\(a,A=\dfrac{2\cdot2-4}{2-1}=0\\ b,B=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ B=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\\ c,AB=\dfrac{2\sqrt{x}-4}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{2\sqrt{x}-4}{\sqrt{x}+1}=\dfrac{5\left(\sqrt{x}+1\right)-3\left(\sqrt{x}+3\right)}{\sqrt{x}+1}\\ AB=5-\dfrac{3\left(\sqrt{x}+3\right)}{\sqrt{x}+1}\)

Vì \(\dfrac{3\left(\sqrt{x}+3\right)}{\sqrt{x}+1}>0\) nên \(AB< 5\)

a. \(x=4\Rightarrow A=\dfrac{2.\sqrt{4}-4}{\sqrt{4}-1}=0\)

b. \(\Rightarrow B=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-\left(6\sqrt{x}-4\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(\Rightarrow B=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

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

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

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

\(\Rightarrow\sqrt{x}=\sqrt{\left(\sqrt{3}-1\right)^2}=\left|\sqrt{3}-1\right|=\sqrt{3}-1\)do \(\sqrt{3}-1>0\)

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

b, Với \(x\ge0;x\ne1\)

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

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

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

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

Vậy biểu thức ko phụ thuộc biến x 

c, Ta có : \(\frac{2A}{B}\)hay \(\frac{2}{\sqrt{x}-1}\)để biểu thức nhận giá trị nguyên 

thì \(\sqrt{x}-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

a) Thay x=25 vào B ta có:

\(B=\dfrac{\sqrt{25}+2}{\sqrt{25}-2}=\dfrac{7}{3}\)

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

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

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

\(A=\dfrac{x-9-x+4+2\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(A=\dfrac{2\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

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

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

c) Ta có: \(A>B\) Khi:

\(\dfrac{2}{\sqrt{x}-2}>\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\)

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

\(\Leftrightarrow\dfrac{2-\sqrt{x}-2}{\sqrt{x}-2}>0\)

\(\Leftrightarrow\dfrac{-\sqrt{x}}{\sqrt{x}-2}>0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}-\sqrt{x}< 0\\\sqrt{x}-2< 0\end{matrix}\right.\\\left\{{}\begin{matrix}-\sqrt{x}>0\\\sqrt{x}-2>0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x< 4\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x>4\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow0< x< 4\) 

sao khó vậy,mình học lớp 9 mà tính mãi chẳng ra đáp án bài này từ lâu rùi

Bài 1 : 

\(2+\sqrt{9}=2+3=5\)

Bài 2 : 

Với \(x\ge0\)

\(B=\left(\frac{1}{\sqrt{x}+2}-\frac{1}{\sqrt{x}+7}\right):\frac{5}{\sqrt{x}+7}\)

\(=\frac{\sqrt{x}+7-\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+7\right)}:\frac{5}{\sqrt{x}+7}\)

\(=\frac{5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+7\right)}.\frac{\sqrt{x}+7}{5}=\frac{1}{\sqrt{x}+2}\)

Bài 3 : 

\(\hept{\begin{cases}x+2y=4\left(1\right)\\x-2y=0\left(2\right)\end{cases}}\)Lấy (1) - (2) ta được : 

\(4y=4\Leftrightarrow y=1\)

Thay y = 1 vào (1) ta được : \(x+2=4\Leftrightarrow x=2\)

Vậy \(\left(x;y\right)=\left(2;1\right)\)

\(a,A=4\sqrt{3}-5\sqrt{3}+2-\sqrt{3}=2-2\sqrt{3}\\ B=\dfrac{x+2\sqrt{x}+8+2\sqrt{x}-8}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-4}\\ b,B-\dfrac{1}{2}A=\dfrac{\sqrt{x}}{\sqrt{x}-4}-\dfrac{1}{2}\left(2-2\sqrt{3}\right)=0\\ \Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-4}=1+\sqrt{3}\\ \Leftrightarrow\sqrt{x}=\left(1+\sqrt{3}\right)\left(\sqrt{x}-4\right)\Leftrightarrow\sqrt{x}=\sqrt{x}-4\sqrt{3}+\sqrt{3x}-4\\ \Leftrightarrow\sqrt{3x}=4\sqrt{3}+4\\ \Leftrightarrow\sqrt{x}=\dfrac{4\sqrt{3}+4}{\sqrt{3}}\\ \Leftrightarrow\sqrt{x}=\dfrac{12+4\sqrt{3}}{3}\\ \Leftrightarrow x=\dfrac{192+96\sqrt{3}}{9}=\dfrac{64+32\sqrt{3}}{3}\)

\(\dfrac{\sqrt{x}}{\sqrt{x}-4}=1-\sqrt{3}\)
Nhỉ???

1) Sửa đề: x=0,09

Thay x=0,09 vào A, ta được:

\(A=\dfrac{\sqrt{0.09}}{\sqrt{0.09}-1}=\dfrac{0.3}{0.3-1}=\dfrac{0.3}{-0.7}=\dfrac{-3}{7}\)

1. \(x=\frac{1}{9}\) thỏa mãn đk: \(x\ge0;x\ne9\)

Thay \(x=\frac{1}{9}\) vào A ta có:

\(A=\frac{\sqrt{\frac{1}{9}}+1}{\sqrt{\frac{1}{9}}-3}=-\frac{1}{2}\)

2. \(B=...\)

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

    \(B=\frac{3x-9\sqrt{x}+x+3\sqrt{x}-4x-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

     \(B=\frac{-6\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

3. \(P=A:B=\frac{\sqrt{x}+1}{\sqrt{x}-3}:\frac{-6\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(P=\frac{\sqrt{x}+3}{-6}\)

Vì \(\sqrt{x}+3\ge3\forall x\)\(\Rightarrow\frac{\sqrt{x}+3}{-6}\le\frac{3}{-6}=-\frac{1}{2}\)

hay \(P\le-\frac{1}{2}\)

Dấu "=" xảy ra <=> x=0

toán lớp 9 khó zậy em đọc k hỉu 1 phân số

mik cần gấp ạ^^

1) ĐKXĐ: \(x\notin\left\{0;1\right\}\)

2) Ta có: \(A=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\left(1-\dfrac{3-\sqrt{x}}{\sqrt{x}+1}\right)\)

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

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

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