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![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt $A=99-97+95-93+91-89+...+7-5+3-1$ (sửa đề)
$=(99-97)+(95-93)+(91-89)+...+(7-5)+(3-1)$
$=2+2+2+...+2+2$
Số các số 2 trong dãy số trên là: $[(99-1):2+1]:2=25$ (số)
Do đó: $A=2.25=50$
Sửa đề: 99-97+95-93+...+7-5+3-1
=(99-97)+(95-93)+...+(7-5)+(3-1)
=2+2+...+2
=2x25=50
![](https://rs.olm.vn/images/avt/0.png?1311)
$24.25+2.38.12+3.8.37$
$=24.25+(2.12).38+(3.8).37$
$=24.25+24.38+24.37$
$=24.(25+38+37)$
$=24.(63+37)$
$=24.100=2400$
24 . 25 + 2 . 38 . 12 + 3 . 8 . 37
= 24 . 25 + ( 2 . 12 ) . 38 + ( 3 . 8 ) . 37
= 24 . 25 + 24 . 38 + 24 . 37
= 24 . ( 25 + 38 + 37 )
= 24 . 100
= 2400
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Xét ΔABE có: \(\widehat{BAE}+\widehat{ABE}+\widehat{AEB}=180^o\)
\(\Rightarrow90^o+x+x=180^o\Rightarrow2x=180^o-90^o=90^o\)
\(\Rightarrow x=\dfrac{90^o}{2}=45^o\)
Xét ΔABC có: \(\widehat{ABC}+\widehat{BAC}+\widehat{CAB}=180^o\)
\(\Rightarrow\left(x+y\right)+90^o+30^o=180^o\)
\(\Rightarrow\left(x+y\right)+120^o=180^o\)
\(\Rightarrow45^o+y=180^o-120^o\)
\(\Rightarrow45^o+y=60^o\)
\(\Rightarrow y=60^o-45^o=15^o\)
∆ABE vuông tại A (gt)
⇒ ∠ABE + ∠AEB = 90⁰
⇒ x + x = 90⁰
⇒ x = 90⁰ : 2
= 45⁰
∆ABC vuông tại A (gt)
⇒ ∠ABC + ∠ACB = 90⁰
⇒ ∠ABC = 90⁰ - ∠ACB
= 90⁰ - 30⁰
= 60⁰
⇒ y = ∠ABC - x
= 60⁰ - 45⁰
= 15⁰
![](https://rs.olm.vn/images/avt/0.png?1311)
\(4\left(x-2\right)-3\left(x+1\right)=5\)
\(\Leftrightarrow4x-8-3x-3=5\)
\(\Leftrightarrow\left(4x-3x\right)=5+8+3\)
\(\Leftrightarrow x=16\)
Vậy \(x=16\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\dfrac{1}{2}=\dfrac{1\times2}{2\times2}=\dfrac{2}{4};\dfrac{1}{4}=\dfrac{1}{4}\)
Vì \(\dfrac{2}{4}>\dfrac{1}{4}\) nên \(\dfrac{1}{2}>\dfrac{1}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{1}{2\cdot6}+\dfrac{1}{3\cdot8}+...+\dfrac{1}{2023\cdot4048}\)
\(=\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{4046\cdot4048}\)
\(=\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{4046}-\dfrac{1}{4048}\)
\(=\dfrac{1}{4}-\dfrac{1}{4048}=\dfrac{1012-1}{4048}=\dfrac{1011}{4048}\)
\(A=\dfrac{1}{2\cdot6}+\dfrac{1}{3\cdot8}+\dfrac{1}{4\cdot10}+...+\dfrac{1}{2023\cdot4048}\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{2023\cdot2024}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2023}-\dfrac{1}{2024}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2024}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{1012-1}{2024}\)
\(=\dfrac{1011}{4048}\)
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
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