\(M=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\left(\text{đ}k\text{x}\text{đ}:x\ge3\right)\\ =\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\\ =\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{2\sqrt{x}-9-\left(x-9\right)-\left(2x-4\sqrt{x}+\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{2\sqrt{x}-9-x+9-2x+4\sqrt{x}-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ =\dfrac{5\sqrt{x}-3x+2}{x-5\sqrt{x}+6}\)

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Để \(M\in Z\) thì \(x-5\sqrt{x}+6\) thuộc ước của \(5\sqrt{x}-3x+2\)

\(\Rightarrow x-5\sqrt{x}+6=-5\sqrt{x}-3x+2\\ \Leftrightarrow x-5\sqrt{x}+6+5\sqrt{x}+3x-2=0\\ \Leftrightarrow4x-4=0\\ \Leftrightarrow4x=4\\ \Leftrightarrow x=1\)

a: \(=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{x-9}\)

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

b: Khi x=11+6 căn 2 thì \(M=\dfrac{3\left(3+\sqrt{2}\right)}{3+\sqrt{2}-3}=\dfrac{9+3\sqrt{2}}{\sqrt{2}}=\dfrac{9\sqrt{2}+6}{2}\)

c: M<1
=>\(\dfrac{3\sqrt{x}-\sqrt{x}+3}{\sqrt{x}-3}< 0\)

=>căn x-3<0

=>0<x<9

`a,` \(M=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{3-11\sqrt{x}}{9-x}\) \(\left(x\ne\pm3;x>0\right)\)

\(M=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{x-9}-\dfrac{3+11\sqrt{x}}{x-9}\)

\(M=\dfrac{2x-6\sqrt{x}}{x-9}+\dfrac{x+3\sqrt{x}+\sqrt{x}+3}{x-9}-\dfrac{3+11\sqrt{x}}{x-9}\)

\(M=\dfrac{3x+9\sqrt{x}}{x-9}\)

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

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

`b,`Ta có :

 \(M=\dfrac{3\sqrt{11+6\sqrt{2}}}{\sqrt{11+6\sqrt{2}}-3}\)

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

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

\(M=\dfrac{9+3\sqrt{2}}{\sqrt{2}}\)

\(M=\dfrac{6+9\sqrt{2}}{2}\)

`c,`  Để `M<1` Ta có :

 \(\dfrac{3\sqrt{x}}{\sqrt{x}-3}< 1\)

\(\dfrac{3\sqrt{x}}{\sqrt{x}-3}-1< 0\)

\(\dfrac{3\sqrt{x}}{\sqrt{x}-3}-\dfrac{\sqrt{x}-3}{\sqrt{x}-3}< 0\)

\(\dfrac{2\sqrt{x}+3}{\sqrt{x}-3}< 0\)

\(\sqrt{x}-3< 0\) ( vì \(2\sqrt{x}+3>0\) )

\(\sqrt{x}< 3\)

\(x< 9\)

Đối chiếu ĐKXĐ ta có : `0<x<9`

a) ĐKXĐ: \(x\ge0;x\ne9;x\ne4\)

\(M=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)

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

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

\(M=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

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

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

\(M=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

b) Ta có M ϵ Z thì \(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{\sqrt{x}-3+4}{\sqrt{x}-3}=\dfrac{\sqrt{x}-3}{\sqrt{x}-3}+\dfrac{4}{\sqrt{x}-3}=1+\dfrac{4}{\sqrt{x}-3}\)

Phải thuộc Z vậy:

4 ⋮ \(\sqrt{x}-3\)

\(\Rightarrow\sqrt{x}-3\inƯ\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)

Mà: \(x\ge0,x\ne4,x\ne9\) nên \(\sqrt{x}-3\in\left\{1;2;-2;4\right\}\)

\(\Rightarrow x\in\left\{16;25;1;49\right\}\)

3:

a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >9\end{matrix}\right.\)

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

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

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

b: M>1/3

=>M-1/3>0

=>\(\dfrac{2}{\sqrt{x}+3}-\dfrac{1}{3}>0\)

=>\(\dfrac{6-\sqrt{x}-3}{3\left(\sqrt{x}+3\right)}>0\)

=>\(3-\sqrt{x}>0\)

=>\(\sqrt{x}< 3\)

=>0<=x<9

c: \(\sqrt{x}+3>=3\) với mọi x thỏa mãn ĐKXĐ

=>\(M=\dfrac{2}{\sqrt{x}+3}< =\dfrac{2}{3}\) với mọi x thỏa mãn ĐKXĐ

Dấu = xảy ra khi x=0

a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)

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

\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

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

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

b: Thay x=16 vào A, ta được:

\(A=\dfrac{3}{4+3}=\dfrac{3}{7}\)

c)\(A=\dfrac{3}{\sqrt{x}+3}=\dfrac{1}{3}\)

\(\Rightarrow\sqrt{x}+3=9\\ \Rightarrow\sqrt{x}=6\\ \Rightarrow x=36\)

d) \(A=\dfrac{3}{\sqrt{x}+3}\)

Vì \(3>0;\sqrt{x}+3>0\Rightarrow\dfrac{3}{\sqrt{x}+3}>0\)

e) \(2A\in Z\Rightarrow\dfrac{6}{\sqrt{x}+3}\in Z \Rightarrow6⋮x+3\\\Rightarrow\sqrt{x}+3\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\Rightarrow x=\left\{0;9\right\}\)

Câu 2:

a,

diện tích nhựa là: 2π. (0,4:2). 16= 6,4π (cm²)

b,

gọi chữ số hàng chục là a  (a>0, a ∈N) 

hàng đơn vị là b (b∈N)

hiệu 2 chữ số là: a-b=3 (1)

tổng bình phương 2 chữ số là: a²+b²=45 (2) 

từ (1) và (2) ta có hpt:

\(\left\{{}\begin{matrix}a-b=3\\a^2+b^2=45\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}a=6\\b=3\end{matrix}\right.\)

vậy chữ số đó là 63

Câu 1

a, Thay x=25 vào biểu thức B ta có

B=\(\dfrac{\sqrt{25}-3}{\sqrt{25}-1}=\dfrac{5-3}{5-1}=\dfrac{2}{4}=\dfrac{1}{2}\)

b, Ta có M=\(A\cdot B\)

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

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

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

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

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

c,  Để M<\(\sqrt{M}\)

Thì\(\text{​​}\text{​​}\text{​​}\text{​​}\dfrac{3\sqrt{x}}{\sqrt{x}+3}< \sqrt{\dfrac{3\sqrt{x}}{\sqrt{x}+3}}\)

⇔\(\text{​​}\text{​​}\text{​​}\text{​​}\dfrac{3\sqrt{x}}{\sqrt{x}+3}< \dfrac{\sqrt{3\sqrt{x}\left(\sqrt{x}+3\right)}}{\sqrt{x}+3}\)

⇔\(\text{​​}\text{​​}\text{​​}\text{​​}3\sqrt{x}< \sqrt{3\sqrt{x}\left(\sqrt{x}+3\right)}\)

⇔\(\text{​​}\text{​​}\text{​​}\text{​​}9x< 3\sqrt{x}\left(\sqrt{x}+3\right)\)

⇔\(\text{​​}\text{​​}\text{​​}\text{​​}3\sqrt{x}< \sqrt{x}+3\)

⇔\(\text{​​}\text{​​}\text{​​}\text{​​}2\sqrt{x}< 3\)

⇔\(\text{​​}\text{​​}\text{​​}\text{​​}\sqrt{x}< \dfrac{3}{2}\)

⇒\(\left\{{}\begin{matrix}x\ge0\\x< \dfrac{9}{4}\end{matrix}\right.\)

⇒\(0\le x< \dfrac{9}{4}\)

Câu 2:

a,

diện tích nhựa là: 2π. (0,4:2). 16= 6,4π (cm²)

b,

gọi chữ số hàng chục là a  (a>0, a ∈N) 

hàng đơn vị là b (b∈N)

hiệu 2 chữ số là: a-b=3 (1)

tổng bình phương 2 chữ số là: a²+b²=45 (2) 

từ (1) và (2) ta có hpt:

{a−b=3a2+b2=45{a−b=3a2+b2=45

=> {a=6b=3

a: ĐKXĐ: x>0; x<>4

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

b: P=2/3

=>(4-căn x)/(căn x-2)=2/3

=>2căn x-4=12-3căn x

=>5căn x=16

=>x=256/25

c: Khi x=8-2căn 7 thì \(P=\dfrac{4-\sqrt{7}+1}{\sqrt{7}-1-2}=\dfrac{5-\sqrt{7}}{\sqrt{7}-3}=-4-\sqrt{7}\)

a,\(ĐK:x>0,x\ne1,x\ne4\)

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

\(A=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{3}=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)

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

\(=>A=\dfrac{\sqrt{2}-3}{3\sqrt{2}-3}\)

a) ĐKXĐ: \(\left\{{}\begin{matrix}\sqrt{x}\ge0\\\sqrt{x}-1>0\\\sqrt{x}-2>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x>1\\x>4\end{matrix}\right.\) \(\Leftrightarrow x>4\)

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

\(=\dfrac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\left(x-1\right)-\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

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

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

b) Ta có \(x=3-2\sqrt{2}=2-2\sqrt{2}+1=\left(2-1\right)^2=1\)

Thay \(x=1\) vào \(A\), ta được:

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

a: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\notin\left\{1;4\right\}\end{matrix}\right.\)

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

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

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

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

1: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\notin\left\{4;9\right\}\end{matrix}\right.\)

Ta có: \(A=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

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

\(1,A=\dfrac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ A=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ A=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\left(x\ge0;x\ne4;x\ne9\right)\\ 2,A< 1\Leftrightarrow\dfrac{\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}-3}< 0\\ \Leftrightarrow\dfrac{4}{\sqrt{x}-3}< 0\Leftrightarrow\sqrt{x}-3< 0\Leftrightarrow0\le x< 9\)