\(a,\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{x+\sqrt{x}+2}{x-1}\right):\dfrac{1}{\sqrt{x}-1}\left(dkxd:x\ge0;x\ne1\right)\)

\(=\left[\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{x+\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\cdot\left(\sqrt{x}-1\right)\)

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

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

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

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

\(b,\) Thay \(x=4-2\sqrt{3}\) vào biểu thức trên, ta được:

\(\sqrt{4-2\sqrt{3}}+1\)

\(=\sqrt{\left(\sqrt{3}\right)^2-2\cdot\sqrt{3}\cdot1+1^2}+1\)

\(=\sqrt{\left(\sqrt{3}-1\right)^2}+1\)

\(=\left|\sqrt{3}-1\right|+1\)

\(=\sqrt{3}-1+1\)

\(=\sqrt{3}\)

Vậy: ...

\(\text{#}Toru\)

\(a\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{x+\sqrt{x}+2}{x-1}\right):\dfrac{1}{\sqrt{x}-1}\\ =\left(\dfrac{\sqrt{x}-1}{x-1}+\dfrac{x+\sqrt{x}+2}{x-1}\right).\sqrt{x}-1\\ =\dfrac{x+\sqrt{2}+1}{x-1}.\sqrt{x}-1\\ =\sqrt{x}+1\\ b,tacóx=4-2\sqrt{3}=\left(\sqrt{3}-\sqrt{1}\right)^2thãy=\sqrt{3}-\sqrt{1}vàobiểuthức,tađược\\ \sqrt{\left(\sqrt{3}-\sqrt{1}\right)^2}-1=\sqrt{3}-1-1=\sqrt{3}-2\)

Đề có phải là \(\dfrac{1}{2\sqrt{x}}-\dfrac{1}{2\sqrt{x}}+\dfrac{2-\sqrt{x}}{1-x}\) không bạn?

a. ĐK: \(x\ne\pm2\)
\(M=\left[\dfrac{x\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{\left(x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x+7}{\left(x-2\right)\left(x+2\right)}\right]:\dfrac{3-x+x-2}{x-2}\)

\(=\dfrac{x^2+2x-\left(x^2-2x+x-2\right)-2x-7}{\left(x-2\right)\left(x+2\right)}.\left(x-2\right)\)

\(=\dfrac{x-5}{x+2}\)

b. \(\dfrac{x-5}{x+2}< 1\Leftrightarrow\dfrac{x-5}{x+2}-1< 0\)

\(\Leftrightarrow\dfrac{-7}{x+2}< 0\Leftrightarrow x+2>0\)

\(\Leftrightarrow x>-2\)
Vậy \(x>-2,x\ne2\)

Câu 1:

b: ĐKXĐ: \(x\notin\left\{3;-3\right\}\)

\(\dfrac{1}{x-3}-\dfrac{1}{x+3}+\dfrac{2x}{9-x^2}\)

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

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

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

c: ĐKXĐ: \(x\notin\left\{2;0\right\}\)

Sửa đề: \(\dfrac{x+1}{x-2}+\dfrac{4-5x}{x^3+4x}:\dfrac{x-2}{x^2+4}\)

\(=\dfrac{x+1}{x-2}+\dfrac{4-5x}{x\left(x^2+4\right)}\cdot\dfrac{x^2+4}{x-2}\)

\(=\dfrac{x+1}{x-2}+\dfrac{4-5x}{x\left(x-2\right)}\)

\(=\dfrac{x\left(x+1\right)+4-5x}{x\left(x-2\right)}=\dfrac{x^2+x-5x+4}{x\left(x-2\right)}\)

\(=\dfrac{x^2-4x+4}{x\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{x\left(x-2\right)}=\dfrac{x-2}{x}\)

a: \(P=1:\left(\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right)\)

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

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

b: \(P-3=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>0\)

=>P>3

Lời giải:

a. 

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

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

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

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

b.

\(P-3=\frac{x+\sqrt{x}+1}{\sqrt{x}}-3=\frac{x-2\sqrt{x}+1}{\sqrt{x}}=\frac{(\sqrt{x}-1)^2}{\sqrt{x}}>0\) với mọi $x>0; x\neq 1$

$\Rightarrow P>3$

a: \(Q=\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{1}{a+\sqrt{a}}\right):\dfrac{\sqrt{a}-1}{a+2\sqrt{a}+1}\)

\(=\dfrac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\cdot\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}-1}\)

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

\(a,A=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{x\sqrt{x}-\sqrt{x}+x-1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-1}\right)\left(dk:x\ge0,x\ne1\right)\)

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

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

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

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

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

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

\(b,x-3\sqrt{x}+2=0\Leftrightarrow x-\sqrt{x}-2\sqrt{x}+2=0\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-1=0\\\sqrt{x}-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(ktm\right)\\x=4\left(tm\right)\end{matrix}\right.\)

Thay \(x=4\) vào A :

\(A=\dfrac{\sqrt{4}-1}{\sqrt{4}+1}=\dfrac{2-1}{2+1}=\dfrac{1}{3}\)