Xét ΔOAB và ΔOCD có

\(\widehat{OAB}=\widehat{OCD}\)(AB//CD)

\(\widehat{AOB}=\widehat{COD}\)(hai góc đối đỉnh)

Do đó: ΔOAB~ΔOCD

=>\(\dfrac{S_{OAB}}{S_{OCD}}=\left(\dfrac{AB}{CD}\right)^2=\dfrac{1}{16}\)

=>\(S_{OCD}=16\cdot S_{OBA}\)

ta có: \(S_{OCD}-S_{OAB}=1995\)

=>\(16\cdot S_{OAB}-S_{OAB}=1995\)

=>\(15\cdot S_{OAB}=1995\)

=>\(S_{OAB}=1995:15=133\left(cm^2\right)\)

=>\(S_{OCD}=133+1995=2128\left(cm^2\right)\)

AB//CD

=>\(\dfrac{OA}{OC}=\dfrac{OB}{OD}=\dfrac{AB}{CD}=\dfrac{1}{4}\)

\(\dfrac{OA}{OC}=\dfrac{1}{4}\)

=>\(\dfrac{S_{BOA}}{S_{BOC}}=\dfrac{1}{4}\)

=>\(S_{BOC}=4\cdot S_{BOA}=4\cdot133=532\left(cm^2\right)\)

Vì OB/OD=1/4

nên \(\dfrac{S_{AOB}}{S_{AOD}}=\dfrac{1}{4}\)

=>\(S_{AOD}=532\left(cm^2\right)\)

\(S_{ABCD}=S_{ABO}+S_{BOC}+S_{COD}+S_{AOD}\)

\(=532+532+133+2128=3325\left(cm^2\right)\)

\(3\left(\dfrac{1}{2}x-1\right)=-\dfrac{3}{4}\)

=>\(\dfrac{1}{2}x-1=-\dfrac{3}{4}:3=-\dfrac{1}{4}\)
=>\(\dfrac{1}{2}x=-\dfrac{1}{4}+1=\dfrac{3}{4}\)

=>\(x=\dfrac{3}{4}\cdot2=\dfrac{3}{2}\)

a: \(3\cdot2,25-0,75=6,75-0,75=6\)

b: \(\left(-1,25\right)+3,5+1,25+36,5\)

\(=\left(-1,25+1,25\right)+\left(3,5+36,5\right)\)

=0+40

=40

mik camon^^

\(3x-x=20140+\left(-3\right)^2\\ \Rightarrow2x=20140+9\\ \Rightarrow2x=20149\\ \Rightarrow x=\dfrac{20149}{2}.\)

xin mng giúp tôi 

\(B=\left(1+\dfrac{1}{1\cdot3}\right)\left(1+\dfrac{1}{2\cdot4}\right)\cdot...\cdot\left(1+\dfrac{1}{2024\cdot2026}\right)\)

\(=\left(1+\dfrac{1}{\left(2-1\right)\left(2+1\right)}\right)\left(1+\dfrac{1}{\left(3-1\right)\left(3+1\right)}\right)\cdot...\cdot\left(1+\dfrac{1}{\left(2025-1\right)\left(2025+1\right)}\right)\)

\(=\left(1+\dfrac{1}{2^2-1}\right)\left(1+\dfrac{1}{3^2-1}\right)\cdot...\cdot\left(1+\dfrac{1}{2025^2-1}\right)\)

\(=\dfrac{2^2}{2^2-1}\cdot\dfrac{3^2}{3^2-1}\cdot...\cdot\dfrac{2025^2}{2025^2-1}\)

\(=\dfrac{2\cdot3\cdot...\cdot2025}{1\cdot2\cdot...\cdot2024}\cdot\dfrac{2\cdot3\cdot...\cdot2025}{3\cdot4\cdot...\cdot2026}\)

\(=\dfrac{2025}{1}\cdot\dfrac{2}{2026}=\dfrac{2025}{1013}\)

kO biết

\(A=\dfrac{4}{7\cdot31}+\dfrac{6}{7\cdot41}+\dfrac{9}{10\cdot41}+\dfrac{7}{10\cdot57}\)

=>\(A=\dfrac{20}{31\cdot35}+\dfrac{30}{35\cdot41}+\dfrac{45}{41\cdot50}+\dfrac{35}{50\cdot57}\)

\(=5\left(\dfrac{4}{31\cdot35}+\dfrac{6}{35\cdot41}+\dfrac{9}{41\cdot50}+\dfrac{7}{50\cdot57}\right)\)

\(=5\left(\dfrac{1}{31}-\dfrac{1}{35}+\dfrac{1}{35}-\dfrac{1}{41}+\dfrac{1}{41}-\dfrac{1}{50}+\dfrac{1}{50}-\dfrac{1}{57}\right)\)

\(=5\left(\dfrac{1}{31}-\dfrac{1}{57}\right)\)

\(B=\dfrac{7}{19\cdot31}+\dfrac{5}{19\cdot43}+\dfrac{3}{23\cdot43}+\dfrac{11}{23\cdot57}\)

\(=\dfrac{14}{31\cdot38}+\dfrac{10}{38\cdot43}+\dfrac{6}{43\cdot46}+\dfrac{22}{46\cdot57}\)

\(=2\left(\dfrac{7}{31\cdot38}+\dfrac{5}{38\cdot43}+\dfrac{3}{43\cdot46}+\dfrac{11}{46\cdot57}\right)\)

\(=2\left(\dfrac{1}{31}-\dfrac{1}{38}+\dfrac{1}{38}-\dfrac{1}{43}+\dfrac{1}{43}-\dfrac{1}{46}+\dfrac{1}{46}-\dfrac{1}{57}\right)\)

\(=2\left(\dfrac{1}{31}-\dfrac{1}{57}\right)\)

=>\(\dfrac{A}{B}=\dfrac{5}{2}\)

a)  Số lần xúc xắc là số chẵn là :

20 + 22 + 15 = 57 (lần)

Xác suất thực nghiệm của sự kiện 'số chấm xuất hiện là số chẵn ' là:

57 : 100 x 100 = 57%

b) Số lần xúc xắc là số lớn hơn 2 là :

18 + 22 + 10 + 15 = 65 (lần)

Xác suất thực nghiệm của sự kiện 'số chấm là số lớn hơn 2 ' là:

65 : 100 x 100 = 65%

\(B=\dfrac{36}{1.3.5}+\dfrac{36}{3.5.7}+\dfrac{36}{5.7.9}+...+\dfrac{36}{25.27.29}\\ \Rightarrow B=36\left(\dfrac{1}{1.3.5}+\dfrac{1}{3.5.7}+\dfrac{1}{5.7.9}+...+\dfrac{1}{25.27.29}\right)\\ \Rightarrow B=36\left(1-\dfrac{1}{29}\right)\\ \Rightarrow B=\dfrac{36.28}{29}\\ \Rightarrow B=\dfrac{1008}{29}.\)

5x5x5=125

a: Số học sinh khối 6 là: \(1800\cdot25\%=450\left(bạn\right)\)

Số học sinh khối 7 là \(1800\cdot\dfrac{3}{10}=540\left(bạn\right)\)

Số học sinh khối 8 là: \(540:\dfrac{6}{5}=540\cdot\dfrac{5}{6}=450\left(bạn\right)\)

Số học sinh khối 9 là:

1800-450-450-540=360(bạn)

b: Tỉ số phần trăm giữa tổng số học sinh khối 8 và 9 so với tổng số học sinh toàn trường là:

\(\dfrac{450+360}{1800}=\dfrac{810}{1800}=45\%\)

Lời giải:

$S^2=\frac{(2.4.6...4046)^2}{(3.5.7...4047)^2}$
$=\frac{2.4}{3^2}.\frac{4.6}{5^2}.\frac{6.8}{7^2}.....\frac{4044.4046}{4045^2}.\frac{2.4046}{4047^2}$

Xét thừa số tổng quát $\frac{n(n+2)}{(n+1)^2}=\frac{n^2+2n}{n^2+2n+1}< 1$

$\Rightarrow \frac{2.4}{3^2}< 1; \frac{4.6}{5^2}<1,...., \frac{4044.4046}{4045^2}<1$

$\Rightarrow S^2< 1.\frac{2.4046}{4047^2}$

Giờ chỉ cần cm: $\frac{2.4046}{4047^2}< \frac{1}{2024}$

Thật vậy:

$\frac{2.4046}{4047^2}-\frac{1}{2024}=\frac{4048.4046-4047^2}{2024.4047^2}=\frac{(4047+1)(4047-1)-4047^2}{2024.4047^2}=\frac{4047^2-1-4047^2}{2024.4047^2}=\frac{-1}{2024.4047^2}< 0$

$\Rightarrow \frac{2.4046}{4047^2}< \frac{1}{2024}$

$\Rightarrow S^2< \frac{1}{2024}$