)\cdot\dfrac{x\left(x-3\right)}{2\left(x+2\right)}\)
\(=\dfrac{x^2+x-2+x^2-x+2-4}{x\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x-3\right)}{2\left(x+2\right)}\)
\(=\dfrac{2x^2-4}{x\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x-3\right)}{2\left(x+2\right)}\)
\(=\dfrac{2x\left(x^2-2\right)\left(x-3\right)}{2x\left(x-2\right)\cdot\left(x+2\right)^2}=\dfrac{\left(x^2-2\right)\left(x-3\right)}{\left(x-2\right)\left(x+2\right)^2}\)
TQ
14 tháng 6 2021
\(ĐKXĐ\) \(x\) khác 9
\(B=\dfrac{3\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\left(2x+3\right)}{9-x}\)
\(B=\dfrac{3\sqrt{x}\left(\sqrt{x-3}\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{2\left(2x+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(B=\dfrac{3\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-2\left(2x+3\right)}{x-9}\)
\(B=\dfrac{3x-9\sqrt{x}+x+3\sqrt{x}-4x-6}{x-9}\)
\(B=\dfrac{-6\sqrt{x}-6}{x-9}\)
14 tháng 6 2021
ĐK: x ≥ 0, x ≠ 9
Ta có: \(B=\dfrac{3\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\left(2x+3\right)}{9-x}\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-2\left(2x+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3x-9\sqrt{x}+x+3\sqrt{x}-4x-6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-6\sqrt{x}-6}{x-9}\)
\(=\dfrac{-6\left(\sqrt{x}+1\right)}{x-9}\)
NH
18 tháng 12 2022
a)
\(DKXD:\left[{}\begin{matrix}x^2+x\ne0\\x\ne0\\x+1\ne0\end{matrix}\right.< =>\left[{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)
b)
\(\left(\dfrac{2x^2-1}{x^2+x}-\dfrac{x-1}{x}+\dfrac{3}{x+1}\right)\cdot\dfrac{x+1}{3}\)
\(=\left(\dfrac{2x^2-1}{x\left(x+1\right)}-\dfrac{x-1}{x}+\dfrac{3}{x+1}\right)\cdot\dfrac{x+1}{3}\)
\(=\left(\dfrac{2x^2-1}{x\left(x+1\right)}-\dfrac{\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}+\dfrac{3x}{x\left(x+1\right)}\right)\cdot\dfrac{x+1}{3}\)
\(=\left(\dfrac{2x^2-1-x^2+1+3x}{x\left(x+1\right)}\right)\cdot\dfrac{x+1}{3}\)
\(=\dfrac{x^2+3x}{x\left(x+1\right)}\cdot\dfrac{x+1}{3}\\ =\dfrac{x\left(x+3\right)\cdot\left(x+1\right)}{x\left(x+1\right)\cdot3}\\ =\dfrac{x+3}{3}\)
\(\dfrac{x^3+2x^2+x}{x^3-x}\left(x\ne\left\{0;\pm1\right\}\right)\\ =\dfrac{x\left(x^2+2x+1\right)}{x\left(x^2-1\right)}\\ =\dfrac{x\left(x+1\right)^2}{x\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{x-1}\)