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29 tháng 7 2019

\(a,A=\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{3}{\sqrt{x}+2}-\frac{9\sqrt{x}-10}{x-4}\left(x\ge0;x\ne16\right)\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{9\sqrt{x}-10}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\frac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{3\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{9\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{x+2\sqrt{x}+3\sqrt{x}-6-9\sqrt{x}+10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{x-4\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\left(\sqrt{x}\right)^2-2.\sqrt{x}.2+2^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}-2}{\sqrt{x}+2}\)

Vây...

\(b,\)Ta có:\(x=4-2\sqrt{3}=\left(1-\sqrt{3}\right)^2\)

Thay \(x=\left(1-\sqrt{3}\right)^2\)vào A ta được:

\(A=\frac{\sqrt{\left(1-\sqrt{3}\right)^2}-2}{\sqrt{\left(1-\sqrt{3}\right)^2}+2}=\frac{\sqrt{3}-1-2}{\sqrt{3}-1+2}=\frac{\sqrt{3}-3}{\sqrt{3}-1}=\frac{-\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}=-\sqrt{3}\)

2 tháng 8 2019

\(a,đkxđ\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)

\(A=\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{3}{\sqrt{x}+2}-\frac{9\sqrt{x}-10}{x-4}.\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)\(-\frac{9\sqrt{x}-10}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\frac{x+2\sqrt{x}+3\sqrt{x}-6-9\sqrt{x}+10}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

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

\(=\frac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}-2}{\sqrt{x}+2}\)

\(b,x=4-2\sqrt{3}=3-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2\)

\(\Rightarrow x=\sqrt{3}-1\)

\(\Rightarrow A=\frac{\sqrt{3}-1-2}{\sqrt{3}-1+2}=\frac{\sqrt{3}-3}{\sqrt{3}-1}\)

\(b,A=\frac{\sqrt{x}-2}{\sqrt{x}+2}=\frac{\sqrt{x}+2-4}{\sqrt{x}+2}\)\(=1-\frac{4}{\sqrt{x}+2}\)

\(A\in Z\Leftrightarrow1-\frac{4}{\sqrt{x}+2}\in Z\Rightarrow\frac{4}{\sqrt{x}+2}\in Z\)

\(\Rightarrow\sqrt{x}+2\inƯ_4\)

Mà \(Ư_4=\left\{\pm1;\pm2;\pm4\right\}\)Nhưng \(\sqrt{x}+2\ge2\)\(\Rightarrow\sqrt{x}+2\in\left\{2;4\right\}\)

\(Th1:\sqrt{x}+2=2\Rightarrow\sqrt{x}=0\Rightarrow x=0\)

\(Th2:\sqrt{x}+2=4\Rightarrow\sqrt{x}=2\Rightarrow x=4\)

\(KL:x\in\left\{0;4\right\}\)

19 tháng 8 2021
Bài 1. a) A=7/6
19 tháng 8 2021
b) √x+1 /(√x +2)(√x-1)

Bài 1 : 

a )\(A=\frac{3-\sqrt{3}}{\sqrt{3}-1}+\frac{\sqrt{35}-\sqrt{15}}{\sqrt{5}}-\sqrt{28}\)

\(A=\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+\frac{\sqrt{5}\left(\sqrt{7}-\sqrt{3}\right)}{\sqrt{5}}-\sqrt{28}\)

\(A=\sqrt{3}+\sqrt{7}-\sqrt{3}-\sqrt{28}\)

\(A=\sqrt{7}-\sqrt{28}\)

\(A=\sqrt{7}-2\sqrt{7}=-\sqrt{7}\)

Vậy \(A=-\sqrt{7}\)

b)\(B=\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\frac{\sqrt{a}+\sqrt{b}}{a-b}\left(a,b>0;a\ne b\right)\)

\(B=\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}:\frac{\sqrt{a}+\sqrt{b}}{a-b}\)

\(B=\left(\sqrt{a}+\sqrt{b}\right).\frac{a-b}{\sqrt{a}+\sqrt{b}}\)

\(B=a-b\)

Vậy \(B=a-b\left(a,b>0;a\ne b\right)\)

_Minh ngụy_

Bài 2 :

a )\(B=\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{x+\sqrt{x}}\left(x>0\right)\)

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

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

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

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

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

\(B=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)

Vậy \(B=\frac{\sqrt{x}-1}{\sqrt{x}+1}\left(x>0\right)\)

b) \(B=\frac{\sqrt{x}-1}{\sqrt{x}+1}\left(x>0\right)\)

Ta có : \(B>0\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+1}>0\)

Vì : \(\sqrt{x}\ge0\forall x\Rightarrow\)để \(B>O\)cần \(\sqrt{x}-1>0\Leftrightarrow\sqrt{x}>1\Leftrightarrow x>1\)( thỏa mãn \(x>0\))

Vậy \(x>1\)thì \(B>0\)

_Minh ngụy_