Ta có \(-\dfrac{4ab^2}{4b^2+1}\ge-\dfrac{4ab^2}{2\sqrt{4b^2}}=\dfrac{4ab^2}{4b}=ab\)

\(-\dfrac{4a^2b}{4a^2+1}\ge-\dfrac{4a^2b}{2\sqrt{4a^2}}=\dfrac{4a^2b}{4a}=ab\)

Mà \(\dfrac{a}{4b^2+1}+\dfrac{b}{4a^2+1}=\dfrac{a\left(4b^2+1\right)}{4b^2+1}-\dfrac{4ab^2}{4b^2+1}+\dfrac{b\left(4a^2+1\right)}{4a^2+1}-\dfrac{4ab^2}{4a^2+1}\ge a-ab+b-ab=4ab-2ab=2ab\)

Mà \(a+b=4ab\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=4\ge\dfrac{2}{2\sqrt{ab}}\Rightarrow4\sqrt{ab}\ge2\Rightarrow ab\ge\dfrac{1}{4}\)

\(\Rightarrow2ab\ge\dfrac{1}{2}\Rightarrow\dfrac{a}{4b^2+1}+\dfrac{b}{4a^2+1}\ge\dfrac{1}{2}\)

Dấu "=" \(\Leftrightarrow a=b=\dfrac{1}{2}\)

20

a) \(A=4x-x^2+3=-\left(x-2\right)^2+7\le7\)

Dấu "=" \(\Leftrightarrow x=2\)

b) \(A=x-x^2=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le-\dfrac{1}{4}\)

Dấu "=" \(\Leftrightarrow x=\dfrac{1}{2}\)

c) \(N=2x-2x^2-5=-2\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{9}{2}=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}\le-\dfrac{9}{2}\)

Dấu "=" \(\Leftrightarrow x=\dfrac{1}{2}\)

19

a) \(P=x^2-2x+5=\left(x-1\right)^2+4\ge4\)

Dấu "=" \(\Leftrightarrow x=1\)

b) \(Q=2x^2-6x=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)

Dấu "=" \(\Leftrightarrow x=\dfrac{3}{2}\)

c) \(M=x^2+y^2-x+6y+10=\left(x^2-x+\dfrac{1}{4}\right)+\left(y^2+6y+9\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

Dấu "=" \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-3\end{matrix}\right.\)

18.

a) \(x^2-6x+10=\left(x^2-6x+9\right)+1=\left(x-3\right)^2+1\ge1\)

Dấu "="\(\Leftrightarrow x=3\)

b)  \(4x-x^2-10=-\left(x^2-4x+4\right)-6=-\left(x-2\right)^2-6\le-6\)

Dấu "=" \(\Leftrightarrow x=2\)

2. 

a) This film is more popular than that one.

b) A bird can fly slower that a plane.

c) Ba is more friendly than Nam.

d) Your house is farther from school than my house.

e) My father drives more careful than I.

1.

a) Mary is as thin as her brother.

b) A lemon is not as sweet as an orange.

c) A donkey cannot run as quick as a horse.

d) This dress is twice as pretty as that one.

e)They don't drive a car as dangerous as us.

\(\sqrt{\left(1-\sqrt{2020}\right)^2\cdot\left(\sqrt{2021-2\sqrt{2020}}\right)}\)

\(=\sqrt{\left(1-\sqrt{2020}\right)^2\cdot\sqrt{\left(1-\sqrt{2020}\right)^2}}\)

\(=\sqrt{\left(1-\sqrt{2020}\right)^2\cdot\left(1-\sqrt{2020}\right)}\)

\(=\sqrt{\left(1-\sqrt{2020}\right)^3}=\left(1-\sqrt{2020}\right)\sqrt{1-\sqrt{2020}}\)

Ta có \(\left(\sqrt{2}+\sqrt{3}\right)^2=5+2\sqrt{6};\left(\sqrt{10}\right)^2=10=5+5\)

Mà \(\left(2\sqrt{6}\right)^2=24;5^2=25\)

\(\Rightarrow2\sqrt{6}< 5\Rightarrow\left(\sqrt{2}+\sqrt{3}\right)^2< \left(\sqrt{10}\right)^2\)

\(\Rightarrow\sqrt{2}+\sqrt{3}< \sqrt{10}\)

Vì tổng 2 cạnh luôn lớn hơn 1 cạnh nên ∆ cân tại C \(\Rightarrow\)CA = 10 
chu vi = 10+10+4=24

Gọi h là đường cao kẻ từ đỉnh C thì h cũng là trung tuyến
h² =10² - 2² suy ra h=\(4\sqrt{6}\)
Diện tích = \(\dfrac{4\cdot4\sqrt{6}}{2}=8\sqrt{6}\)