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6 tháng 4 2019

Cái này là toán lớp 9 chứ.

a)
ĐKXĐ : \(x\ne\pm4\)

\(A=\left(\frac{x-\sqrt{x}+7}{x-4}+\frac{\sqrt{x}+2}{x-4}\right):\left(\frac{\left(\sqrt{x}+2\right)^2}{x-4}-\frac{\left(\sqrt{x}-2\right)^2}{x-4}-\frac{2\sqrt{x}}{x-4}\right)\)

\(=\left(\frac{x-\sqrt{x}+7+\sqrt{x}+2}{x-4}\right):\left(\frac{x+4\sqrt{x}+4-x+4\sqrt{x}-4-2\sqrt{x}}{x-4}\right)\)

\(=\frac{x+9}{x-4}\cdot\frac{x-4}{6\sqrt{x}}=\frac{x+9}{6\sqrt{x}}\)

b)

Ta có

\(x+9-6\sqrt{x}=\left(\sqrt{x}-3\right)^2\ge0\)
\(\Rightarrow x+9\ge6\sqrt{x}\)

\(\Rightarrow\frac{x+9}{6\sqrt{x}}\ge1\)

\(\Leftrightarrow A\ge1\)

\(\Leftrightarrow\frac{1}{A}\le1\)

\(\Rightarrow A\ge\frac{1}{A}\)

30 tháng 4 2020

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

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

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

\(A=\left[\frac{x+2\sqrt{x}}{\left(x-4\right)\left(\sqrt{x}+2\right)}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)

\(A=\left[\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(x-4\right)\left(\sqrt{x+2}\right)}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)

\(A=\frac{\sqrt{x}}{x-4}\cdot\frac{\sqrt{x}-2}{\sqrt{x}}\)

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

\(A=\frac{\sqrt{x}-2}{x-4}\)

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 : \(\frac{3}{3+1}=\frac{3}{4}\)

28 tháng 3 2018

\(A=\left(\frac{2+\sqrt{x}}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{2-\sqrt{x}}-\frac{\sqrt{x}+2}{\sqrt{x}-3}\right)\) \(:\left(2-\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)

\(A=\left[\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+2}{\sqrt{x}-3}\right]\) 

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

\(A=\left[\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+2}{\sqrt{x}-3}\right]\)

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

\(A=\left[\frac{\sqrt{x}+2+x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\right]\)  \(:\left[\frac{\sqrt{x}+2}{\sqrt{x}+1}\right]\)

\(A=\left[\frac{\sqrt{x}+x-7-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]:\frac{\sqrt{x}+2}{\sqrt{x}+1}\)

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

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

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