\(\dfrac{M}{N}=\left(\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{3-\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\right)\) (ĐKXĐ: \(x\ge0;x\ne4;x\ne9\))

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

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

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

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

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

\(\Rightarrow P=\dfrac{M}{N}+1=\dfrac{2}{\sqrt{x}+2}+1\)

Ta thấy: \(\sqrt{x}\ge0\forall x\)

\(\Rightarrow\sqrt{x}+2\ge2\forall x\)

\(\Rightarrow\dfrac{2}{\sqrt{x}+2}\le1\forall x\)

\(\Rightarrow\dfrac{2}{\sqrt{x}+2}+1\le2\forall x\)

\(\Rightarrow Max_P=2\Leftrightarrow\dfrac{2}{\sqrt{x}+2}+1=2\)

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

\(\Leftrightarrow\sqrt{x}+2=2\)

\(\Leftrightarrow\sqrt{x}=0\)

\(\Leftrightarrow x=0\left(tm\right)\)

#Urushi☕

Bạn tự rút gọn nha .

c) Ta có : \(P\text{=}\dfrac{M}{N}+1\text{=}\dfrac{2}{\sqrt{x}+2}+1\)

Để P có giá trị lớn nhất.

\(\Leftrightarrow\dfrac{2}{\sqrt{x}+2}cóGTLN\)

\(\Leftrightarrow\sqrt{x}+2cóGTNN\)

Mà : \(\sqrt{x}+2\ge2\)

\(\Rightarrow\) Để : \(\left(\sqrt{x}+2\right)_{min}\) \(\Leftrightarrow\sqrt{x}\text{=}0\Leftrightarrow x\text{=}0\)

Vậy............

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

a) \(P=\dfrac{3x+3\sqrt{x}-9}{x+\sqrt{x}-2}+\dfrac{\sqrt{x}+3}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\left(x\ge0,x\ne1\right)\)

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

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

\(=\dfrac{3x+5\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}\)

b) \(P=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}=\dfrac{3\sqrt{x}+6+2}{\sqrt{x}+2}=3+\dfrac{2}{\sqrt{x}+2}\)

Để \(P\in Z\Rightarrow2⋮\sqrt{x}+2\Rightarrow\sqrt{x}+2=2\left(\sqrt{x}+2\ge2\right)\)

\(\Rightarrow x=0\)

c) Ta có: \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+2\ge2\Rightarrow\dfrac{2}{\sqrt{x}+2}\le1\Rightarrow3+\dfrac{2}{\sqrt{x}+2}\le4\)

\(\Rightarrow P_{max}=4\) khi \(x=0\)

Ta có: \(x=9-4\sqrt{5}\)

⇔ \(\sqrt{x}=\sqrt{9-4\sqrt{5}}=\sqrt{5-4\sqrt{5}+4}\)

⇔ \(\sqrt{x}=\sqrt{\left(\sqrt{5}-2\right)^2}=\left|\sqrt{5}-2\right|\)

⇔ \(\sqrt{x}=\sqrt{5}-2\)   

Khi đó:    \(P=\dfrac{1-\sqrt{5}+2}{\sqrt{5}-2+2}=\dfrac{3-\sqrt{5}}{\sqrt{5}}\)

\(P=\sqrt{x}+\sqrt{9-x}+\sqrt{x\left(9-x\right)}\ge\sqrt{x}+\sqrt{9-x}\)

\(\Rightarrow P^2\ge\left(\sqrt{x}+\sqrt{9-x}\right)^2=9+2\sqrt{x\left(9-x\right)}\ge9\)

\(\Rightarrow P\ge3\)

\(P_{\min}=3\) khi x=0 hoặc x=9

\(P=\sqrt{x}+\sqrt{9-x}+\sqrt{x\left(9-x\right)}\le\sqrt{2\left(x+9-x\right)}+\frac12\left(x+9-x\right)=\frac92+3\sqrt2\)

\(P_{max}=\frac92+3\sqrt2\) khi \(x=9-x\Rightarrow x=\frac92\)

\(P = \frac{x}{9 - x} + x \left(\right. 9 - x \left.\right)\)

\(P = \frac{x}{9 - x} + x \left(\right. 9 - x \left.\right)\)

Đạo hàm từng phần:

\(u = x , v = 9 - x \Rightarrow u^{'} = 1 , v^{'} = - 1\)\(\left(\left(\right. \frac{u}{v} \left.\right)\right)^{'} = \frac{u^{'} v - u v^{'}}{v^{2}} = \frac{1 \cdot \left(\right. 9 - x \left.\right) - x \cdot \left(\right. - 1 \left.\right)}{\left(\right. 9 - x \left.\right)^{2}} = \frac{9 - x + x}{\left(\right. 9 - x \left.\right)^{2}} = \frac{9}{\left(\right. 9 - x \left.\right)^{2}}\)

\(9 - 2 x\)

Vậy đạo hàm của \(P\) là:

\(P^{'} = \frac{9}{\left(\right. 9 - x \left.\right)^{2}} + 9 - 2 x\)

\(\frac{9}{\left(\right. 9 - x \left.\right)^{2}} + 9 - 2 x = 0\)

Chuyển vế:

\(\frac{9}{\left(\right. 9 - x \left.\right)^{2}} = 2 x - 9\)

Lưu ý: Để vế phải \(2 x - 9\) dương (vì vế trái luôn dương), ta có:

\(2 x - 9 > 0 \Rightarrow x > \frac{9}{2} = 4.5\)

Nhân hai vế với \(\left(\right. 9 - x \left.\right)^{2}\):

\(9 = \left(\right. 2 x - 9 \left.\right) \left(\right. 9 - x \left.\right)^{2}\)

Đặt \(t = 9 - x\), khi \(x > 4.5 \Rightarrow t = 9 - x < 4.5\).

Thay \(x = 9 - t\):

\(9 = \left(\right. 2 \left(\right. 9 - t \left.\right) - 9 \left.\right) \cdot t^{2} = \left(\right. 18 - 2 t - 9 \left.\right) t^{2} = \left(\right. 9 - 2 t \left.\right) t^{2}\)

Ta có:

\(9 = \left(\right. 9 - 2 t \left.\right) t^{2} = 9 t^{2} - 2 t^{3}\)

Chuyển hết về một phía:

\(9 t^{2} - 2 t^{3} - 9 = 0\)

Hay:

\(- 2 t^{3} + 9 t^{2} - 9 = 0\)

Nhân cả phương trình với -1 để thuận tiện:

\(2 t^{3} - 9 t^{2} + 9 = 0\)

Thử các nghiệm nguyên hoặc hữu tỉ:

\(2 \left(\right. 1 \left.\right)^{3} - 9 \left(\right. 1 \left.\right)^{2} + 9 = 2 - 9 + 9 = 2 \neq 0\)

\(2 \left(\right. 27 \left.\right) - 9 \left(\right. 9 \left.\right) + 9 = 54 - 81 + 9 = - 18 \neq 0\)

\(2 \left(\right. 4.5 \left.\right)^{3} - 9 \left(\right. 4.5 \left.\right)^{2} + 9 = 2 \cdot 91.125 - 9 \cdot 20.25 + 9 = 182.25 - 182.25 + 9 = 9 \neq 0\)

\(2 \left(\right. 8 \left.\right) - 9 \left(\right. 4 \left.\right) + 9 = 16 - 36 + 9 = - 11 \neq 0\)

Không tìm được nghiệm nguyên, dùng phương pháp đồ thị hoặc nghiệm gần đúng.

Ta có hàm:

\(f \left(\right. t \left.\right) = 2 t^{3} - 9 t^{2} + 9\)

Vậy nghiệm nằm trong khoảng \(\left(\right. 4 , 5 \left.\right)\).

Tiếp tục thử \(t = 4.5\):

\(f \left(\right. 4.5 \left.\right) = 2 \cdot 91.125 - 9 \cdot 20.25 + 9 = 182.25 - 182.25 + 9 = 9 > 0\)

Có vẻ trước đó tính sai, ta kiểm tra lại:

\(t = 4.25 \Rightarrow f \left(\right. 4.25 \left.\right) = 2 \cdot \left(\right. 4.25 \left.\right)^{3} - 9 \cdot \left(\right. 4.25 \left.\right)^{2} + 9\)\(\left(\right. 4.25 \left.\right)^{3} = 76.765625 , \left(\right. 4.25 \left.\right)^{2} = 18.0625\)\(f \left(\right. 4.25 \left.\right) = 2 \cdot 76.765625 - 9 \cdot 18.0625 + 9 = 153.53125 - + 9 = - 0.03125\)

Gần bằng 0, nghiệm ở gần \(4.25\).

\(t \approx 4.25 \Rightarrow x = 9 - t = 9 - 4.25 = 4.75\)

\(P = \frac{4.75}{9 - 4.75} + 4.75 \left(\right. 9 - 4.75 \left.\right) = \frac{4.75}{4.25} + 4.75 \times 4.25\)\(\frac{4.75}{4.25} \approx 1.1176 , 4.75 \times 4.25 = 20.1875\)\(P \approx 1.1176 + 20.1875 = 21.3051\)

\(P \rightarrow \frac{0}{9} + 0 \times 9 = 0\)

\(\frac{x}{9 - x} \rightarrow + \infty , x \left(\right. 9 - x \left.\right) \rightarrow 0\)

Nên \(P \rightarrow + \infty\).

Vì \(P \rightarrow + \infty\) gần biên \(x \rightarrow 9^{-}\), nên không có GTLN hữu hạn trên khoảng \(\left(\right. 0 , 9 \left.\right)\).

Còn GTNN là khoảng \(x \rightarrow 0\) hoặc tại cực trị \(x = 4.75\).

Tham khảo

giải bằng Bunhiaskopki nha bạn, search gg

Ta có P \(\le\dfrac{1^2+\left(\sqrt{x-1}\right)^2}{2}+\dfrac{2^2+\left(\sqrt{y-4}\right)^2}{2}+\dfrac{3^2+\left(\sqrt{z-9}\right)^2}{2}\)

\(=\dfrac{1+x-1+4+y-4+9+z-9}{2}=\dfrac{x+y+z}{2}=\dfrac{28}{2}=14\)

Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}1=\sqrt{x-1}\\2=\sqrt{y-4}\\3=\sqrt{z-9}\end{matrix}\right.\Leftrightarrow x=2;y=8;z=18\)(tm)