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Bài 1: Sửa đề: \(B=\left(\frac{x-2}{x+2\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
a) Thay x=49 vào biểu thức \(A=\frac{\sqrt{x}+3}{\sqrt{x}-1}\), ta được:
\(A=\frac{\sqrt{49}+3}{\sqrt{49}-1}=\frac{7+3}{7-1}=\frac{10}{6}=\frac{5}{3}\)
Vậy: Khi x=49 thì \(A=\frac{5}{3}\)
b) Sửa đề: Rút gọn biểu thức B
Ta có: \(B=\left(\frac{x-2}{x+2\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\left(\frac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\frac{x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\frac{x+2\sqrt{x}-\sqrt{x}-2}{\sqrt{x}\cdot\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)-\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}}\)
c) Ta có: \(\frac{B}{A}=\frac{\sqrt{x}+1}{\sqrt{x}}:\frac{\sqrt{x}+3}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}}\cdot\frac{\sqrt{x}-1}{\sqrt{x}+3}\)
\(=\frac{x-1}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
Để \(\frac{B}{A}< \frac{3}{4}\) thì \(\frac{x-1}{\sqrt{x}\left(\sqrt{x}+3\right)}-\frac{3}{4}< 0\)
\(\Leftrightarrow\frac{4\left(x-1\right)-3\sqrt{x}\left(\sqrt{x}+3\right)}{4\sqrt{x}\left(\sqrt{x}+3\right)}< 0\)
mà \(4\sqrt{x}\left(\sqrt{x}+3\right)>0\forall x\) thỏa mãn ĐKXĐ
nên \(4\left(x-1\right)-3\sqrt{x}\left(\sqrt{x}+3\right)< 0\)
\(\Leftrightarrow4x-4-3x-9\sqrt{x}< 0\)
\(\Leftrightarrow x-9\sqrt{x}-4< 0\)
\(\Leftrightarrow x^2-9x-4< 0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\frac{9}{2}+\frac{81}{4}-\frac{97}{4}< 0\)
\(\Leftrightarrow\left(x-\frac{9}{2}\right)^2< \frac{97}{4}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-\frac{9}{2}>-\frac{\sqrt{97}}{2}\\x-\frac{9}{2}< \frac{\sqrt{97}}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>\frac{9-\sqrt{97}}{2}\\x< \frac{9+\sqrt{97}}{2}\end{matrix}\right.\)
Kết hợp ĐKXĐ, ta được:
\(3< x< \frac{9+\sqrt{97}}{2}\)
Bài 1 : Với : \(x>0;x\ne1\)
\(P=\left(1+\frac{1}{\sqrt{x}-1}\right)\frac{1}{x-\sqrt{x}}=\left(\frac{\sqrt{x}}{\sqrt{x}-1}\right).\sqrt{x}\left(\sqrt{x}-1\right)=x\)
Thay vào ta được : \(P=x=25\)
Bài 2 :
a, Với \(x\ge0;x\ne1\)
\(A=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2}{\sqrt{x}+1}-\frac{2}{x-1}=\frac{x+\sqrt{x}-2\sqrt{x}+2-2}{x-1}\)
\(=\frac{x-\sqrt{x}}{x-1}=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}}{\sqrt{x}+1}\)
Thay x = 9 vào A ta được : \(\frac{3}{3+1}=\frac{3}{4}\)
a) \(\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)
\(=\sqrt{4-4\sqrt{3}+3}-\sqrt{4+4\sqrt{3}+3}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=\left|2-\sqrt{3}\right|-\left|2+\sqrt{3}\right|\)
\(=2-\sqrt{3}-2-\sqrt{3}\)
\(=-2\sqrt{3}\)
mk làm luôn
a)\(A=\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}-1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}:\left(\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right).\)
=\(\frac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}-1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\frac{3\sqrt{x}+1}{3}\)
=\(\frac{\left(3x+3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right).3}\)
=\(\frac{3x+3\sqrt{x}-1}{9\sqrt{x}-3}\)
=
a/ \(A=\frac{\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}}{1-\frac{3\sqrt{x}-2}{3\sqrt{x}+1}}\)
\(A=\frac{\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}-1\right)-\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}-\frac{8\sqrt{x}}{9x-1}}{1-\frac{3\sqrt{x}+1-3}{3\sqrt{x}+1}}\)
\(A=\frac{\frac{3x-4\sqrt{x}+1-3\sqrt{x}-1}{\left(3\sqrt{x}\right)^2-1}-\frac{8\sqrt{x}}{9x-1}}{1-1-\frac{3}{3\sqrt{x}+1}}\)
\(A=\frac{\frac{3x-7\sqrt{x}}{9x-1}-\frac{8\sqrt{x}}{9x-1}}{-\frac{3}{3\sqrt{x}+1}}\)
\(A=\frac{3x-7\sqrt{x}-8\sqrt{x}}{9x-1}\left(\frac{-\left(3\sqrt{x}+1\right)}{3}\right)\)
\(A=\frac{3x-15\sqrt{x}}{9x-1}\left(\frac{-3\sqrt{x}-1}{3}\right)\)
\(A=\frac{3\left(x-3\sqrt{x}\right)}{9x-1}\left(\frac{-3\sqrt{x}-1}{3}\right)\)
\(A=\frac{\left(x-3\sqrt{x}\right)\left(-3\sqrt{x}-1\right)}{9x-1}\)
\(A=\frac{3x\sqrt{x}+8x+3\sqrt{x}}{9x-1}\)
\(A=\frac{3x\sqrt{x}}{9x-1}+\frac{8x}{9x-1}+\frac{3\sqrt{x}}{9x-1}\)
\(A=\frac{x\sqrt{x}}{x-\frac{1}{3}}+\frac{8x}{9x-1}+\frac{\sqrt{x}}{x-\frac{1}{3}}\)
\(A=\frac{\sqrt{x}\left(x-1\right)}{x-\frac{1}{3}}+\frac{\frac{8}{3}x}{x-\frac{1}{3}}\)
\(A=\frac{\sqrt{x}\left(x-1\right)+\frac{8}{3}x}{x-\frac{1}{3}}\)
A=?;B=?