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10 tháng 8 2021

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 : \(P=x=25\)

10 tháng 8 2021

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 : \(\frac{3}{3+1}=\frac{3}{4}\)

7 tháng 7 2018

mk làm luôn

a)\(A=\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}-1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}:\left(\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right).\)

=\(\frac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}-1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\frac{3\sqrt{x}+1}{3}\)

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

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

=

6 tháng 7 2018

a/ \(A=\frac{\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}}{1-\frac{3\sqrt{x}-2}{3\sqrt{x}+1}}\)

\(A=\frac{\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}-1\right)-\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}-\frac{8\sqrt{x}}{9x-1}}{1-\frac{3\sqrt{x}+1-3}{3\sqrt{x}+1}}\)

\(A=\frac{\frac{3x-4\sqrt{x}+1-3\sqrt{x}-1}{\left(3\sqrt{x}\right)^2-1}-\frac{8\sqrt{x}}{9x-1}}{1-1-\frac{3}{3\sqrt{x}+1}}\)

\(A=\frac{\frac{3x-7\sqrt{x}}{9x-1}-\frac{8\sqrt{x}}{9x-1}}{-\frac{3}{3\sqrt{x}+1}}\)

\(A=\frac{3x-7\sqrt{x}-8\sqrt{x}}{9x-1}\left(\frac{-\left(3\sqrt{x}+1\right)}{3}\right)\)

\(A=\frac{3x-15\sqrt{x}}{9x-1}\left(\frac{-3\sqrt{x}-1}{3}\right)\)

\(A=\frac{3\left(x-3\sqrt{x}\right)}{9x-1}\left(\frac{-3\sqrt{x}-1}{3}\right)\)

\(A=\frac{\left(x-3\sqrt{x}\right)\left(-3\sqrt{x}-1\right)}{9x-1}\)

\(A=\frac{3x\sqrt{x}+8x+3\sqrt{x}}{9x-1}\)

\(A=\frac{3x\sqrt{x}}{9x-1}+\frac{8x}{9x-1}+\frac{3\sqrt{x}}{9x-1}\)

\(A=\frac{x\sqrt{x}}{x-\frac{1}{3}}+\frac{8x}{9x-1}+\frac{\sqrt{x}}{x-\frac{1}{3}}\)

\(A=\frac{\sqrt{x}\left(x-1\right)}{x-\frac{1}{3}}+\frac{\frac{8}{3}x}{x-\frac{1}{3}}\)

\(A=\frac{\sqrt{x}\left(x-1\right)+\frac{8}{3}x}{x-\frac{1}{3}}\)

7 tháng 7 2018

bạn huy hoàng làm sai rồi

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 \(\frac{x-1}{\sqrt{x}\left(\sqrt{x}+3\right)}-\frac{3}{4}< 0\)

\(\Leftrightarrow\frac{4\left(x-1\right)-3\sqrt{x}\left(\sqrt{x}+3\right)}{4\sqrt{x}\left(\sqrt{x}+3\right)}< 0\)

\(4\sqrt{x}\left(\sqrt{x}+3\right)>0\forall x\) 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:

\(3< x< \frac{9+\sqrt{97}}{2}\)