côg thức về tính j -;

1I haven't eaten Chinese food for 4 years

2I don't mind going anywhere you like

-4/5 : ( -9)/8

= 32/45

\(-\dfrac{4}{5}\div\left(\dfrac{5}{8}-\dfrac{7}{4}\right)=-\dfrac{4}{5}\div\left(-\dfrac{9}{8}\right)=\dfrac{32}{45}\)

= -4/5 : (-9)/8

= 32/45

4x^2 − y^2 + 10y − 25

= (2x)^2 − ( y^2 − 2.y.5 + 5^2)

= (2x)^2− (y−5)^2

= ( 2x−y+5 ) ( 2x+y−5 ) 

b: Xét tứ giác ADEF có 

AD//EF

AD=EF

Do đó: ADEF là hình bình hành

mà \(\widehat{EAF}=90^0\)

nên ADEF là hình chữ nhật

\(a,\dfrac{a+b}{a+b+c}+\dfrac{b+c}{a+b+c}+\dfrac{c+a}{a+b+c}=\dfrac{a+b+b+c+c+a}{a+b+c}=\dfrac{2a+2b+2c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)

\(b,\dfrac{6x-3}{x}:\dfrac{4x^2-1}{3x^2}=\dfrac{3x^2.3\left(2x-1\right)}{x\left(4x^2-1\right)}=\dfrac{9x\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{9x}{2x+1}\)

a)

\(=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)

b) \(=\dfrac{3\left(2x-1\right)}{x}.\dfrac{3x^2}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{9x}{2x+1}\)

Bài 3:

a: Ta có: \(4\left(x+3\right)-5\left(x-1\right)=-7\)

\(\Leftrightarrow4x+12-5x+5=-7\)

\(\Leftrightarrow-x=-24\)

hay x=24

b: Ta có: \(x\cdot\left(x-5\right)-x\left(x+2\right)=9\)

\(\Leftrightarrow x^2-5x-x^2-2x=9\)

\(\Leftrightarrow x=-\dfrac{9}{7}\)

a: =>3x=-16

hay x=-16/3

b: =>3x+6-18=2x-10

=>3x-2x=-10+12

=>x=2

c: =>\(x^2+10x+26=0\)

\(\Leftrightarrow\left(x+5\right)^2+1=0\)(vô lý)