Thấy gì đâu??

._.?

Bài 2:

a: Thay x=-2 và y=-1 vào (d), ta được:

-2(m+1)+m+2=-1

=>-2m-2+m+2=-1

=>-m=-1

=>m=1

b: (d): y=2x+3

Tọa độ A là:

y=0 và 2x+3=0

=>x=-3/2 và y=0

=>OA=1,5

Tọa độ B là:

x=0 và y=2*0+3=3

=>OB=3

\(AB=\sqrt{1.5^2+3^2}=1.5\sqrt{5}\)

=>\(C=1.5+3+1.5\sqrt{5}=1.5\sqrt{5}+4.5\)

\(S=\dfrac{1}{2}\cdot OA\cdot OB=2.25\)

Ai giúp mình zới, khó quá huhuhuhuhuh

Bài 22:

a: \(\frac{x}{x-1}-\frac{2}{x+1}-\frac{2}{x^2-1}\)

\(=\frac{x\left(x+1\right)-2\left(x-1\right)-2}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{x^2-x-2x+2-2}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{x^2-3x}{x^2-1}\)

b: \(\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{3x-6}{4-9x^2}\)

\(=\frac{1}{3x-2}-\frac{4}{3x+2}+\frac{3x-6}{\left(3x-2\right)\left(3x+2\right)}\)

\(=\frac{3x+2-4\left(3x-2\right)+3x-6}{\left(3x-2\right)\left(3x+2\right)}=\frac{6x-4-12x+8}{\left(3x-2\right)\left(3x+2\right)}\)

\(=\frac{-6x+4}{\left(3x-2\right)\left(3x+2\right)}=\frac{-2\left(3x-2\right)}{\left(3x-2\right)\left(3x+2\right)}=\frac{-2}{3x+2}\)

c: \(\frac{x^2}{x^2-x}-\frac{x^2}{x+1}-\frac{2x}{x^2-1}\)

\(=\frac{x}{x-1}-\frac{x^2}{x+1}-\frac{2x}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{x\left(x+1\right)-x^2\left(x-1\right)-2x}{\left(x-1\right)\left(x+1\right)}=\frac{x^2+x-x^3+x^2-2x}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{-x^3+2x^2-x}{\left(x-1\right)\left(x+1\right)}=\frac{-x\left(x^2-2x+1\right)}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{-x\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\frac{-x\left(x-1\right)}{x+1}\)

d: \(\frac{4x^2-3x+5}{x^3-1}-\frac{1-2x}{x^2+x+1}-\frac{6}{x-1}\)

\(=\frac{4x^2-3x+5+\left(2x-1\right)\left(x-1\right)-6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\frac{4x^2-3x+5+2x^2-3x+1-6x^2-6x-6}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{-12x}{x^3-1}\)

e: \(\frac{5}{2x^2+6x}-\frac{4-3x^2}{x^2-9}-3\)

\(=\frac{5}{2x\left(x+3\right)}+\frac{3x^2-4}{\left(x-3\right)\left(x+3\right)}-3\)

\(=\frac{5\left(x-3\right)+2x\left(3x^2-4\right)-3\cdot2x\left(x-3\right)\left(x+3\right)}{2x\left(x-3\right)\left(x+3\right)}\)

\(=\frac{5x-15+6x^3-8x-6x\left(x^2-9\right)}{2x\left(x-3\right)\left(x+3\right)}=\frac{6x^3-3x-15-6x^3+54x}{2x\left(x-3\right)\left(x+3\right)}\)

\(=\frac{51x-15}{2x\left(x-3\right)\left(x+3\right)}\)

f: \(\frac{5}{x+1}-\frac{10}{x-x^2-1}-\frac{15}{x^3+1}\)

\(=\frac{5}{x+1}+\frac{10}{x^2-x+1}-\frac{15}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(=\frac{5\left(x^2-x+1\right)+10\left(x+1\right)-15}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{5x^2-5x+5+10x+10-15}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(=\frac{5x^2+5x}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{5x}{x^2-x+1}\)

Bài 24:

\(\frac{1}{x}-\frac{1}{x+1}\)

\(=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x\left(x+1\right)}\)

a: \(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}\)

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

\(=\frac{1}{x}-\frac{1}{x+4}=\frac{x+4-x}{x\left(x+4\right)}=\frac{4}{x\left(x+4\right)}\)

b: \(\frac{1}{x^2+x}+\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}+\frac{1}{x+5}\)

\(=\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{x+5}\)

\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}\)

\(=\frac{1}{x}\)

a, Vì ABCD là hbh nên AB//CD

Do đó \(\widehat{A}+\widehat{D}=180^0\Rightarrow3\widehat{D}=180^0\Rightarrow\widehat{D}=60^0\Rightarrow\widehat{A}=120^0\)

Mà ABCD là hbh nên \(\left\{{}\begin{matrix}\widehat{A}=\widehat{C}=120^0\\\widehat{D}=\widehat{B}=60^0\end{matrix}\right.\)

b, Vì CE=CB nên tam giác CEB cân tại C

Do đó \(\widehat{B}=\widehat{CEB}\)

\(\Rightarrow\widehat{D}=\widehat{CEB}\left(1\right)\)

Mà ABCD là hbh nên AB//CD hay AE//CD

Do đó AECD là hình thang

Kết hợp (1) ta được AECD là hthang cân