20/23 x 4/5 + 30/23 x 4/5 - 8/46 x 4/5

= 20/23 x 4/5 + 30/23 x 4/5 - 4/23 x 4/5

= ( 20/23 + 30/23 - 4/23 ) x 4/5

=                 2            x         4/5

= 8/5

\(1=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}\)

\(1=\dfrac{7}{10}+\dfrac{1}{10}+\dfrac{1}{5}\)

\(1=\dfrac{1}{15}+\dfrac{2}{15}+\dfrac{4}{5}\)

\(1=\dfrac{10}{30}+\dfrac{15}{30}+\dfrac{5}{30}\)

\(1=\dfrac{1}{60}+\dfrac{11}{12}+\dfrac{1}{15}\)

\(18=\dfrac{x+\left(x+2\right)+\left(x+4\right)+\left(x+6\right)+\left(x+8\right)+\left(x+10\right)+\left(x+12\right)}{7}\\ 18=\dfrac{7x+42}{7}\\ \Rightarrow126=7x+42\\ \Rightarrow18=x+6\\ \Rightarrow x=12\)

vậy 7 số chẵn liên tiếp đó là: \(12;14;16;18;20;22;24\)

0,32 dm^{3  =}  320 cm³

C.320

a: Xét ΔABI và ΔACI có

AB=AC

BI=CI

AI chung

Do đó; ΔABI=ΔACI

b: Ta có: ΔAIB=ΔAIC

=>\(\widehat{AIB}=\widehat{AIC}\)

mà \(\widehat{AIB}+\widehat{AIC}=180^0\)(hai góc kề bù)

nên \(\widehat{AIB}=\widehat{AIC}=\dfrac{180^0}{2}=90^0\)

=>AI\(\perp\)BC

c: Xét ΔABC có

AI là đường trung tuyến

G là trọng tâm

Do đó: \(AG=\dfrac{2}{3}AI=\dfrac{2}{3}\cdot12=8\left(cm\right)\)

Ta có: AG+GI=AI

=>GI+8=12

=>GI=4(cm)

50 cm²