a: \(\dfrac{1}{2}\cdot2^n+4\cdot2^n=9\cdot5^n\)

=>\(2^n\cdot\left(\dfrac{1}{2}+4\right)=5^n\cdot9\)

=>\(2^n\cdot\dfrac{9}{2}=5^n\cdot9\)

=>\(2^{n-1}=5^n\)

=>\(n-1=n\cdot log_25\)

=>\(n\left(1-log_25\right)=1\)

=>\(n=\dfrac{1}{1-log_25}\)

\(\dfrac{x-10}{30}+\dfrac{x-14}{43}+\dfrac{x-5}{95}+\dfrac{x-148}{8}=0\\ \Rightarrow\left(\dfrac{x-10}{30}-3\right)+\left(\dfrac{x-14}{43}-2\right)+\left(\dfrac{x-5}{95}-1\right)+\left(\dfrac{x-148}{8}+3\right)=0\\ \Rightarrow\dfrac{x-100}{30}+\dfrac{x-100}{43}+\dfrac{x-100}{95}+\dfrac{x-100}{8}=0\\ \Rightarrow\left(x-100\right)\left(\dfrac{1}{30}+\dfrac{1}{43}+\dfrac{1}{95}+\dfrac{1}{8}\right)=0\\ \Rightarrow x-100=0\\ \Rightarrow x=100\)

a)

 \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\\ \Rightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\\ \Rightarrow\left(x+1\right)\cdot\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)

Vì \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\) nên:

\(x+1=0\\ \Rightarrow x=-1\)

Vậy...

b) \(\dfrac{315-x}{101}+\dfrac{313-x}{103}+\dfrac{311-x}{105}+\dfrac{309-x}{107}=-4\\ \Rightarrow\dfrac{315-x}{101}+\dfrac{313-x}{103}+\dfrac{311-x}{105}+\dfrac{309-x}{107}+4=0\\ \Rightarrow\left(\dfrac{315-x}{101}+1\right)+\left(\dfrac{313-x}{103}+1\right)+\left(\dfrac{311-x}{105}+1\right)+\left(\dfrac{309-x}{107}+1\right)=0\\ \Rightarrow\dfrac{416-x}{101}+\dfrac{416-x}{103}+\dfrac{416-x}{105}+\dfrac{416-x}{107}=0\\ \Rightarrow\left(416-x\right)\left(\dfrac{1}{101}+\dfrac{1}{103}+\dfrac{1}{105}+\dfrac{1}{107}\right)=0\)

Vì \(\dfrac{1}{101}+\dfrac{1}{103}+\dfrac{1}{105}+\dfrac{1}{107}\ne0\) nên:

\(416-x=0\\ \Rightarrow x=416\)

Vậy...

14 cỉrcles

15 starts

16 doesn't study

17 teaches

18 don't have

19 is studying

20 are making

21 Is Tom driving

22 are not preparing

23 is she doing

viết công thức thôi

Bạn lưu ý, khi đăng đề thì đăng đầy đủ đề (bao gồm cả điều kiện và yêu cầu).

Đề yêu cầu tìm $x,y$?

$x,y$ là số như thế nào? Số nguyên? Số tự nhiên?

Bạn nên ghi rõ ra để mọi người hỗ trợ nhanh hơn nhé. 

\(x+2xy-2y=5\)

\(x+2y\times\left(x-1\right)=5\)

\(\left(x-1\right)+2y\times\left(x-1\right)=5-1\)

\(\left(x-1\right)\times\left(2y+1\right)=4\)

Ta có: 4 = (-1) x (-4) = (-2) x (-2) = 2 x 2 = 1 x 4

Ta lập bảng:

⇒ (x; y) ϵ {(5; 0); (-3; -1)}

1 They are a big fan of watching movie

2 She is interested in cooking

3 My father fancies going jogging

4 He is keen on dancing

He has an enthusiasm for dancing

5 She is fond of watching cartoons

She is a big fan of watching cartoons

Lời giải:
\(B=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}....\frac{-99}{100}\\ =-\frac{3.8.15...99}{4.9...100}\) (do $B$ có lẻ các thừa số)

\(=-\frac{(1.3)(2.4)(3.5)...(9.11)}{2^2.3^2.4^2...10^2}\)

\(=-\frac{(1.2.3...9)(3.4.5...11)}{(2.3....10)(2.3.4...10)}\\ =-\frac{1.2.3...9}{2.3.4...10}.\frac{3.4.5...11}{2.3.4...10}\\ =-\frac{1}{10}.\frac{11}{2}=\frac{-11}{20}< \frac{-11}{21}\)

a: Xét ΔMFA vuông tại F và ΔMFC vuông tại F có

MF chung

FA=FC

Do đó: ΔMFA=ΔMFC

=>\(\hat{FMA}=\hat{FMC}\)

=>MF là phân giác của góc AMC

=>\(\hat{AMC}=2\cdot\hat{FMC}\)

ΔFMC vuông tại F

=>\(\hat{FMC}+\hat{FCM}=90^0\)

=>\(\hat{FMC}=90^0-\hat{ACB}\)

=>\(\hat{AMC}=2\cdot\left(90^0-\hat{ACB}\right)=180^0-2\cdot\hat{ACB}=\hat{BAC}\)

b:

Ta có: \(\hat{ABM}+\hat{ABC}=180^0\) (hai góc kề bù)

\(\hat{CAN}+\hat{CAM}=180^0\) (hai góc kề bù)

mà \(\hat{ABC}=\hat{CAM}\left(=\hat{MCA}\right)\)

nên \(\hat{ABM}=\hat{CAN}\)

Xét ΔABM và ΔCAN có

AB=CA

\(\hat{ABM}=\hat{CAN}\)

BM=AN

Do đó: ΔABM=ΔCAN

=>AM=CN

c: Xét ΔABC có \(\frac{AE}{AB}=\frac{AD}{AC}\)

nên ED//BC

=>ED//BI

Xét ΔDEB và ΔIBE có

DE=IB

\(\hat{DEB}=\hat{IBE}\) (hai góc so le trong, ED//BI)

BE chung

Do đó: ΔDEB=ΔIBE

=>\(\hat{DBE}=\hat{IEB}\)

mà hai góc này là hai góc ở vị trí so le trong

nên EI//BD