Bài 5 :

a, ĐKXĐ ; \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

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

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

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

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

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

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

\(\Rightarrow P>3\)

\(P=1:\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{x-1}\right)\) (Đk:\(x\ge0;x\ne1\))

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

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

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

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

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

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

b) Áp dụng AM-GM có:

\(\sqrt{x}+\dfrac{1}{\sqrt{x}}\ge2\sqrt{\sqrt{x}.\dfrac{1}{\sqrt{x}}}=2\)

Dấu "=" xảy ra khi \(\sqrt{x}=\dfrac{1}{\sqrt{x}}\Leftrightarrow x=1\left(ktm\right)\)

\(\Rightarrow\)Dấu "=" không xảy ra

\(\Rightarrow\sqrt{x}+\dfrac{1}{\sqrt{x}}>2\)\(\Rightarrow\sqrt{x}+1+\dfrac{1}{\sqrt{x}}>3\) 

hay P>3

Vậy...

\(P=1:\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{x-1}\right)\left(x>0,x\ne1\right)\)

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

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

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

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

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

b) \(\dfrac{x+\sqrt{x}+1}{\sqrt{x}}=1+\sqrt{x}+\dfrac{1}{\sqrt{x}}\ge1+2\sqrt{\sqrt{x}.\dfrac{1}{\sqrt{x}}}=3\)

\(\Rightarrow P\ge3\)

Xét \(P=3\Rightarrow\sqrt{x}=\dfrac{1}{\sqrt{x}}\Rightarrow x=1\) (trái với ĐKXĐ)

\(\Rightarrow P>3\)

x chia hết cho 5 và 2 hả bn?

Ta có : \(\hept{\begin{cases}x⋮5\\x⋮2\end{cases}}\Rightarrow x\in BC\left(2;5\right)\)

Lại có : BCNN(2;5) = 2.5 = 10

mà BC(2;5) = B(10) = {0;10;20;30;40;50;60;70;80;90;100...}

Vì x < 90

=> x \(\in\){0;10;20;30;40;50;60;70;80}

\( S = 1-1/5 +1/5-1/9+1/9-1/13+1/13-1/17+1/17-1/21+1/21-1/25+1/25-1/29. \)

\(S= 1- 1/29 \)

\(S=\frac{28}{29}\)

Nếu mình ko nhầm!

6x=38-8

6x=30

x=30:6

x=5

5*x\(+8+x=38\)

6*x=38-8

6*x=30

x=30/6

x=5

5 * x + 8 + x = 38

5 * x + x + 8 = 38

6 * x + 8       = 38

6 * x             = 38 - 8

6 * x             = 30

     x             = 30 : 6

     x             = 5

cái bài dó mk chỉ giải cho bạn theo kiểu bấm máy thôi chứ trình bày sợ bạn không hiểu

Bạn ghi cách giải giúp mk đi ko sao đâu. Cám ơn bạn trước

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