2:

a: Xét ΔBAD có BA=BD

nên ΔBAD cân tại B

ΔBAD cân tại B có BE là phân giác

nên BE vuông góc AD

b: Xét ΔBAE và ΔBDE có

BA=BD

góc ABE=góc DBE

BE chung

=>ΔBAE=ΔBDE

=>EA=ED

=>ΔEAD cân tại E

c: Xét ΔEAF vuông tại A và ΔEDC vuông tại D có

EA=ED

AF=DC

=>ΔEAF=ΔEDC

=>EF=EC

=>ΔECF cân tại E

d: ΔEAF=ΔEDC

=>góc AEF=góc DEC

=>góc AEF+góc AED=180 độ

=>D,E,F thẳng hàng

g: \(=\dfrac{x^2+2x-x^2-4x-2x+4}{x\left(x-2\right)\left(x+2\right)}=\dfrac{-4x+4}{x\left(x-2\right)\left(x+2\right)}\)

h: \(=\dfrac{2x^2+1-x^2+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}\)

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

\(e,=\dfrac{1}{x-1}-\dfrac{2x}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{x^2-2x+1}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{\left(x-1\right)^2}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{x-1}{x^2+1}\\ f,=\dfrac{3x-1}{2\left(3x+1\right)}+\dfrac{3x+1}{2\left(3x-1\right)}-\dfrac{6x}{\left(3x-1\right)\left(3x+1\right)}\\ =\dfrac{9x^2-6x+1+9x^2+6x+1-12x}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{2\left(3x-1\right)^2}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{3x-1}{3x+1}\)

\(g,=\dfrac{x}{x\left(x-2\right)}-\dfrac{x^2+4x}{x\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x\left(x+2\right)}\\ =\dfrac{x^2+2x-x^2-4x-2x+4}{x\left(x-2\right)\left(x+2\right)}=\dfrac{-4x+4}{x\left(x-2\right)\left(x+2\right)}\\ h,=\dfrac{2x^2+1-x^2+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{1}{x^2-x+1}\)

Bài 2.8
Cửa hàng lãi:\(\left(1000000:800000\right)-100\%=25\%\left(giavon\right)\)
Bài 2.9
Số tiền phải bán để lãi 25% giá vốn:\(740000+\left(740000\times25\%\right)==925000\left(đồng\right)\)

a) \(\Rightarrow\left|\dfrac{3}{4}+x\right|=0\Rightarrow\dfrac{3}{4}+x=0\Rightarrow x=-\dfrac{3}{4}\)

b) \(\Rightarrow x+0,4=\dfrac{4}{9}:\dfrac{2}{3}=\dfrac{2}{3}\Rightarrow x=\dfrac{2}{3}-0,4=\dfrac{4}{15}\)

thế e hỏi j :)?

Bài 1:

\(a,\dfrac{25}{14x^2y}=\dfrac{75y^4}{42x^2y^5};\dfrac{14}{21xy^5}=\dfrac{28x}{42x^2y^5}\\ b,\dfrac{3x+1}{12xy^4}=\dfrac{3x\left(3x+1\right)}{36x^2y^4};\dfrac{y-2}{9x^2y^3}=\dfrac{4y\left(y-2\right)}{36x^2y^4}\\ c,\dfrac{1}{6x^3y^2}=\dfrac{6y^2}{36x^3y^4};\dfrac{x+1}{9x^2y^4}=\dfrac{4x\left(x+1\right)}{36x^3y^4};\dfrac{x-1}{4xy^3}=\dfrac{9x^2y\left(x-1\right)}{36x^3y^4}\\ d,\dfrac{3+2x}{10x^4y}=\dfrac{12y^4\left(3+2x\right)}{120x^4y^5};\dfrac{5}{8x^2y^2}=\dfrac{75x^2y^3}{120x^4y^5};\dfrac{2}{3xy^5}=\dfrac{80x^3}{120x^4y^5}\)

Câu e,d à bạn