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25 tháng 12 2018

a)Q=\(\dfrac{1+x}{x}\)

b)x không tính được hoặc đề sai

c)?

12 tháng 12 2022

a: \(Q=\dfrac{1+x}{x\left(x+1\right)}\cdot\dfrac{x+1}{1}=\dfrac{x+1}{x}\)

b: Để Q=1 thì x+1=x(loại)

c: \(Q-\dfrac{1}{2}=\dfrac{x+1}{x}-\dfrac{1}{2}=\dfrac{2x+2-x}{2x}=\dfrac{x+2}{2x}\)

TH1: x>0 hoặc x<-2

=>Q>0

TH2: -2<x<0

=>Q<0

19 tháng 1 2019

a ) ĐK : \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)\(P=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^{^2}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)

\(=\dfrac{x-1-2\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+3}\)

\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{x-2\sqrt{x}+1}{x+4\sqrt{x}+3}\)

6 tháng 4 2018

Bài 2:

a, ĐKXĐ: \(x\ne\pm1;x\ne\dfrac{-1}{2}\)

\(P=\left(\dfrac{x-1}{x+1}-\dfrac{x}{x-1}-\dfrac{3x+1}{1-x^2}\right):\dfrac{2x+1}{x^2-1}\)

\(P=\left(\dfrac{x-1}{x+1}-\dfrac{x}{x-1}+\dfrac{3x+1}{x^2-1}\right).\dfrac{x^2-1}{2x+1}\)

\(P=\dfrac{\left(x-1\right)^2-x\left(x+1\right)+3x+1}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\)

\(P=\dfrac{x^2-2x+1-x^2-x+3x+1}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\)

\(P=\dfrac{2}{2x+1}\)

b, ĐKXĐ: \(x\ne\pm1;x\ne\dfrac{-1}{2}\)

Để \(P=\dfrac{3}{x-1}\Leftrightarrow\dfrac{2}{2x+1}=\dfrac{3}{x-1}\Leftrightarrow2\left(x-1\right)=3\left(2x+1\right)\)

\(\Leftrightarrow2x-2=6x+3\)\(\Leftrightarrow-4x=5\Leftrightarrow x=\dfrac{-5}{4}\)(TMĐK)

c, \(ĐKXĐ:x\ne\pm1;x\ne\dfrac{-1}{2}\)

Để \(P\in Z\Leftrightarrow\dfrac{2}{2x+1}\in Z\Leftrightarrow2x+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

+) Với \(2x+1=1\Leftrightarrow x=0\left(TMĐK\right)\)

+) Với \(2x+1=-1\Leftrightarrow x=-1\left(KTMĐK\right)\)

+) Với \(2x+1=2\Leftrightarrow x=\dfrac{1}{2}\left(TMĐK\right)\)

+) Với \(2x+1=-2\Leftrightarrow x=\dfrac{-3}{2}\left(TMĐK\right)\)

Vậy để \(P\in Z\Leftrightarrow x\in\left\{0;\dfrac{1}{2};\dfrac{-3}{2}\right\}\)

24 tháng 12 2018

a) Điều kiện xác định :

x ≠ 3; x ≠ -3; x ≠ 0

M = \(\dfrac{x}{x^2-9}\) - \(\dfrac{1}{x+3}\): ( \(\dfrac{x}{x\left(x-3\right)}\) - \(\dfrac{x-3}{x\left(x-3\right)}\) )

M = \(\dfrac{x}{x^2-9}\) - \(\dfrac{1}{x+3}\) : ( \(\dfrac{x-x+3}{x\left(x-3\right)}\) )

M = \(\dfrac{x}{x^2-9}\) - \(\dfrac{1}{x+3}\) : \(\dfrac{3}{x\left(x-3\right)}\)

M = \(\dfrac{x}{x^2-9}\) - \(\dfrac{x\left(x-3\right)}{3\left(x+3\right)}\) = \(\dfrac{x}{\left(x-3\right)\left(x+3\right)}\) - \(\dfrac{x\left(x-3\right)}{3\left(x+3\right)}\)

M = \(\dfrac{3x}{3\left(x-3\right)\left(x+3\right)}\) - \(\dfrac{x\left(x-3\right)^2}{3\left(x-3\right)\left(x+3\right)}\)

M = \(\dfrac{3x-x\left(x-3\right)^2}{3\left(x-3\right)\left(x+3\right)}\) = \(\dfrac{3x-x\left(x^2-6x+9\right)}{3\left(x-3\right)\left(x+3\right)}\)

M = \(\dfrac{3x-x^3+6x^2-9x}{3\left(x-3\right)\left(x+3\right)}\) = \(\dfrac{-x^3+6x^2-6x}{3\left(x-3\right)\left(x+3\right)}\)

Mk đang mệt sai thì bạn thông cảm cho mk.

12 tháng 12 2022

a: \(M=\dfrac{x}{\left(x-3\right)\left(x+3\right)}-\dfrac{1}{x+3}:\dfrac{x-x+3}{x\left(x-3\right)}\)

\(=\dfrac{x}{\left(x-3\right)\left(x+3\right)}-\dfrac{1}{x+3}\cdot\dfrac{x\left(x-3\right)}{3}\)

\(=\dfrac{x}{\left(x-3\right)\left(x+3\right)}-\dfrac{x\left(x-3\right)}{3\left(x+3\right)}\)

\(=\dfrac{3x-x\left(x^2-6x+9\right)}{3\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{3x-x^3+6x^2-9x}{3\left(x-3\right)\left(x+3\right)}=\dfrac{-x^3+6x^2-6x}{3\left(x-3\right)\left(x+3\right)}\)

b: Để M>1/2 thì M-1/2>0

=>\(\dfrac{-x^3+6x^2-6x}{3\left(x^2-9\right)}-\dfrac{1}{2}>0\)

=>\(\dfrac{-2x^3+12x^2-12x-3x^2+9}{6\left(x^2-9\right)}>0\)

=>\(\dfrac{-2x^3+9x^2-12x+9}{x^2-9}>0\)

TH1: \(\left\{{}\begin{matrix}-2x^3+9x^2-12x+9>0\\x^2-9>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 3\\\left[{}\begin{matrix}x>3\\x< -3\end{matrix}\right.\end{matrix}\right.\Leftrightarrow x< -3\)

TH2: \(\left\{{}\begin{matrix}-2x^3+9x^2-12x+9< 0\\x^2-9< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>3\\-3< x< 3\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

12 tháng 5 2017

a/ ĐKXĐ: \(x\ge0,x\ne1\)

\(P=\left(\dfrac{3}{\sqrt{x}-1}+\dfrac{\sqrt{x}-3}{x-1}\right):\left(\dfrac{x+2}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\)

= \(\dfrac{3\left(\sqrt{x}+1\right)+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{x+2-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

= \(\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

= \(\dfrac{4\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

= \(\dfrac{4\sqrt{x}}{\sqrt{x}+1}\)

b/ Với \(x\ge0,x\ne1\)

Để \(P=\sqrt{x}-1\Leftrightarrow\dfrac{4\sqrt{x}}{\sqrt{x}+1}=\sqrt{x}-1\)

\(\Leftrightarrow4\sqrt{x}=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)

\(\Leftrightarrow x-4\sqrt{x}-1=0\)

\(\Leftrightarrow\left(\sqrt{x}-2+\sqrt{5}\right)\left(\sqrt{x}-2-\sqrt{5}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-2+\sqrt{5}=0\\\sqrt{x}-2-\sqrt{5}=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2-\sqrt{5}\left(ktm\right)\\\sqrt{x}=2+\sqrt{5}\left(tm\right)\end{matrix}\right.\)

\(\Leftrightarrow x=9+4\sqrt{5}\)

Vậy để \(P=\sqrt{x}-1\) thì \(x=9+4\sqrt{5}\)

a: Sửa đề; \(P=\left(\dfrac{3x+3\sqrt{x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\right)\cdot\left(\dfrac{1}{1-\sqrt{x}}-1\right)\)

\(=\dfrac{3x+3\sqrt{x}-3-x+1+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{1-1+\sqrt{x}}{1-\sqrt{x}}\)

\(=\dfrac{3x+3\sqrt{x}-6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}}{1-\sqrt{x}}=\dfrac{3\sqrt{x}}{1-\sqrt{x}}\)

b: Để \(P=\sqrt{x}\) thì \(3\sqrt{x}=\sqrt{x}-x\)

\(\Leftrightarrow x+2\sqrt{x}=0\)

hay x=0

7 tháng 10 2018

Khôi Bùi , DƯƠNG PHAN KHÁNH DƯƠNG, Mysterious Person, Phạm Hoàng Giang, Phùng Khánh Linh, TRẦN MINH HOÀNG, Dũng Nguyễn, Nhã Doanh, hattori heiji, ...

a: \(A=\dfrac{\sqrt{x}+x\sqrt{y}+\sqrt{y}+y\sqrt{x}+\sqrt{x}-x\sqrt{y}-\sqrt{y}+y\sqrt{x}}{1-xy}:\dfrac{1-xy+x+y+2xy}{1-xy}\)

\(=\dfrac{2\sqrt{x}+2y\sqrt{x}}{x+y+xy+1}\)

\(=\dfrac{2\sqrt{x}\left(y+1\right)}{\left(x+1\right)\left(y+1\right)}=\dfrac{2\sqrt{x}}{x+1}\)

b: \(x=\dfrac{1}{\sqrt{2}+1}=\sqrt{2}-1\)

\(A=\dfrac{2\sqrt{\sqrt{2}-1}}{\sqrt{2}-1+1}=\sqrt{2\left(\sqrt{2}-1\right)}\)

16 tháng 10 2018

Cho \(5\sqrt{x}7\) mk viet nham

Sua lai thanh \(5\sqrt{x}-7\)

19 tháng 10 2022

a: \(A=\left(\dfrac{2}{\sqrt{x}-2}+\dfrac{3}{2\sqrt{x}+1}-\dfrac{5\sqrt{x}-7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\right)\cdot\dfrac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)

\(=\dfrac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)

\(=\dfrac{2\sqrt{x}+3}{\left(2\sqrt{x}+1\right)}\cdot\dfrac{5\sqrt{x}}{2\sqrt{x}+3}=\dfrac{5\sqrt{x}}{2\sqrt{x}+1}\)

b: Để A là số nguyên thì \(5\sqrt{x}⋮2\sqrt{x}+1\)

=>10 căn x+5-5 chia hết cho 2 căn x+1

=>\(2\sqrt{x}+1\in\left\{1;5\right\}\)

hay \(x\in\varnothing\)