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10 tháng 10 2021

giúp mk vs

10 tháng 10 2021

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

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

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

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

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

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

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

5 tháng 11 2023

\(a,\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{2}{\sqrt{x}+2}\right):\dfrac{x+4}{x+2\sqrt{x}}\left(dkxd:x>0;x\ne4\right)\)

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

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

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

\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

\(---\)

\(b,\) Để biểu thức trên bằng $-x$

thì \(\dfrac{\sqrt{x}}{\sqrt{x}-2}=-x\)

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

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

\(\Leftrightarrow\sqrt{x}\left(x-2\sqrt{x}+1\right)=0\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Kết hợp với ĐKXĐ của $x$, ta được:

\(x=1\)

Vậy biểu thức bằng $-x$ khi $x=1$

\(\text{#}Toru\)

29 tháng 12 2021

\(a,ĐK:x\ne\pm3\\ Sửa:M=\dfrac{x}{x+3}+\dfrac{2x}{x-3}+\dfrac{9-3x^2}{x^2-9}\\ M=\dfrac{x^2-3x+2x^2+6x+9-3x^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x-3}\\ b,x=2\Leftrightarrow M=\dfrac{3}{2-3}=-3\\ c,M\in Z\Leftrightarrow x-3\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow x\in\left\{0;2;4;6\right\}\left(tm\right)\)

Đề có phải là \(\dfrac{1}{2\sqrt{x}}-\dfrac{1}{2\sqrt{x}}+\dfrac{2-\sqrt{x}}{1-x}\) không bạn?

9 tháng 4 2021

2cănx-2 2cănx+2 nha bạn mik nghi thiếu

a: \(P=\left(\dfrac{x+2}{\left(x-2\right)\left(x-3\right)}+\dfrac{x+3}{x-2}-\dfrac{x+2}{x-3}\right):\dfrac{\left(2x+5\right)\left(x-3\right)+9}{x-3}\)

\(=\dfrac{x+2+\left(x+3\right)\left(x-3\right)-\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\cdot\dfrac{x-3}{2x^2-6x+5x-15+9}\)

\(=\dfrac{x+2+x^2-9-x^2+4}{\left(x-2\right)}\cdot\dfrac{1}{2x^2-x-6}\)

\(=\dfrac{x-3}{x-2}\cdot\dfrac{1}{2x^2-4x+3x-6}\)

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

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

 

6 tháng 8 2023

\(M=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\left(\text{đ}k\text{x}\text{đ}:x\ge3\right)\\ =\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\\ =\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{2\sqrt{x}-9-\left(x-9\right)-\left(2x-4\sqrt{x}+\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{2\sqrt{x}-9-x+9-2x+4\sqrt{x}-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ =\dfrac{5\sqrt{x}-3x+2}{x-5\sqrt{x}+6}\)

__

Để \(M\in Z\) thì \(x-5\sqrt{x}+6\) thuộc ước của \(5\sqrt{x}-3x+2\)

\(\Rightarrow x-5\sqrt{x}+6=-5\sqrt{x}-3x+2\\ \Leftrightarrow x-5\sqrt{x}+6+5\sqrt{x}+3x-2=0\\ \Leftrightarrow4x-4=0\\ \Leftrightarrow4x=4\\ \Leftrightarrow x=1\)

 

 

6 tháng 8 2023

Điều kiện có sai k v? Xem lại giúp mình với

7 tháng 8 2023

a) ĐKXĐ: \(x\ge0;x\ne9;x\ne4\)

\(M=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)

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

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

\(M=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

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

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

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

b) Ta có M ϵ Z thì \(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{\sqrt{x}-3+4}{\sqrt{x}-3}=\dfrac{\sqrt{x}-3}{\sqrt{x}-3}+\dfrac{4}{\sqrt{x}-3}=1+\dfrac{4}{\sqrt{x}-3}\)

Phải thuộc Z vậy:

4 ⋮ \(\sqrt{x}-3\)

\(\Rightarrow\sqrt{x}-3\inƯ\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)

Mà: \(x\ge0,x\ne4,x\ne9\) nên \(\sqrt{x}-3\in\left\{1;2;-2;4\right\}\)

\(\Rightarrow x\in\left\{16;25;1;49\right\}\)

12 tháng 11 2023

a: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x< >1\end{matrix}\right.\)

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

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

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

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

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

b: Để A<0 thì \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}< 0\)

=>\(\sqrt{x}-1< 0\)

=>\(\sqrt{x}< 1\)

=>0<x<1

12 tháng 11 2023

\(A=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{2\left(x-2\sqrt{x}+1\right)}{x-1}\left(đkxđ:x>0;x\ne1\right)\\ =\left(\dfrac{\left(\sqrt{x}\right)^3-1^3}{\sqrt{x}\left(\sqrt{x-1}\right)}-\dfrac{\left(\sqrt{x}\right)^3+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{x-1}{2\left(\sqrt{x}-1\right)^2}\\ =\left(\dfrac{\left(\sqrt{x}-1\right)\left(x+2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}\\ =\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}\\ =\dfrac{4\sqrt{x}}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}\\ =\dfrac{2\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)

`b,` Để `A<0` thì :

\(\dfrac{2\left(\sqrt{x}+1\right)}{\sqrt{x}-1}< 0\\ \Leftrightarrow\sqrt{x}-1< 0\left(vì.2\left(\sqrt{x}+1\right)>0\right)\\ \Leftrightarrow\sqrt{x}< 1\\ \Leftrightarrow0\le x< 1\)

Kết hợp với điều kiện xác định ta có : \(0< x< 1\)