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Xét tam giác ABH vuông tại H có:
\(sinB=\dfrac{AH}{AB}=0,5\Rightarrow AB=\dfrac{AH}{0,5}=\dfrac{5}{0,5}=10\)
Xét tam giác ABH vuông tại H có:
\(AB^2=AH^2+BH^2\left(Pytago\right)\)
\(\Rightarrow BH=\sqrt{AB^2-AH^2}=\sqrt{10^2-5^2}=5\sqrt{3}\)
1. a) 6² : 4 . 3 + 2 . 5²
= 36 : 4 . 3 + 2 . 25
= 9 . 3 + 50
= 27 + 50
= 77
b) 5 . 4² - 32 : 4²
= 5 . 16 - 32 : 16
= 80 - 2
= 78
c) 80 - (4 . 5² - 3 .2³)
= 80 - (4 . 25 - 3 . 8)
= 80 - (100 - 24)
= 80 - 76
= 4
d) 2448 : [119 - (23 - 6)]
= 2448 : (119 - 17)
= 2448 : 102
= 24
e) 23 . 85 + 15 . 23 + 170
= 23 . (85 + 15) + 170
= 23 . 100 + 170
= 2300 + 170
= 2470
f) 3³ . 2 - 5 . 2023⁰ + 36 : 3²
= 27 . 2 - 5 . 1 + 36 : 9
= 54 - 5 + 4
= 53
2. a) 123 - 5.(x + 4) = 38
5.(x + 4) = 123 - 38
5.(x + 4) = 85
x + 4 = 85 : 5
x + 4 = 17
x = 17 - 4
x = 13
b) (3.x - 2⁴) . 7³ = 2 . 7⁴
3x - 16 = 2 . 7⁴ : 7³
3x - 16 = 2 . 7
3x = 14 + 16
3x = 30
x = 30 : 3
x = 10
c) (2600 + 6400) - 3.x = 1200
9000 - 3x = 1200
3x = 9000 - 1200
3x = 7800
x = 7800 : 3
x = 2600
d) [(6.x - 72) : 2 - 84] . 28 = 5628
(6x - 72) : 2 - 84 = 5628 : 28
(6x - 72) : 2 - 84 = 201
(6x - 72) : 2 = 201 + 84
(6x - 72) : 2 = 285
6x - 72 = 285 . 2
6x - 72 = 570
6x = 570 + 72
6x = 642
x = 642 : 6
x = 107
e) 70 - 5(x - 3) = 45
5(x - 3) = 70 - 45
5(x - 3) = 25
x - 3 = 25 : 5
x - 3 = 5
x = 5 + 3
x = 8
f) 10 + 2.x = 4⁷ : 4⁵
10 + 2x = 4²
10 + 2x = 16
2x = 16 - 10
2x = 6
x = 6 : 2
x = 3
g) 2.x - 138 = 2³.2²
2x - 138 = 2⁵
2x - 138 = 32
2x = 32 + 138
2x = 170
x = 170 : 2
x = 85
h) 231 - (x - 6) = 1339 : 13
231 - (x - 6) = 103
x - 6 = 231 - 103
x - 6 = 128
x = 128 + 6
x = 134
1 are
2 am
3 is
4 are
5 are
6 are
7 is
8 is
9 is
10 are
IV
1 is writing
2 are losing
3 is having
4 is staying
5 am not lying
6 is always using
7 are having
8 Are you playing
9 are not touching
10 Is - listening
11 Is- winning
12 am not staying
13 is not working
14 is not reading
15 isn't raining
16 am not listening
17 Are they making
18 Are you doing
19 Is - sitting
20 is - doing
21 are-putting
22 are-wearing
23 is-studying
2, am
3, is
4,are
5,are
6,are
7,is
8,is
9,is
10,are
IV
1,2,7 OK
3,is having
4,has stayed
5,am not lying
6,always uses
8,Are-playing
9,not to touch
10,Is-listening
11,Are-winning
12,am not staying
13,isn't working
14,isn't reading
15,isn't raining
16,am not listening
17,Are-making
18,Are-doing
19,Is-sitting
20,is-doing
21,do-putting
22,do-wear
23,is-studying
a, mong bn tự vẽ vì nó dễ nhất
b, hai đèn // => \(I_A=I_1+I_2=1,1\left(A\right)\)
c, 2 đèn // \(\Rightarrow U_1=U_2=U=4,5\left(V\right)\)
d, khi K ngắt chỉ mỗi vôn kế gắn ở nguồn giữ nguyên giá trị 4,5 v còn lại về 0
Bạn ơi giúp thêm 3 câu nữa đc ko mai mik thi rồi á
Câu 3:
a: Ta có: \(2x\left(3x-1\right)-\left(x-3\right)\left(6x+2\right)\)
\(=6x^2-2x-6x^2-2x+18x+6\)
=14x+6
b: Ta có: \(2x\left(x+7\right)-3x\left(x+1\right)\)
\(=2x^2+14x-3x^2-3x\)
\(=-x^2+11x\)
Câu 2:
a: Ta có: \(\left(-8x^5+12x^3-16x^2\right):4x^2\)
\(=-8x^5:4x^2+12x^3:4x^2-16x^2:4x^2\)
\(=-2x^3+3x-4\)
b: Ta có: \(\left(12x^3y^3-18x^2y+9xy^2\right):6xy\)
\(=12x^3y^3:6xy-18x^2y:6xy+9xy^2:6xy\)
\(=2x^2y^2-3x+\dfrac{3}{2}y\)
c: Ta có: \(\dfrac{x^3-11x^2+27x-9}{x-3}\)
\(=\dfrac{x^3-3x^2-8x^2+24x+3x-9}{x-3}\)
\(=x^2-8x+3\)
d: Ta có: \(\dfrac{6x^4-13x^3+7x^2-x-5}{3x+1}\)
\(=\dfrac{6x^4+2x^3-15x^3-5x^2+12x^2+4x-5x-\dfrac{5}{3}-\dfrac{10}{3}}{3x+1}\)
\(=2x^3-5x^2+4x-\dfrac{5}{3}-\dfrac{\dfrac{10}{3}}{3x+1}\)
Kẻ \(AH\perp BC\)
Có \(sinB=\dfrac{AH}{AB}\)
\(\Rightarrow AB.sinB=AH\)\(\Leftrightarrow\dfrac{1}{2}AB.BC.sinB=\dfrac{1}{2}AH.BC=S_{ABC}\)
Có \(sinC=\dfrac{AH}{AC}\)
\(\Rightarrow AC.sinC=AH\)\(\Leftrightarrow\dfrac{1}{2}AC.BC.sinC=\dfrac{1}{2}AH.BC=S_{ABC}\)
\(\Rightarrow\dfrac{1}{2}AB.BC.sinB=\dfrac{1}{2}AC.BC.sinC=S_{ABC}\)
Áp dụng \(S_{ABC}=\dfrac{1}{2}AB.BC.sinB=\dfrac{1}{2}.10.15.sin60^0=\dfrac{75\sqrt{3}}{2}\left(cm^2\right)\)
Kẻ đường cao AH của tam giác ABC
Áp dụng tslg trong tam giác ABH vuông tại H và tam giác AHC vuông tại H:
\(sinB=\dfrac{AH}{AB},sinC=\dfrac{AH}{AC}\)
\(S_{ABC}=\dfrac{1}{2}BC.AH=\dfrac{1}{2}AB.BC.\dfrac{AH}{AB}=\dfrac{1}{2}AB.BC.sinB\)
\(S_{ABC}=\dfrac{1}{2}BC.AH=\dfrac{1}{2}AC.BC.\dfrac{AH}{AC}=\dfrac{1}{2}AC.BC.sinC\)
\(S_{ABC}=\dfrac{1}{2}AB.BC.sinB=\dfrac{1}{2}.10.15.sin60^0=\dfrac{1}{2}.10.15.\dfrac{\sqrt{3}}{2}=\dfrac{75\sqrt{3}}{2}\left(cm^2\right)\)