M nằm trên cạnh BCcap B cap C
𝐵𝐶
sao cho BM=2MC⇒BM⃗=23BC⃗=23AD⃗cap B cap M equals 2 cap M cap C implies modified cap B cap M with right arrow above equals two-thirds modified cap B cap C with right arrow above equals two-thirds modified cap A cap D with right arrow above
𝐵𝑀=2𝑀𝐶⇒𝐵𝑀⃗=23𝐵𝐶⃗=23𝐴𝐷⃗
.
N nằm trên cạnh ADcap A cap D
𝐴𝐷
sao cho AN=3ND⇒AN⃗=34AD⃗cap A cap N equals 3 cap N cap D implies modified cap A cap N with right arrow above equals three-fourths modified cap A cap D with right arrow above
𝐴𝑁=3𝑁𝐷⇒𝐴𝑁⃗=34𝐴𝐷⃗
. 

2. Phân tích AE⃗modified cap A cap E with right arrow above
𝐴𝐸⃗
theo AB⃗modified cap A cap B with right arrow above
𝐴𝐵⃗
và AD⃗modified cap A cap D with right arrow above
𝐴𝐷⃗
 
Vì Ecap E
𝐸
nằm trên AMcap A cap M
𝐴𝑀
, ta có: AE⃗=k⋅AM⃗=k(AB⃗+BM⃗)=k⋅AB⃗+2k3AD⃗modified cap A cap E with right arrow above equals k center dot modified cap A cap M with right arrow above equals k open paren modified cap A cap B with right arrow above plus modified cap B cap M with right arrow above close paren equals k center dot modified cap A cap B with right arrow above plus 2 k over 3 end-fraction modified cap A cap D with right arrow above
𝐴𝐸⃗=𝑘⋅𝐴𝑀⃗=𝑘(𝐴𝐵⃗+𝐵𝑀⃗)=𝑘⋅𝐴𝐵⃗+2𝑘3𝐴𝐷⃗
(với 0<k<10 is less than k is less than 1
0<𝑘<1
).
Vì Ecap E
𝐸
nằm trên BNcap B cap N
𝐵𝑁
, ta có: BE⃗=m⋅BN⃗⇒AE⃗−AB⃗=m(AN⃗−AB⃗)=m(34AD⃗−AB⃗)modified cap B cap E with right arrow above equals m center dot modified cap B cap N with right arrow above implies modified cap A cap E with right arrow above minus modified cap A cap B with right arrow above equals m open paren modified cap A cap N with right arrow above minus modified cap A cap B with right arrow above close paren equals m open paren three-fourths modified cap A cap D with right arrow above minus modified cap A cap B with right arrow above close paren
𝐵𝐸⃗=𝑚⋅𝐵𝑁⃗⇒𝐴𝐸⃗−𝐴𝐵⃗=𝑚(𝐴𝑁⃗−𝐴𝐵⃗)=𝑚34𝐴𝐷⃗−𝐴𝐵⃗
. ⇒AE⃗=(1−m)AB⃗+3m4AD⃗implies modified cap A cap E with right arrow above equals open paren 1 minus m close paren modified cap A cap B with right arrow above plus 3 m over 4 end-fraction modified cap A cap D with right arrow above
⇒𝐴𝐸⃗=(1−𝑚)𝐴𝐵⃗+3𝑚4𝐴𝐷⃗
. 

Vì cách biểu diễn vectơ theo AB⃗modified cap A cap B with right arrow above
𝐴𝐵⃗
và AD⃗modified cap A cap D with right arrow above
𝐴𝐷⃗
là duy nhất, ta có hệ phương trình:
{k=1−m 2k3=3m41 cases; Case 1: k equals 1 minus m space 2 k over 3 end-fraction equals 3 m over 4 end-fraction end-cases;
𝑘=1−𝑚 2𝑘3=3𝑚4

Giải hệ: 
Thay k=1−mk equals 1 minus m
𝑘=1−𝑚
vào phương trình dưới: 2(1−m)3=3m4⇔8−8m=9m⇔17m=8⇔m=817the fraction with numerator 2 open paren 1 minus m close paren and denominator 3 end-fraction equals 3 m over 4 end-fraction implies and is implied by 8 minus 8 m equals 9 m implies and is implied by 17 m equals 8 implies and is implied by m equals 8 over 17 end-fraction
2(1−𝑚)3=3𝑚4⇔8−8𝑚=9𝑚⇔17𝑚=8⇔𝑚=817
.
Suy ra k=1−817=917k equals 1 minus 8 over 17 end-fraction equals 9 over 17 end-fraction
𝑘=1−817=917
. 

Thay kk
𝑘
vào biểu thức AE⃗modified cap A cap E with right arrow above
𝐴𝐸⃗
:
AE⃗=917AB⃗+2⋅(9/17)3AD⃗=917AB⃗+617AD⃗modified cap A cap E with right arrow above equals 9 over 17 end-fraction modified cap A cap B with right arrow above plus the fraction with numerator 2 center dot open paren 9 / 17 close paren and denominator 3 end-fraction modified cap A cap D with right arrow above equals 9 over 17 end-fraction modified cap A cap B with right arrow above plus 6 over 17 end-fraction modified cap A cap D with right arrow above
𝐴𝐸⃗=917𝐴𝐵⃗+2⋅(9/17)3𝐴𝐷⃗=917𝐴𝐵⃗+617𝐴𝐷⃗
Vậy x=917x equals 9 over 17 end-fraction
𝑥=917
và y=617y equals 6 over 17 end-fraction
𝑦=617
. 

3. Tính giá trị S=4x−5ycap S equals 4 x minus 5 y
𝑆=4𝑥−5𝑦
 
S=4(917)−5(617)=3617−3017=617cap S equals 4 open paren 9 over 17 end-fraction close paren minus 5 open paren 6 over 17 end-fraction close paren equals 36 over 17 end-fraction minus 30 over 17 end-fraction equals 6 over 17 end-fraction
𝑆=4917−5617=3617−3017=617

Kết quả: x=917x equals 9 over 17 end-fraction
𝑥=917
, y=617y equals 6 over 17 end-fraction
𝑦=617
và S=617cap S equals 6 over 17 end-fraction
𝑆=617
.