a)\(đkx\ge1,x\ne-1\)

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

\(\Leftrightarrow\dfrac{x-1}{x+1}=4\)

\(\Leftrightarrow x-1=4x-4\)

\(\Leftrightarrow x=1\)(nhận)

Vậy S=\(\left\{1\right\}\)

c)đk\(25x^2-10x+1=\) \(\left(5x-1\right)^2\ge0\Leftrightarrow x\ge\dfrac{1}{5}\)

\(\sqrt{25x^2-10x+1}+2x=1\)

\(\Leftrightarrow\sqrt{\left(5x-1\right)^2}+2x=1\)

\(\Leftrightarrow5x-1+2x=1\)

\(\Leftrightarrow x=\dfrac{2}{7}\)(nhận)

Vậy S=\(\left\{\dfrac{2}{7}\right\}\)

c: Ta có: \(\sqrt{25x^2-10x+1}+2x=1\)

\(\Leftrightarrow\left|5x-1\right|=1-2x\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-1=1-2x\left(x\ge\dfrac{1}{5}\right)\\5x-1=2x-1\left(x< \dfrac{1}{5}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{7}\left(nhận\right)\\x=0\left(nhận\right)\end{matrix}\right.\)

ĐIều kiện:`x^2-7x+8>=0`

`<=>x^2-2*x*7/2+49/4-17/4>=0`

`<=>(x-7/2)^2-17/4>=0`

`<=>(x-7/2)^2>=17/4`

`<=>|x-7/2|>=sqrt{17}/2`

`<=>` \(\left[ \begin{array}{l}x \ge \dfrac{7+\sqrt{17}}{2}\\x \le \dfrac{-\sqrt{17}+7}{2}\end{array} \right.\) 

`pt<=>x^2-7x+sqrt{x^2-7x+8}-12=0`

`<=>x^2-7x+8+sqrt{x^2-7x+8}-20=0`

Đặt `a=sqrt{x^2-7x+8}(a>=0)`

`pt<=>a^2+a-20=0`

`<=>a=4(tm),a=-5(l)`

`<=>x^2-7x+8=16`

`<=>x^2-7x-8=0`

`a-b+c=0`

`=>x_1=-1(tm),x_2=8(tm)`

Vậy `S={-1,8}`

a khác 0

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

b: \(P-1=\dfrac{\sqrt{a}-1-\sqrt{a}}{\sqrt{a}}=\dfrac{-1}{\sqrt{a}}< 0\)

\(19,ĐKXĐ:\dfrac{-2\sqrt{6}+\sqrt{23}}{-x+5}\ge0;x\ne5\\ \Leftrightarrow5-x< 0\left(-2\sqrt{6}+\sqrt{23}< 0\right)\\ \Leftrightarrow x>5\)

\(21,ĐKXĐ:x\ne7;\dfrac{2\sqrt{15}-\sqrt{59}}{x-7}\ge0\\ \Leftrightarrow x-7>0\left(2\sqrt{15}-\sqrt{59}>0\right)\\ \Leftrightarrow x>7\)

\(23,ĐKXĐ:49x^2-24x+4\ge0\Leftrightarrow\left(49x^2-14\cdot\dfrac{12}{7}x+\dfrac{144}{49}\right)+\dfrac{52}{49}\ge0\\ \Leftrightarrow\left(7x-\dfrac{12}{7}\right)^2+\dfrac{52}{49}\ge0\\ \Leftrightarrow x\in R\)

Đkxđ:

y≥0

x-1≠0 => x≠1

a: ĐKXĐ: x>0; x<>1

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

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

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

b: Q>2

=>Q-2>0

=>\(\dfrac{x-\sqrt{x}+1}{\sqrt{x}-1}>0\)

=>căn x-1>0

=>x>1

a) ĐK: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

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

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

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

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

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

b) Q>2 <=> \(\dfrac{x}{\sqrt{x}-1}>2\Leftrightarrow x>2\sqrt{x}-2\)

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

\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+1>0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)^2\ge0\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-1\le0\\\sqrt{x}-1\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ge1\\x\le1\end{matrix}\right.\)

KL:.....

a + b , \(N=\left(\frac{2}{x^2+x}+\frac{1}{x+1}\right):\frac{1}{x+1}\)ĐK : \(x\ne0;-1\)

\(=\left(\frac{2}{x\left(x+1\right)}+\frac{x}{x\left(x+1\right)}\right):\frac{1}{x+1}=\frac{x+2}{x\left(x+1\right)}.\frac{x+1}{1}=\frac{x+2}{x}\)

c, Ta có : \(\frac{x+2}{x}=1+\frac{2}{x}\)

Để N nguyên khi \(2⋮x\Rightarrow x\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

Vậy \(x=\pm1;\pm2\)thì N nguyên 

d, ta có : \(N< 1\Rightarrow\frac{x+2}{x}< 1\Leftrightarrow\frac{x+2-x}{x}< 0\Rightarrow x< 0\)vì 2 > 0 

 bổ sung hộ mình 

c, Kết hợp với đk vậy \(x=1;\pm2\)thì N nguyên 

d, Kết hợp với đk vậy \(x< 0;x\ne-1\)

a) đk: \(\hept{\begin{cases}x\ne0\\x\ne-1\end{cases}}\)

b) \(N=\left(\frac{2}{x^2+x}+\frac{1}{x+1}\right)\div\frac{1}{x+1}\)

\(N=\frac{2+x}{x\left(x+1\right)}\cdot\left(x+1\right)=\frac{2+x}{x}\)

c) \(N=\frac{2+x}{x}=\frac{2}{x}+1\)

Để N nguyên \(x\inƯ\left(2\right)\Rightarrow x\in\left\{1;-2;2\right\}\)

d) \(N< 1\Leftrightarrow\frac{2}{x}+1< 1\Rightarrow\frac{2}{x}< 0\Rightarrow x< 0\)

Đk: \(x\ge4\)

\(A=\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}\)

\(=\sqrt{\left(x-4\right)+4\sqrt{x-4}+4}+\sqrt{\left(x-4\right)-4\sqrt{x-4}+4}\)

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

\(=\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|\)

TH1:\(\sqrt{x-4}>2\Leftrightarrow x>8\)

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

TH2:\(\sqrt{x-4}\le2\Leftrightarrow4\le x\le8\)

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

Vậy...

Bài 1 

Mình làm mẫu một số câu thôi nhé

\(9,\sqrt{5}=\left(\sqrt{5}\right)^2=5\\ \sqrt{6}=\left(\sqrt{6}\right)^2=6\)

Vì \(5< 6\)

\(\Rightarrow\sqrt{5}< \sqrt{6}\)

\(10,2\sqrt{5}=\left(2\sqrt{5}\right)^2=20\\ \sqrt{7}=\left(\sqrt{7}\right)^2=7\)

Vì \(20>7\)

\(\Rightarrow2\sqrt{5}>\sqrt{7}\)

\(11,5\sqrt{2}=\left(5\sqrt{2}\right)^2=50\\ 2\sqrt{3}=\left(2\sqrt{3}\right)^2=12\)

Vì \(50>12\Rightarrow5\sqrt{2}>2\sqrt{3}\)

\(12,2\sqrt{6}=\left(2\sqrt{6}\right)^2=24\\ 5=5^2=25\)

Vì \(25>24\Rightarrow5>2\sqrt{6}\)

\(13,\sqrt{7}=\left(\sqrt{7}\right)^2=7\\ 2=2^2=4\)

Vì \(7>4\Rightarrow\sqrt{7}>2\)

\(14,3=3^2=9\\ \sqrt{5}=\left(\sqrt{5}\right)^2=5\)

Vì \(9>5\Rightarrow3>\sqrt{5}\)

\(15,3\sqrt{6}=\left(3\sqrt{6}\right)^2=54\)

Vì \(54>1\Rightarrow3\sqrt{6}>1\)

\(16,2\sqrt{2}=\left(2\sqrt{2}\right)^2=8\\ 3=3^2=9\)

Vì \(8< 9\Rightarrow2\sqrt{2}< 3\)

Phương pháp làm dạng bài này là bình phương hai vế rồi so sánh 

Bài 2

Gợi ý : Biểu thức dưới dấu căn \(\ge\) 0

Lưu ý : Nếu biểu thức dưới dấu căn ở dưới mẫu thì \(>0\)

\(21,ĐK:4x^2-12x+9>0\\ \Rightarrow\left(2x-3\right)^2>0\\ \Leftrightarrow x\ne\dfrac{3}{2}\)

\(22,ĐK:x^2-8x+15\ge0\\ \Rightarrow\left[{}\begin{matrix}x\le3\\x\ge5\end{matrix}\right.\)

\(23,ĐK:\left\{{}\begin{matrix}x-2\ge0\\x-5\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge2\\x\ne5\end{matrix}\right.\)

\(24,ĐK:\left\{{}\begin{matrix}\dfrac{2+x}{5-x}\ge0\\5-x\ne0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}2+x\ge0\\5-x\ge0\\x\ne5\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x\ge-2\\x\le5\\x\ne5\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x\ge-2\\x< 5\end{matrix}\right.\left(t/m\right)\)

Hoặc

\(\left\{{}\begin{matrix}2+x\le0\\5-x\le0\\5-x\ne0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x\le-2\\x\ge5\\x\ne5\end{matrix}\right.\left(loại\right)\)

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