12 số

@kurumicute mik cam on!

a) \(S_{EAG}=\dfrac{1}{2}\times AG\times ED=\dfrac{1}{2}\times2\times3=3\left(cm^2\right)\)

\(S_{PBC}=\dfrac{1}{2}\times BC\times DC=\dfrac{1}{2}\times5\times5=12,5\left(cm^2\right)\)

b) Ta có:

\(S_{EBC}=\dfrac{1}{2}\times BC\times EC=\dfrac{1}{2}\times5\times8=20\left(cm^2\right)\)

\(S_{PEC}=S_{ECB}-S_{PBC}=20-12,5=7,5\left(cm^2\right)\)

Vậy nên:

\(PD=\dfrac{2\times S_{PEC}}{EC}=\dfrac{2\times7,5}{8}=1,875\left(cm\right)\)

c) Ta thấy:

\(\dfrac{IM}{IP}=\dfrac{S_{MIG}}{S_{IPG}}=\dfrac{S_{MIE}}{S_{IPE}}\) nên \(\dfrac{IM}{IP}=\dfrac{S_{MGE}}{S_{GPE}}=\dfrac{\dfrac{1}{2}\times MG\times3}{\dfrac{1}{2}\times GP\times3}=\dfrac{MG}{GP}\)

Kéo dài AD cắt EF tại K.

Ta có \(S_{AKM}=\dfrac{1}{2}\times3\times2=3\left(cm^2\right)\)

nên \(S_{EKM}=S_{AKE}-S_{AKM}=\dfrac{1}{2}\times3\times5-3=4,5\left(cm^2\right)\)

Vậy \(FM=\dfrac{2\times S_{EKM}}{KE}=1,8\left(cm\right)\)

Thế thì \(MG=3-1,8=1,2\left(cm\right)\)

Lại có \(GP=3-1,875=1,125\left(cm\right)\)

Vậy nên:

\(\dfrac{IM}{IP}=\dfrac{MG}{GP}=\dfrac{1,2}{1,125}=\dfrac{16}{15}\).

Tttt

Ggghghnkhg

a: 11+13+15+17+19

=(11+19)+(13+17)+15

=30+30+15

=75

b: \(122+2116+278+84\)

=(122+278)+(2116+84)

=400+2200

=2600

c: \(12\times125\times54\)

=1500x54

=81000

d: \(27\times36+27\times64=27\times\left(36+64\right)=27\times100=2700\)

e: \(25\text{x}37+25\text{x}63-150\)

=25x(37+63)-150

=25x100-150

=2500-150

=2350

f: \(425\text{x}74-27\text{x}425-1\)

=425x(74-27-1)

=425x46=19550

g: \(8\text{x}9\text{x}14+6\text{x}17\text{x}12+19\text{x}4\text{x}18\)

=14x72+17x72+19x72

=72x(14+17+19)

=50x72=3600

a) $11+13+15+17+19$

$=(11+19)+(13+17)+15$

$=30+30+15$

$=60+15=75$

b) $122+2116+278+84$

$=(122+278)+(2116+84)$

$=400+2200=2600$

c) $12\times125\times54$

$=3\times4\times125\times2\times27$

$=(4\times2)\times125\times(3\times27)$

$=8\times125\times81$

$=1000\times81=81000$

d) $27\times36+27\times64$

$=27\times(36+64)$

$=27\times100=2700$

e) $25\times37+25\times63-150$

$=25\times(37+63)-150$

$=25\times100-150$

$=2500-150=2350$

f) $425\times74-27\times425-425$

$=425\times(74-27-1)$

$=425\times(47-1)$

$=425\times46=19550$

g) $8\times9\times14+6\times17\times12+19\times4\times18$

$=(8\times9)\times14+(6\times12)\times17+(4\times18)\times19$

$=72\times14+72\times17+72\times19$

$=72\times(14+17+19)$

$=72\times(31+19)$

$=72\times50=3600$

$\mathtt{Toru}$

1 cach:

3+2+1+3+1

Ta có :

\(\dfrac{1300}{1500}=\dfrac{13}{15}=1-\dfrac{2}{15}\)

\(\dfrac{1333}{1555}=1-\dfrac{222}{1555}\)

Vì \(\dfrac{222}{1555}>\dfrac{2}{15}\)

\(\Rightarrow1-\dfrac{222}{1555}< 1-\dfrac{2}{15}\)

\(\dfrac{\Rightarrow1333}{1555}< \dfrac{1300}{1500}\)

hi

\(\dfrac{29}{2}=\dfrac{28+1}{2}=14+\dfrac{1}{2}=14\dfrac{1}{2}\)

\(\dfrac{15}{4}=\dfrac{12+3}{4}=3+\dfrac{3}{4}=3\dfrac{3}{4}\)

\(\dfrac{31}{2}=\dfrac{30+1}{2}=15\dfrac{1}{2}\)

\(\dfrac{29}{3}=\dfrac{27+2}{3}=9\dfrac{2}{3}\)

\(\dfrac{125}{8}=\dfrac{120+5}{8}=15+\dfrac{5}{8}=15\dfrac{5}{8}\)

\(\dfrac{36}{27}=\dfrac{27+9}{27}=1+\dfrac{9}{27}=1\dfrac{9}{27}\)

\(\dfrac{124}{15}=\dfrac{120+4}{15}=8+\dfrac{4}{15}=8\dfrac{4}{15}\)

\(\dfrac{96}{3}=\dfrac{93+3}{3}=31\dfrac{3}{3}\)

\(\dfrac{129}{24}=\dfrac{120+9}{24}=5+\dfrac{9}{24}=5\dfrac{9}{24}\)

\(\dfrac{78}{13}=\dfrac{65+13}{13}=5+\dfrac{13}{13}=5\dfrac{13}{13}\)

\(\dfrac{91}{4}=\dfrac{88+3}{4}=22+\dfrac{3}{4}=22\dfrac{3}{4}\)

\(\dfrac{115}{8}=\dfrac{112+3}{8}=14+\dfrac{3}{8}=14\dfrac{3}{8}\)

a: \(B=2021\times2025=\left(2023-2\right)\times\left(2023+2\right)=2023\times2023-2\times2\)

=>\(B=A-4\)

=>A lớn hơn B 4 đơn vị

b: \(C=35\times53-18=35\times35+35\times18-18\)

\(=35\times35+18\times\left(35-1\right)\)

\(=35\times35+18\times34\)

\(D=35+53\times34\)

\(=35+\left(35-1\right)\times\left(35+18\right)\)

\(=35+35\times35+35\times18-35\times1-18\)

\(=35\times35+35\times17+17=35\times35+36\times17\)

\(=35\times35+18\times34\)

=C

=>C=D