Bài 6: Diện tích đa giác

Bài 4: 

a: \(A=\dfrac{x+3-x+3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{\left(x-3\right)^2}{6}=x-3\)

b: Thay x=2003 vào A, ta được:

A=2003-3=2000

b3:

a, 3x(x-1)+x-1=0

  (x-1)(3x+1)=0

=>x=1 hoặc x=\(-\dfrac{1}{3}\)

b,

\(4x^2+y^2-20x-2y+26=0\\ 4x^2+y^2-20x-2y+25+1=0\\ \left(4x^2-20x+25\right)+\left(y^2-2y+1\right)=0\\ \left(2x-5\right)^2+\left(y-1\right)^2=0\)

\(\left\{{}\begin{matrix}2x-5=0\\y-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=1\end{matrix}\right.\)

b4:

a,

\(A=\left(\dfrac{1}{x-3}-\dfrac{1}{x+3}\right):\dfrac{6}{x^2-6x+9}\\\left(\dfrac{\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right):\dfrac{6}{x^2-6x+9}\)

\(=\dfrac{x+3-x+3}{\left(x-3\right)\left(x+3\right)}:\dfrac{6}{x^2-6x+9}\\ =\dfrac{6}{\left(x-3\right)\left(x+3\right)}:\dfrac{6}{\left(x-3\right)\left(x-3\right)}=\dfrac{6\left(x-3\right)\left(x-3\right)}{6\left(x-3\right)\left(x+3\right)}\\ =\dfrac{x-3}{x+3}\)

b, thay x=2003 vào bt A ta được :

\(A=\dfrac{x-3}{x+3}=\dfrac{2003-3}{2003+3}=\dfrac{2000}{2006}=\dfrac{1000}{1003}\)

Lời giải:
$\frac{S_{ADE}}{S_{ABD}}=\frac{AE}{AB}=\frac{1}{2}$

$\frac{S_{ABD}}{S_{ABC}}=\frac{AD}{AC}=\frac{1}{3}$

Nhân theo vế thì: $\frac{S_{ADE}}{S_{ABC}}=\frac{1}{6}$

$\Rightarrow S_{ADE}=\frac{1}{6}.S_{ABC}=\frac{1}{6}.30=5$ (cm²)

$S_{BEDC}=S_{ABC}-S_{ADE}=30-5=25$ (cm²)

Hình vẽ:

Lời giải:

a. Diện tích: $3.4=12$ (cm²)

b. Gọi đường cao của hình bình hành là $h$ thì $h=3$ (cm)

$S_{ADM}=\frac{h.AM}{2}=\frac{h.DC}{4}=\frac{3.4}{4}=3$ (cm²)

c. Áp dụng định lý Talet:

$\frac{DN}{NM}=\frac{DC}{AM}=\frac{AB}{AM}=2$

$\Rightarrow DN=2NM$

d. Vì $\frac{DN}{NM}=2\Rightarrow \frac{NM}{DM}=\frac{1}{3}$

$\frac{S_{AMN}}{S_{ADM}}=\frac{NM}{DM}=\frac{1}{3}$

$\Rightarrow S_{AMN}=\frac{1}{3}.S_{ADM}=\frac{1}{3}.3=1$ (cm²)

Hình vẽ: