6565và 5536

m khác 3 nhá chứ ko phải -3 đâu bạn ạ 

Để đths trên song song <=> \(\hept{\begin{cases}m-1=3-m\\2\ne1\end{cases}}\Leftrightarrow2m=4\Leftrightarrow m=2\)( tm ) 

\(A=\left(\frac{2\sqrt{x}+1}{x-1}-\frac{1}{\sqrt{x}+1}\right):\frac{1}{\sqrt{x}-1}\)ĐK : \(x\ge0;x\ne1\)

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

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

\(B=\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{1-\sqrt{x}}{\sqrt{x}-2}-\frac{\sqrt{x}+4}{x-\sqrt{x}-2}\)ĐK : \(x\ge0;x\ne4\)

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

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

A =( \(\frac{2\sqrt{x}+1}{x-1}\)\(-\) \(\frac{1}{\sqrt{x}+1}\)) \(\div\) \(\frac{1}{\sqrt{x}-1}\)ĐK: x\(\ge0\)và x\(\ne1\)

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

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

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

   =\(\frac{\sqrt{x}+2}{\sqrt{x}+1}\)

Vậy A = \(\frac{\sqrt{x}+2}{\sqrt{x}+1}\)với x\(\ge0\)và x\(\ne1\)

à..........cái đó thì mình chưa tính ra được

Bài 2 : 

a, \(\sqrt{4a^2}+5a\)với a >= 0 

\(=\left|2a\right|+5a=7a\)

b, \(\sqrt{25x^2}+3x=\left|5x\right|+3x=-5x+3x=-2x\)với x =< 0 

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

\(=x-2-2+x=-4\)với x =< 2 

d, \(3-x+\sqrt{9+9x+x^2}=3-x+\sqrt{\left(3+x\right)^2}=3-x+\left|x+3\right|\)

\(=3-x+x+3=6\)với x =< -3 

BẠn viết ra giấy đc ko mình ko nhìn thấy

a) \(a=2\sqrt{5}=\sqrt{2^2.5}=\sqrt{20}< \sqrt{21}=b\).

b) \(a=4\sqrt{5}=\sqrt{4^2.5}=\sqrt{80}< \sqrt{90}=\sqrt{3^2.10}=3\sqrt{10}=b\)

c) \(a=\sqrt{10}+\sqrt{5}>\sqrt{9}+\sqrt{4}=3+2=5=b\)

d) \(a-b=\sqrt{15}+\sqrt{13}-2\sqrt{14}\)

Có \(\left(\sqrt{15}+\sqrt{13}\right)^2=15+13+2\sqrt{15.13}=28+2\sqrt{\left(14+1\right)\left(14-1\right)}\)

\(=28+2\sqrt{14^2-1}< 28+2\sqrt{14^2}=56=4.14=\left(2\sqrt{14}\right)^2\)

Do đó \(\sqrt{15}+\sqrt{13}< 2\sqrt{14}\)suy ra \(a< b\). 

e) \(a=\sqrt{199}+\sqrt{999}>\sqrt{196}+\sqrt{961}=14+31=45=\sqrt{2025}>\sqrt{1998}=b\)

 vghnghb gttyhbbygh bvi vn u

Điều kiện xác định của biểu thức \(A\)là: 

\(\hept{\begin{cases}x\ge0\\\sqrt{x}-1\ne0\\x-1\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\).

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

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

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

Khi \(x=9\): \(A=\frac{\sqrt{9}}{\sqrt{9}+1}=\frac{3}{4}\).

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

\(=x-\sqrt{x}=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)

Dấu \(=\)khi \(\sqrt{x}-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{4}\).