Câu 44: C

Câu 45: C

Xem lại đề.

\(7x-2x=6^{20}:36^9+44:11\)

\(=>x\left(7-2\right)=6^{20}:6^{18}+4\)

\(=>x5=6^2+4=36+4=40\)

\(=>x=40:5\)

\(=>x=8\)

đề đúng rồi á bạn.

 
 a.chia cho 15 dư 1
 
b. chia cho 15 dư 14
 
 c.chia hết cho 81

k mình nha

chia 15 dư 1

chia cho 15 dư 14

chia hết cho 81 

tích mk đi dù chỉ 1 cái

a) Chia hết cho 15 dư 1

b) chia 15 dư 14

c) chia hết 81

4: Đặt \(x=\dfrac{a+b}{a-b};y=\dfrac{b+c}{b-c};z=\dfrac{c+a}{c-a}\).

Ta có \(\left(x+1\right)\left(y+1\right)\left(z+1\right)=\dfrac{2a.2b.2c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\left(x-1\right)\left(y-1\right)\left(z-1\right)\)

\(\Rightarrow xy+yz+zx=-1\).

Bất đẳng thức đã cho tương đương:

\(x^2+y^2+z^2\ge2\Leftrightarrow\left(x+y+z\right)^2-2\left(xy+yz+zx\right)-2\ge0\Leftrightarrow\left(x+y+z\right)^2\ge0\) (luôn đúng).

Vậy ta có đpcm

mình xí câu 45,47,51 :>

45. a) Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel ta có :

\(\dfrac{1}{a}+\dfrac{2}{b}=\dfrac{1}{a}+\dfrac{4}{2b}\ge\dfrac{\left(1+2\right)^2}{a+2b}=\dfrac{9}{a+2b}\left(đpcm\right)\)

Đẳng thức xảy ra <=> a=b

b) Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel ta có :

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}\ge\dfrac{\left(1+1+1\right)^2}{a+b+b}=\dfrac{9}{a+2b}\)(1)

\(\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}\ge\dfrac{\left(1+1+1\right)^2}{b+c+c}=\dfrac{9}{b+2c}\)(2)

\(\dfrac{1}{c}+\dfrac{1}{a}+\dfrac{1}{a}\ge\dfrac{\left(1+1+1\right)^2}{c+a+a}=\dfrac{9}{c+2a}\)(3)

Cộng (1),(2),(3) theo vế ta có đpcm

Đẳng thức xảy ra <=> a=b=c

47. Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel ta có :

\(\dfrac{\left(a+b\right)^2}{c}+\dfrac{\left(b+c\right)^2}{a}+\dfrac{\left(c+a\right)^2}{b}\ge\dfrac{\left(a+b+b+c+c+a\right)^2}{a+b+c}=\dfrac{\left[2\left(a+b+c\right)\right]^2}{a+b+c}=\dfrac{4\left(a+b+c\right)^2}{a+b+c}=4\left(a+b+c\right)\)(đpcm)

Đẳng thức xảy ra <=> a=b=c

Ta có 33^44=(33^4)^11=1185921^11

          44^33=(44^3)^11=85184^11

Vì 1185921^11>85184^11 nên 33^44>44^33 

33⁴⁴ = 11^3.44 = 11¹³²

44³³= 11^4.33  = 11¹³²

vậy 33⁴⁴ = 44³³

0,2 x 4 x 44 + 0,8 x 20 + 2,4 x 12

= 0,8 x 44 + 0,8 x 20 + 0,8 x 3 x 12

= 0,8 x 44 + 0,8 x 20 + 0,8 x 36

= 0,8 x (44 + 20 + 36)

= 0,8 x 100

= 80

44 If I had known your name, I could have introduced you with my friends

45 If the driver in front hadn't stopped so quickly, the accident wouldn't have happened

46 If I had known Geogre wanted to get up early, I would have waken him 

47 If Jim hadn't lent me the money, I wouldn't have been able to buy the motorbike 

48 If she hadn't been wearing the seat belt, she wouldn't have injured (câu này sao ta)

49 If she had had enough money on her, she would have bought the coat

50 If....(tên ai đó) hadn't gone to the bus stop in time, she would have late for the interview

51 If.......hadn't recommened the hotel to us, we wouldn't have stayed at this hotel

52 If you had had breakfast, you wouldn't be hungry now

Bài 44:

a, 1,2 : 3,24 = \(\frac{1,2}{3,24}=\frac{120}{324}=\frac{10}{27}\)

b, \(2\frac{1}{5}:\frac{3}{4}=3\frac{41}{45}=\frac{176}{45}\)

c,\(\frac{2}{7}:0,42=\frac{\frac{2}{7}}{0,42}=\frac{2}{7}.\frac{100}{42}=\frac{100}{147}\)

vào đây nè:Chuyên trang lý thuyết tổng hợp các môn THPT bài nào cũng có

<=> 18 - 3 + x - 44 + x =55

<=> x + x = 55 - 18 + 3 + 44

<=> 2x  =  84

<=> x   =   84  : 2

<=> x   =   42

Trả lời :.....................................................

18 + ( -3 + x ) - ( 44 - x ) = 55

\(\Leftrightarrow x+x=55-18+3+44\)

\(\Leftrightarrow2x=84\)

\(\Leftrightarrow x=84:2\)

\(\Leftrightarrow x=42\)

44-x (3+2) = 7.2

44-x.5=14

x.5=30

x=6

\(44-x\left(3+2\right)=7x2=14\)

\(x\left(3+2\right)=44-14=30\)

\(5x=30\)

\(x=30:5=6\)

𝑥=7.36. Mik Mệt nên tạm ko viết LG