\dfrac{ \dfrac{ 5 }{ 7 } + \dfrac{ 5 }{ 9 } - \dfrac{ 5 }{ 11 } }{ \dfrac{ 15 }{ 7 } + \dfrac{ 5 }{ 9 } - \dfrac{ 15 }{ 11 } }
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a, \(S_{ABC}=\dfrac{1}{2}a.a=\dfrac{a^2}{2}\)
Theo Pytago tam giac ABC vuong tai B
\(AC=\sqrt{a^2+a^2}=\sqrt{2}a\Rightarrow AO=\dfrac{\sqrt{2}a}{2}\)
Theo Pytago tam giac SOA vuong tai O
\(SO=\sqrt{4a^2-\dfrac{2}{4}a^2}=\sqrt{\dfrac{14a^2}{4}}=\sqrt{\dfrac{7}{2}}a\)
\(V_{ABC}=\dfrac{1}{3}.\dfrac{a^2}{2}.\dfrac{\sqrt{7}}{\sqrt{2}}a=\dfrac{a^3\sqrt{7}}{6\sqrt{2}}\)
b, Ta co \(\dfrac{d\left(C;\left(SAB\right)\right)}{d\left(O;\left(SAB\right)\right)}=\dfrac{AC}{OA}=2\Rightarrow d\left(C;\left(SAB\right)\right)=2d\left(O;\left(SAB\right)\right)\)
Ke OH vuong AB, SO vuong AB, SO;OH chua (SOH)
=> AB vuong (SOH)
Ke OK vuong SH => OK la khoang cach
- bn tinh not nhe
c, ((SAB);(ABCD)) = ^SHO
- tinh dc phan b roi ap vao tam giac SHO la ra nhe
![](https://rs.olm.vn/images/avt/0.png?1311)
\(5^{16}:5^{14}+3^2+2000^0\)
\(=5^2+3^2+2000^0\)
\(=25+9+1\)
\(=35\)
\(3^{96}:3^{95}+2.2^3+1^{2024}\)
\(=3^1+2^4+1\)
\(=3+16+1\)
\(=20\)
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Vì \(\widehat A\) và \(\widehat B\) bù nhau nên \(\widehat A+\widehat B=180^\circ\) (1)
Lại có: \(\widehat A-\widehat B=30^\circ\) (2)
Từ (1) và (2) \(\Rightarrow (\widehat A+\widehat B)+(\widehat A-\widehat B)=180^\circ+30^\circ\)
\(\Rightarrow2\widehat{A}=210^{\circ}\)
\(\Rightarrow\widehat{A}=\dfrac{210^{\circ}}{2}=105^{\circ}\)
Khi đó: \(105^{\circ}-\widehat{B}=30^{\circ}\Rightarrow\widehat{B}=105^{\circ}-30^{\circ}=75^{\circ}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a; \(\dfrac{1}{n}\); \(\dfrac{1}{n+1}\) (n > 0; n \(\in\) Z)
\(\dfrac{1}{n}\) - \(\dfrac{1}{n+1}\) = \(\dfrac{n+1-1}{n.\left(n+1\right)}\) = \(\dfrac{1}{n\cdot\left(n+1\right)}\)
⇒ \(\dfrac{1}{n}\) - \(\dfrac{1}{n+1}\) = \(\dfrac{1}{n\left(n+1\right)}\) (đpcm)
Bài 1b
A = \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\) + \(\dfrac{1}{7.8}\) + \(\dfrac{1}{8.9}\)
A = \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) +\(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{9}\)
A = \(\dfrac{1}{2}\) - \(\dfrac{1}{9}\)
A = \(\dfrac{7}{18}\)
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Vì đây là dãy số không cách đều nên ta tính số số hạng ở phần nguyên
số số hạng là:(100 - 2) : 2 + 1 = 50 (số )
Tổng của dãy số trên là: (100,100 + 2,2 ) x 50 : 2 = 2557,5
Đ/s:2557,5
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bài 2: c) (2018/1 - 2018 x 1) / (2018 x 2008 + 2018 x 2002)
= 0 / (2018 x 2008 + 2018 x 2002) = 0
a) 815 - 23 - 77 + 185
= (815 + 185) - (23 + 77)
= 1000 - 100 = 900
d) (9 - 8 - 7 - 2 - 1) x (500 x 9 - 250 x 18)
= (9 - 8 - 7 - 2 - 1) x (250 x 2 x 9 - 250 x 18)
= (9 - 8 - 7 - 2 - 1) x (250 x 18 - 250 x 18)
= (9 - 8 - 7 - 2 - 1) x 0 = 0
b) 3145 - 246 + 2347 - 145 + 4246 - 347
= (3145 - 145) + (4246 - 246) + (2347 - 347)
= 3000 + 4000 + 2000 = 9000
bài 3: b) 27 + 27 x 5 + 27 x 7 - 27 x 2
= 27 x (1 + 5 + 7 - 2)
= 27 x 11 = 297
c) 754,75 - 25 x 2262 + 4568
= 754,75 - 56550 + 4568
= -51227,25
a) 40 x 113 x 25 - 20 x 112 x 50
= 20 x 113 x 50 - 20 x 112 x 50
= 100 x 113 - 100 x 112
= 100 x (113 - 112)
= 100 x 1 = 100
a) 544544 - 444444
= 100100 + 444444 - 444444
= 10010
b) 131313 - 10101 - 20202
= 131313 - (10101 + 20202)
= 131313 - 30303 = 011010
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a: \(BC=\dfrac{3}{4}\cdot12=9\left(cm\right)\)
Chu vi hình chữ nhật ABCD là:
\(C_{ABCD}=\left(AB+BC\right)\cdot2=42\left(cm\right)\)
b: \(BM=\dfrac{2}{3}BC=\dfrac{2}{3}\cdot9=6\left(cm\right)\)
\(S_{ABM}=\dfrac{1}{2}\cdot BA\cdot BM=\dfrac{1}{2}\cdot12\cdot6=36\left(cm^2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Mình đọc chả biết rõ đề bài là như thế nào?
\(\dfrac{\dfrac{5}{7}+\dfrac{5}{9}-\dfrac{5}{11}}{\dfrac{15}{7}+\dfrac{5}{9}-\dfrac{15}{11}}=\dfrac{\dfrac{495}{693}+\dfrac{385}{693}-\dfrac{315}{693}}{\dfrac{1485}{693}+\dfrac{385}{693}-\dfrac{945}{693}}\\ =\dfrac{\dfrac{565}{693}}{\dfrac{925}{693}}=\dfrac{565}{925}=\dfrac{113}{815}\)