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3 tháng 8 2017

1. ĐK \(\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)

a. Ta có \(R=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right).\left(\frac{1}{\sqrt{x}+2}+\frac{4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\)

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

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

b. Với \(x=4+2\sqrt{3}\Rightarrow R=\frac{\sqrt{4+2\sqrt{3}}+2}{\sqrt{4+2\sqrt{3}}\left(\sqrt{4+2\sqrt{3}}-2\right)}=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}+2}{\sqrt{\left(\sqrt{3}+1\right)^2}\left(\sqrt{\left(\sqrt{3}+1\right)^2}-2\right)}\)

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

c. Để \(R>0\Rightarrow\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}>0\Rightarrow\sqrt{x}-2>0\Rightarrow x>4\)

Vậy \(x>4\)thì \(R>0\)

2. Ta có \(A=6+2\sqrt{2}=6+\sqrt{8};B=9=6+3=6+\sqrt{9}\)

Vì \(\sqrt{8}< \sqrt{9}\Rightarrow A< B\)

3. a. \(VT=\frac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}.\left(\sqrt{a}+\sqrt{b}\right)\)

\(=\left(\sqrt{a}-\sqrt{b}\right).\left(\sqrt{a}+\sqrt{b}\right)=a-b=VP\left(đpcm\right)\)

b. Ta có \(VT=\left(2+\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right).\left(2-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\)

\(=\left(2+\sqrt{a}\right)\left(2-\sqrt{a}\right)=4-a=VP\left(đpcm\right)\)

14 tháng 5 2021

1) Khi x = 49 thì:

\(A=\frac{4\sqrt{49}}{\sqrt{49}-1}=\frac{4\cdot7}{7-1}=\frac{28}{6}=\frac{14}{3}\)

2) Ta có:

\(B=\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{x-1}\)

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

\(B=\frac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

c) \(P=A\div B=\frac{4\sqrt{x}}{\sqrt{x}-1}\div\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{4\sqrt{x}}{\sqrt{x}+1}\)

Ta có: \(P\left(\sqrt{x}+1\right)=x+4+\sqrt{x-4}\)

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

\(\Leftrightarrow4\sqrt{x}=x+4+\sqrt{x-4}\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+\sqrt{x-4}=0\)

Mà \(VT\ge0\left(\forall x\ge0,x\ne1\right)\)

\(\Rightarrow\hept{\begin{cases}\left(\sqrt{x}-2\right)^2=0\\\sqrt{x-4}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x}=2\\x-4=0\end{cases}}\Rightarrow x=4\)

Vậy x = 4

12 tháng 8 2021

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

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

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

\(A=\frac{4}{\sqrt{x}+2}\)

b, \(A=\frac{4}{\sqrt{x}+2}=\frac{2}{3}\)

=> 2cawn x + 4 = 12

=> 2.căn x = 8

=> căn x = 4

=> x = 16 (thỏa mãn)

c, có A = 4/ căn x + 2 và B  = 1/căn x - 2

=> A.B = 4/x - 4 

mà AB nguyên

=> 4 ⋮ x - 4

=> x - 4 thuộc Ư(4) 

=> x - 4 thuộc {-1;1;-2;2;-4;4}

=> x thuộc {3;5;2;6;0;8} mà x > 0 và x khác 4

=> x thuộc {3;5;2;6;8}

d, giống c thôi

3 tháng 8 2017

Mới đc câu a ak, thog cảm nha, trih độ mih thấp lắm:

\(\frac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\frac{2b}{a-b}\)

=\(\frac{a+\sqrt{ab}-\sqrt{ab}+b}{a-b}-\frac{2b}{a-b}\)

=\(\frac{a+b-2b}{a-b}=\frac{a-b}{a-b}=1\)

3 tháng 8 2017

bùn ngủ , mai lm câu b cho nha

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_

8 tháng 8 2019

ai giúp mình với ạ ngaingung