a: \(=\dfrac{48}{36}+\dfrac{7}{5}\cdot\dfrac{20}{21}=\dfrac{4}{3}+\dfrac{4}{3}=\dfrac{8}{3}\)

b: \(=\dfrac{7}{3}-\dfrac{1}{3}\cdot\left[-\dfrac{3}{2}+\dfrac{2}{3}+\dfrac{4}{10}\cdot5\right]\)

\(=\dfrac{7}{3}-\dfrac{1}{3}\cdot\dfrac{7}{6}=\dfrac{7}{3}-\dfrac{7}{18}=\dfrac{42-7}{18}=\dfrac{35}{18}\)

c: \(=\left(29+\dfrac{1}{4}\right):\dfrac{9}{4}=\dfrac{117}{4}\cdot\dfrac{4}{9}=\dfrac{117}{9}=13\)

d: \(=\left(4-\dfrac{4}{5}\right)\cdot\dfrac{11}{8}-\dfrac{8}{5}\cdot4\)

\(=\dfrac{16}{5}\cdot\dfrac{11}{8}-\dfrac{32}{5}\)

\(=\dfrac{22}{5}-\dfrac{32}{5}=-\dfrac{10}{5}=-2\)

Cách 2: 

if (b%2==0) cout<<"b la so chan";

else cout<<"b la so le";

nhân chéo

x=11?

a: \(=\left(2+\dfrac{4}{9}+4+\dfrac{5}{9}\right)+\left(-4-\dfrac{13}{21}-2-\dfrac{8}{21}\right)\)

=7-7

=0

b: \(=\dfrac{5}{23}\left(\dfrac{7}{13}+\dfrac{6}{13}\right)+\dfrac{18}{23}=\dfrac{5}{23}+\dfrac{18}{23}=1\)

c: \(=\dfrac{3}{7}\left(\dfrac{12}{19}-\dfrac{4}{19}\right)+\dfrac{4}{7}\cdot\dfrac{8}{19}\)

\(=\dfrac{3}{7}\cdot\dfrac{8}{19}+\dfrac{4}{7}\cdot\dfrac{8}{19}=\dfrac{8}{19}\)

d: \(=\dfrac{2}{5}+\dfrac{3}{5}:\dfrac{9-10}{15}-\dfrac{7}{2}\)

\(=\dfrac{-31}{10}+\dfrac{3}{5}\cdot\dfrac{-15}{1}=\dfrac{-31}{10}-9=\dfrac{-121}{10}\)

1 He lives with his family in a small town near HN

2 They are workers

3 They go there by motorbike 

 khi có ánh sáng từ vật đó truyền vào mắt ta

Khi ánh sáng truyền vào mắt ta

a: x^3+8=(x+2)(x^2-2x+4)

b: =(3x+1)(9x^2-3x+1)

c: =(x+3)(x^2-3x+9)

d: =(4x-3y)(16x^2+24xy+9y^2)

\(a.x^3+8=\left(x+2\right)\left(x^2-2x+4\right)\) 

\(b.27x^3+1=\left(3x+1\right)\left(9x-3x+1\right)\)

\(c.x^3+27=\left(x+3\right)\left(x^2-3x+9\right)\) 

\(d.64x^3-27y^3=\left(4x-3y\right)\left(16x^2+12xy+9y^2\right)\)  

Sửa đề: cắt DC tại M

a: Xét tứ giác ABMD có

AB//MD

AD//MB

Do đó: ABMD là hình bình hành

b: Ta có: ABMD là hình bình hành

=>AD=BM

mà AD=BC(ABCD là hình thang cân)

nên BM=BC

=>ΔBMC cân tại B