xét tứ giác AFCD có EA=EC;ED=EF nên tứ giác AFCD là hình bình hành

\(B=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{110}\)

\(=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{10.11}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{10}-\dfrac{1}{11}\)

\(=\dfrac{1}{2}-\dfrac{1}{11}< \dfrac{1}{2}\)

Bài 6: 

\(A=\left(x+\dfrac{1}{2}\right)^2+\left|2y-\dfrac{3}{4}\right|+\dfrac{175}{3}\ge\dfrac{175}{3}\forall x,y\)

Dấu '=' xảy ra khi \(\left(x,y\right)=\left(-\dfrac{1}{2};\dfrac{3}{8}\right)\)

7:

a: \(\widehat{MHK}=\widehat{MHP}+\widehat{PHK}=25^0+50^0=75^0\)

\(\Leftrightarrow\widehat{MHK}=\widehat{NMH}\)

mà hai góc này là hai góc ở vị trí so le trong

nên MN//HK

b: \(\widehat{QPH}+\widehat{PHK}=180^0\)

mà hai góc này là hai góc ở vị trí trong cùng phía

nên PQ//HK

mà MN//HK

nên MN//PQ

c: MH nằm giữa MN và MP

=>\(\widehat{NMH}+\widehat{HMP}=\widehat{NMP}\)

=>\(\widehat{PMH}=117^0-75^0=42^0\)

Xét ΔPMH có \(\widehat{MPH}+\widehat{PMH}+\widehat{PHM}=180^0\)

=>\(\widehat{MPH}=180^0-42^0-25^0=113^0\)

\(\widehat{MPH}+\widehat{MPQ}+\widehat{HPQ}=360^0\)

=>\(\widehat{MPQ}=360^0-113^0-130^0=117^0\)

d: HK//PQ

d\(\perp\)HK

Do đó: d\(\perp\)PQ

HK//MN

d\(\perp\)HK

Do đó: d\(\perp\)MN

Bài 7:

\(a,A=\dfrac{2a+a-3}{a-3}\cdot\dfrac{\left(a-3\right)\left(a+3\right)}{3}=\dfrac{3\left(a-1\right)\left(a+3\right)}{3}=\left(a-1\right)\left(a+3\right)\\ b,B=\dfrac{b+3-6}{b+3}:\dfrac{b^2-9-b^2+10}{\left(b-3\right)\left(b+3\right)}\\ B=\dfrac{b-3}{b+3}\cdot\left(b-3\right)\left(b+3\right)=\left(b-3\right)^2\)

Bài 8:

\(a,M=\dfrac{4m^2-4mn+n^2}{m^2}:\dfrac{n-2m}{mn}=\dfrac{\left(n-2m\right)^2}{m^2}\cdot\dfrac{mn}{n-2m}=\dfrac{n\left(n-2m\right)}{m}\\ b,N=\dfrac{1}{3}+x:\dfrac{x+3-x}{x+3}=\dfrac{1}{3}+x\cdot\dfrac{x+3}{3}=\dfrac{1+x^2+3x}{3}\)

Bài 8: 

b: \(N=\dfrac{1}{3}+\dfrac{x}{\dfrac{x+3-x}{x+3}}=\dfrac{1}{3}+\dfrac{x}{\dfrac{3}{x+3}}=\dfrac{1}{3}+\dfrac{x+3}{3x}=\dfrac{x+x+3}{3x}=\dfrac{2x+3}{3x}\)

Bài 7:

a)ĐKXĐ:\(\left\{{}\begin{matrix}x\ge m+1\\x\ge\dfrac{m}{4}\end{matrix}\right.\)

TH1: \(m+1< \dfrac{m}{4}\Rightarrow m< -\dfrac{4}{3}\)

\(\Rightarrow x\ge\dfrac{m}{4}\)\(\Rightarrow x\in\)\([\dfrac{m}{4};+\)\(\infty\)\()\)

Để hàm số xác định với mọi x dương \(\Leftrightarrow\)\(\left(0;+\infty\right)\subset\)\([\dfrac{m}{4};+\)\(\infty\)\()\)

\(\Leftrightarrow\dfrac{m}{4}\ge0\Leftrightarrow m\ge0\) kết hợp với \(m< -\dfrac{4}{3}\Rightarrow m\in\varnothing\)

TH2:\(m+1\ge\dfrac{m}{4}\Rightarrow m\ge-\dfrac{4}{3}\)

\(\Rightarrow x\ge m+1\)\(\Rightarrow\)\(x\in\)\([m+1;+\)\(\infty\))

Để hàm số xác định với mọi x dương \(\Leftrightarrow\)\(\left(0;+\infty\right)\subset\)\([m+1;\)\(+\infty\)\()\)

\(\Leftrightarrow m+1\le0\Leftrightarrow m\le-1\) kết hợp với \(m\ge-\dfrac{4}{3}\)

\(\Rightarrow m\in\left[-\dfrac{4}{3};-1\right]\)

Vậy...

b)ĐKXĐ:\(\left\{{}\begin{matrix}x\ge2-m\\x\ne-m\end{matrix}\right.\)\(\Rightarrow x\in\)\([2-m;+\)\(\infty\)) (vì \(-m< 2-m\))

Để hàm số xác ddingj với mọi x dương

\(\Leftrightarrow\left(0;+\infty\right)\subset\)\([2-m;+\)\(\infty\))

\(\Leftrightarrow2-m\le0\Leftrightarrow m\ge2\)

Vậy...

Bài 9:

a)Đặt \(f\left(x\right)=x^2+2x-2\)

TXĐ:\(D=R\)

TH1:\(x\in\left(-\infty;-1\right)\)

Lấy \(x_1;x_2\in\left(-\infty;-1\right)\)\(:x_1\ne x_2\) 

Xét \(I=\dfrac{f\left(x_1\right)-f\left(x_2\right)}{x_1-x_2}=\dfrac{x_1^2+2x_1-2-\left(x_2^2+2x_2-2\right)}{x_1-x_2}=x_1+x_2+2\)

Vì \(x_1;x_2\in\left(-\infty;-1\right)\Rightarrow x_1+x_2< -1+-1=-2\)\(\Leftrightarrow x_1+x_2+2< 0\)

\(\Rightarrow I< 0\)

Suy ra hàm nb trên \(\left(-\infty;-1\right)\)

TH2:\(x\in\left(-1;+\infty\right)\)

Lấy \(x_1;x_2\in\left(-1;+\infty\right)\)\(:x_1\ne x_2\) 

Xét \(I=\dfrac{f\left(x_1\right)-f\left(x_2\right)}{x_1-x_2}=\dfrac{x_1^2+2x_1-2-\left(x_2^2+2x_2-2\right)}{x_1-x_2}=x_1+x_2+2>0\)

Suy ra hàm đb trên \(\left(-1;+\infty\right)\)

Vậy...

b)Đặt \(f\left(x\right)=\dfrac{2}{x-3}\)

TXĐ:\(D=R\backslash\left\{3\right\}\)

TH1:\(x\in\left(-\infty;3\right)\)

Lấy \(x_1;x_2\in\left(-\infty;3\right)\)\(:x_1\ne x_2\) 

Xét \(I=\dfrac{f\left(x_1\right)-f\left(x_2\right)}{x_1-x_2}=\dfrac{\dfrac{2}{x_1-3}-\dfrac{2}{x_2-3}}{x_1-x_2}=\dfrac{-2}{\left(x_1-3\right)\left(x_2-3\right)}\)

Vì \(x_1;x_2\in\left(-\infty;3\right)\Rightarrow x_1-3< 0;x_2-3< 0\Rightarrow\left(x_1-3\right)\left(x_2-3\right)>0\)

\(\Rightarrow I< 0\)

Suy ra hàm nb trên \(\left(-\infty;3\right)\)

TH2:\(x\in\left(3;+\infty\right)\)

Lấy \(x_1;x_2\in\left(3;+\infty\right)\)\(:x_1\ne x_2\) 

Xét \(I=\dfrac{f\left(x_1\right)-f\left(x_2\right)}{x_1-x_2}=\dfrac{\dfrac{2}{x_1-3}-\dfrac{2}{x_2-3}}{x_1-x_2}=\dfrac{-2}{\left(x_1-3\right)\left(x_2-3\right)}\)

Vì \(x_1;x_2\in\left(3;+\infty\right)\Rightarrow x_1-3>0;x_2-3>0\Rightarrow\left(x_1-3\right)\left(x_2-3\right)>0\)

\(\Rightarrow I< 0\)

Suy ra hàm nb trên \(\left(3;+\infty\right)\)

Vậy hàm nb trên \(\left(-\infty;3\right)\) và \(\left(3;+\infty\right)\)

Bài 7:

\(\sqrt{4x^2-4x+1}=2-x\\ \sqrt{\left(2x-1\right)^2}=2-x\\ \left|2x-1\right|=2-x\\ 2x+x=2+1\\ 3x=3\\ x=\dfrac{3}{3}=1\)

Hình 1: 

a: Ta có: AC//BD

AB\(\perp\)AC

Do đó: BD\(\perp\)AB