\(Q=\dfrac{\sqrt{x}+6}{\sqrt{x}-2}\left(đk:x\ge0,x\ne4\right)=\dfrac{\sqrt{x}-2}{\sqrt{x}-2}+\dfrac{8}{\sqrt{x}-2}=1+\dfrac{8}{\sqrt{x}-2}\in Z\)

\(\Rightarrow\sqrt{x}-2\inƯ\left(8\right)=\left\{-8;-4;-2;-1;1;2;4;8\right\}\)

Do \(x\ge0,x\ne4\)

\(\Rightarrow x\in\left\{0;1;9;16;36;100\right\}\)

Đkxđ: x # 4

Q = 1 + 8/(sqrt(x) - 2)

Q nguyên --> sqrt(x) - 2 là ước của 8

Do sqrt(x) >=0 nên sqrt(x) - 2 >= -2

TH1: sqrt(x) - 2 = -2 <=> x = 0 (thỏa)

TH2: sqrt(x) - 2 = -1 <=> x = 1 (thỏa)

Th3: sqrt(x) - 2 = 1 <=> x = 9(thỏa)

TH4: sqrt(x) - 2 = 2<=> x = 16 (thỏa)

Th5: sqrt(x) - 2 = 4 <=> x = 36 (thỏa)

Th6: sqrt(x) - 2 = 8 <=> x = 100 (thỏa)

Đề sai rồi bạn

Câu 2: 

2) Ta có: \(M=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)

\(=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{2\sqrt{x}-9-x+9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

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

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

\(=\dfrac{-\sqrt{x}}{\sqrt{x}-3}\)

Câu 2 : 

Gọi : vận tốc của người đi chậm là : x (km/h) ( x > 0 ) 

Vận tốc của người đi nhanh : x + 4 (km/h) 

Vi : người đi chậm đến muộn hơn : 45 phút \(=\dfrac{3}{4}\left(h\right)\)

Khi đó : 

\(\dfrac{36}{x}-\dfrac{36}{x+4}=\dfrac{3}{4}\)

\(\Leftrightarrow\left[36\cdot\left(x+4\right)-36x\right]\cdot4=3x\cdot\left(x+4\right)\)

\(\Leftrightarrow3x^2+12x-144=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=12\left(n\right)\\x=16\left(l\right)\end{matrix}\right.\)

Câu 1: 

1) Thay x=16 vào N, ta được:

\(N=\dfrac{2\cdot\sqrt{16}+1}{3-\sqrt{16}}=\dfrac{2\cdot4+1}{3-4}=\dfrac{9}{-1}=-9\)

Vậy: Khi x=16 thì N=-9

a: Khi x=25 thì \(A=\dfrac{5-2}{5-3}=\dfrac{3}{2}\)

b: P=A*B

\(=\dfrac{\sqrt{x}-2}{\sqrt{x}-3}\left(\dfrac{6x+6\sqrt{x}-12}{x+5\sqrt{x}+4}-\dfrac{5\sqrt{x}}{\sqrt{x}+4}\right)\)

\(=\dfrac{\sqrt{x}-2}{\sqrt{x}-3}\cdot\left(\dfrac{6x+6\sqrt{x}-12}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+4\right)}-\dfrac{5\sqrt{x}}{\sqrt{x}+4}\right)\)

\(=\dfrac{\sqrt{x}-2}{\sqrt{x}-3}\cdot\dfrac{6x+6\sqrt{x}-12-5x-5\sqrt{x}}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x+\sqrt{x}-12}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}-1}=\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)

c: \(\sqrt{P}< =\dfrac{1}{2}\)

=>0<=P<=1/4

=>\(\left\{{}\begin{matrix}P>=0\\P-\dfrac{1}{4}< =0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{\sqrt{x}-2}{\sqrt{x}-1}>=0\\\dfrac{\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{1}{4}< =0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>=4\\0< =x< 1\end{matrix}\right.\\\dfrac{4\left(\sqrt{x}-2\right)-\sqrt{x}+1}{4\left(\sqrt{x}-1\right)}< =0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>=4\\0< =x< 1\end{matrix}\right.\\\dfrac{3\sqrt{x}-7}{\sqrt{x}-1}< =0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>=4\\0< =x< 1\end{matrix}\right.\\1< \sqrt{x}< =\dfrac{7}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>=4\\0< =x< 1\end{matrix}\right.\\1< x< \dfrac{49}{9}\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>=4\\0< =x< 1\end{matrix}\right.\\x=\dfrac{49}{9}\end{matrix}\right.\)

=>\(4< =x< =\dfrac{49}{9}\)

mà x nguyên

nên \(x\in\left\{4;5\right\}\)

Lời giải:

$5A+B=\frac{5\sqrt{x}+1}{2\sqrt{x}+1}$

$2(5A+B)=\frac{10\sqrt{x}+2}{2\sqrt{x}+1}=\frac{5(2\sqrt{x}+1)-3}{2\sqrt{x}+1}=5-\frac{3}{2\sqrt{x}+1}$

$5A+B$ nguyên 

$\Rightarrow 2(5A+B)$ nguyên 

$\Leftrightarrow 5-\frac{3}{2\sqrt{x}+1}$ nguyên 

$\Leftrightarrow \frac{3}{2\sqrt{x}+1}$ nguyên 

Ta thấy: $\frac{3}{2\sqrt{x}+1}\leq 3$ với mọi $x\geq 0$ và $\frac{3}{2\sqrt{x}+1}>0$ với mọi $x\geq 0$

Do đó $\frac{3}{2\sqrt{x}+1}$ nguyên thì nhận các giá trị $1,2,3$

$\Leftrightarrow x=0; \frac{1}{16}; 1$

a) Ta có: \(M=\left(1-\dfrac{x-3\sqrt{x}}{x-9}\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}-\dfrac{\sqrt{x}-3}{2-\sqrt{x}}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)

\(=\left(1-\dfrac{x-3\sqrt{x}}{x-9}\right):\left(\dfrac{9-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right)\)

\(=\left(1-\dfrac{x-3\sqrt{x}}{x-9}\right):\left(\dfrac{9-x+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\dfrac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right)\)

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

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

\(=\dfrac{-3}{\sqrt{x}-2}\)

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

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

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

\(=\dfrac{\left(\sqrt{x}-1\right)^2.2}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}=\dfrac{2}{x+\sqrt{x}+1}\)

Để A nguyên thì: \(x+\sqrt{x}+1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)

Mà \(x+\sqrt{x}+1=\left(x+\sqrt{x}+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)

\(\Rightarrow x+\sqrt{x}+1\in\left\{1;2\right\}\)

+ Với \(x+\sqrt{x}+1=1\)

\(\Leftrightarrow\sqrt[]{x}\left(\sqrt{x}+1\right)=0\)

\(\Leftrightarrow x=0\left(tm\right)\left(do.\sqrt{x}+1\ge1>0\right)\)

+ Với \(x+\sqrt{x}+1=2\)

\(\Leftrightarrow\left(x+\sqrt{x}+\dfrac{1}{4}\right)=\dfrac{5}{4}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+\dfrac{1}{2}=\dfrac{\sqrt{5}}{2}\\\sqrt{x}+\dfrac{1}{2}=-\dfrac{\sqrt{5}}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{\sqrt{5}-1}{2}\\\sqrt{x}=-\dfrac{\sqrt{5}+1}{2}\left(VLý\right)\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{3-\sqrt{5}}{2}\left(tm\right)\)

Vậy \(S=\left\{1;\dfrac{3-\sqrt{5}}{2}\right\}\)

\(x^2-5x+1=m-2\sqrt{6+5x-x^2}\) (đk: \(x\in\left[-1;6\right]\))

\(\Leftrightarrow7-\left(6+5x-x^2\right)=m-2\sqrt{6+5x-x^2}\)

\(Đặt \) \(a=\sqrt{6+5x-x^2}\left(a\ge0\right)\)

(bình phương cái vừa đặt lên, tìm được \(\Delta_x=49-4a^2\) nên với mỗi \(a\in\left[0;\dfrac{7}{2}\right]\backslash\left\{\dfrac{7}{2}\right\}\) sẽ có 2 nghiệm x phân biệt)

pttt: \(7-a^2=m-2a\)

\(\Leftrightarrow a^2-2a-7=-m\) (*)

BBT \(f\left(x\right)=a^2-2a-7\) với \(a\in\left[0;\dfrac{7}{2}\right]\backslash\left\{\dfrac{7}{2}\right\}\)

nên để pt ban đầu có 2 nghiệm x phân biệt <=>pt (*) có 1 nghiệm <=> \(\left[{}\begin{matrix}-m=-8\\-7< -m< \dfrac{7}{4}\end{matrix}\right.\) hay \(\left[{}\begin{matrix}m=8\\\dfrac{7}{4}< m< 7\end{matrix}\right.\)
Ý A

Lời giải:

ĐKXĐ: $x>0; x\neq 4$
\(A=\frac{\sqrt{x}-2+\sqrt{x}+2}{(\sqrt{x}+2)(\sqrt{x}-2)}.\frac{\sqrt{x}-2}{\sqrt{x}}=\frac{2\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.\frac{\sqrt{x}-2}{\sqrt{x}}=\frac{2}{\sqrt{x}+2}\)

\(B=\frac{7}{3}A=\frac{14}{3(\sqrt{x}+2)}\)

Hiển nhiên $B>0$

Với $x>0; x\neq 4\Rightarrow 3(\sqrt{x}+2)\geq 6$

$\Rightarrow B=\frac{14}{3(\sqrt{x}+2)}\leq \frac{14}{6}<3$

Vậy $0< B< 3$. $B$ nguyên $\Leftrightarrow B\in\left\{1;2\right\}$

$\Leftrightarrow \frac{14}{3(\sqrt{x}+2)}\in\left\{1;2\right\}$

$\Leftrightarrow x\in\left\{\frac{64}{9}; \frac{1}{9}\right\}$ (tm)