A ) Đặt 

\(A=0,1+0,2+...+1,9\\ \Rightarrow10A=1+2+3+..+19\\ =\left(1+19\right)\cdot\dfrac{19}{2}\\ =20\cdot\dfrac{19}{2}\\ =10\cdot19=190\\ \Rightarrow A=19\)

b) \(\left(1999\cdot1998+1998\cdot1997\right)\cdot\left(1+\dfrac{1}{2}:1\dfrac{1}{2}-1\dfrac{1}{3}\right)\)

\(=1998\cdot\left(1999+1997\right)\cdot\left(1+\dfrac{1}{2}:\dfrac{3}{2}-\dfrac{4}{3}\right)\)

\(=1998\cdot3996\cdot\left(1+\dfrac{1}{3}-\dfrac{4}{3}\right)\)

\(=1998\cdot3996\cdot0=0\)

B) Đặt 

\(B=\left(1999x1998+1998+1997\right)x\left(1+\dfrac{1}{2}:1\dfrac{1}{2}-1\dfrac{1}{3}\right)\\ =\left(1999x1998+1998+1997\right)x\left(1+\dfrac{1}{2}:\dfrac{3}{2}-\dfrac{4}{3}\right)\\ =\left(1999x1998+1998+1997\right)x\left(1+\dfrac{1}{2}\cdot\dfrac{2}{3}-\dfrac{4}{3}\right)=\\ =\left(1999x1998+1998+1997\right)x\left(1+\dfrac{1}{3}-\dfrac{4}{3}\right)\\ =\left(1999x1998+1998+1997\right)x\left(\dfrac{4}{3}-\dfrac{4}{3}\right)\\ =\left(1999x1998+1998+1997\right)x0\\ =0\)

`y : 5/11 = 2/5 : 2`

`=> y : 5/11 = 2/5 xx1/2`

`=> y : 5/11 = 2/10`

`=> y : 5/11 =1/5`

`=> y= 1/5 xx 5/11`

`=> y= 5/55`

`=> y=1/11`

\(y:\dfrac{5}{11}=\dfrac{2}{5}:2\)
\(y:\dfrac{5}{11}=\dfrac{2}{5}\times\dfrac{1}{2}\)
\(y:\dfrac{5}{11}=\dfrac{1}{5}\)
\(y=\dfrac{1}{5}\times\dfrac{5}{11}\)
\(y=\dfrac{1}{11}\)

vậy 

\(9x^2-x+\dfrac{1}{36}\)

\(=\left(3x\right)^2-2\cdot3x\cdot\dfrac{1}{6}+\left(\dfrac{1}{6}\right)^2\)

\(=\left(3x-\dfrac{1}{6}\right)^2\)

Chọn B. Thay \(\dfrac{1}{3}\)vào x và \(\dfrac{1}{2}\)vào y 

giải để ra được m

1: \(\Leftrightarrow x^2-6x=x^2-7x+10\)

hay x=10

2.

\(x-2\sqrt{x}=\sqrt{x}(\sqrt{x}-3)+\frac{1}{4}(\sqrt{x}-3)+\frac{3}{4}(\sqrt{x}+1)\)

\(\geq \frac{3}{4}(\sqrt{x}+1)\)

\(\Rightarrow I\leq \frac{\sqrt{x}+1}{\frac{3}{4}(\sqrt{x}+1)}=\frac{4}{3}\)

Vậy $I_{\max}=\frac{4}{3}$ tại $x=9$

1. Với $x\geq \frac{1}{2}$ thì:

\(3x+\sqrt{x}+1=(\sqrt{2x}-1)(\sqrt{\frac{9}{2}x}-1)+(1+\frac{5\sqrt{2}}{2})\sqrt{x}\)

\(\geq (1+\frac{5\sqrt{2}}{2})\sqrt{x}\)

\(\Rightarrow H=\frac{\sqrt{x}}{3x+\sqrt{x}+1}\leq \frac{\sqrt{x}}{(1+\frac{5\sqrt{2}}{2})\sqrt{x}}=\frac{1}{1+\frac{5\sqrt{2}}{2}}=\frac{5\sqrt{2}-2}{23}\)

Đây chính là $H_{\max}$. Giá trị này đạt tại $x=\frac{1}{2}$

Ta có: \(A=\dfrac{x}{\sqrt{y}}\) khi \(y=625\) và \(A< 0,2\)  

Nên: \(\dfrac{x}{\sqrt{625}}< 0,2\)

\(\Leftrightarrow\dfrac{x}{\sqrt{25^2}}< 0,2\)

\(\Leftrightarrow\dfrac{x}{25}< 0,2\)

\(\Leftrightarrow x< 0,2\cdot25\)

\(\Leftrightarrow x< 5\) 

Vậy khi \(y=625\) và \(A< 0,2\) khi và chỉ khi \(x< 5\)

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