Câu 1:

\(\frac{A}{B}\ge\frac{x}{4}+5\Leftrightarrow\frac{\sqrt{x}+4}{\sqrt{x}-1}:\frac{1}{\sqrt{x}-1}\ge\frac{x}{4}+5\)

\(\Rightarrow\sqrt{x}+4\ge\frac{x}{4}+5\Rightarrow x-4\sqrt{x}+4\le0\)

\(\Rightarrow\left(\sqrt{x}-2\right)^2\le0\Rightarrow\sqrt{x}-2=0\Rightarrow x=4\)

Câu 2:

Bạn coi lại đề, biểu thức B không hợp lý

ĐKXĐ: ...

\(A=\left(\frac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-1\right):\left(\frac{25-x+\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}-\frac{\sqrt{x}+3}{\sqrt{x}+5}\right)\)

\(=\left(\frac{\sqrt{x}}{\sqrt{x}+5}-\frac{\sqrt{x}+5}{\sqrt{x}+5}\right):\left(\frac{25-x+x-25}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}-\frac{\sqrt{x}+3}{\sqrt{x}+5}\right)\)

\(=\frac{-5}{\left(\sqrt{x}+5\right)}.\frac{\left(\sqrt{x}+5\right)}{-\left(\sqrt{x}+3\right)}=\frac{5}{\sqrt{x}+3}\)

b/ \(B=\frac{x+16}{\sqrt{x}+3}=\sqrt{x}-3+\frac{25}{\sqrt{x}+3}=\sqrt{x}+3+\frac{25}{\sqrt{x}+3}-6\)

\(\Rightarrow B\ge2\sqrt{\frac{\left(\sqrt{x}+3\right).25}{\sqrt{x}+3}}-6=4\)

\(B_{min}=4\) khi \(\left(\sqrt{x}+3\right)^2=25\Rightarrow x=4\)

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

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

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

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

bùn ngủ , mai lm câu b cho nha

b4 : 

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

\(b,x-5=\left(\sqrt{x}-\sqrt{5}\right)\left(\sqrt{x}+\sqrt{5}\right)\)

\(c,x+2\sqrt{xy}+y=\left(\sqrt{x}+\sqrt{y}\right)^2\)

\(d,x-4\sqrt{x}\sqrt{y}+4y=\left(\sqrt{x}-2\sqrt{y}\right)^2\)

b5:

\(a,ĐK:x\ge1\)

\(\sqrt{9\left(x-1\right)}+\sqrt{4\left(x-1\right)}-\frac{4}{5}\sqrt{25\left(x-1\right)}=1\)

\(\Leftrightarrow3\sqrt{x-1}+2\sqrt{x-1}-4\sqrt{x-1}=1\)

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

\(\Leftrightarrow x=2\left(tm\right)\)

\(b,ĐK:x\ge5\)

\(\frac{1}{3}\sqrt{9\left(x-5\right)}+\frac{1}{2}\sqrt{4\left(x-5\right)}-\frac{7}{5}\sqrt{25\left(x-5\right)}=2\)

\(\Leftrightarrow\sqrt{x-5}+\sqrt{x-5}-7\sqrt{x-5}=2\)

\(\Leftrightarrow-5\sqrt{x-5}=2\)

\(\Leftrightarrow\sqrt{x-5}=-\frac{2}{5}\left(voli\right)\)

\(c,ĐK:x>0\)

\(\sqrt{x}+\frac{9}{\sqrt{x}}=6\)

\(\Leftrightarrow x+9=6\sqrt{x}\)

\(\Leftrightarrow x-6\sqrt{x}+9=0\)

\(\Leftrightarrow\left(\sqrt{x}-3\right)^2=0\)

\(\Leftrightarrow x=9\left(tm\right)\)

Bài 1:

a)  \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)

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

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

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

b)   \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)

\(=\frac{2\left(5+\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)

\(=\frac{2\left(5+\sqrt{3}\right)}{2}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\)

\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)

c)  ĐK:  \(a\ge0;a\ne1\)

  \(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)

\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)

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

\(=1-a+a=1\)