\(a)3^{2x-1}+2\cdot9^{x-1}=405\\ =>3^{2x-1}+2\cdot\left(3^2\right)^{x-1}=405\\ =>3^{2x-1}+2\cdot3^{2x-2}=405\\ =>3^{2x-2}\cdot\left(3+2\right)=405\\ =>3^{2x-2}\cdot5=405\\ =>3^{2x-2}=\dfrac{405}{5}=81\\ =>3^{2x-2}=3^4\\ =>2x-2=4\\ =>2x=4+2=6\\ =>x=\dfrac{6}{2}\\ =>x=3\)

\(b)\left(\dfrac{1}{3}\right)^{x-1}+5\left(\dfrac{1}{3}\right)^{x+1}=\dfrac{14}{9^3}\\ =>\left(\dfrac{1}{3}\right)^{x-1}\left(1+5\cdot\dfrac{1}{3^2}\right)=\dfrac{14}{729}\\ =>\left(\dfrac{1}{3}\right)^{x-1}\cdot\dfrac{14}{9}=\dfrac{14}{729}\\ =>\left(\dfrac{1}{3}\right)^{x-1}=\dfrac{14}{729}:\dfrac{14}{9}\\ =>\left(\dfrac{1}{3}\right)^{x-1}=\dfrac{9}{729}=\dfrac{1}{81}\\ =>\left(\dfrac{1}{3}\right)^{x-1}=\left(\dfrac{1}{3}\right)^4\\ =>x-1=4\\ =>x=1+4\\ =>x=5\)

\(c)\dfrac{3}{5}\left(3x^3-\dfrac{8}{9}\right)-\dfrac{1}{2}\left(\dfrac{3}{2}-1\right)=-\dfrac{1}{4}\\ =>\dfrac{3}{5}\left(3x^3-\dfrac{8}{9}\right)-\dfrac{1}{2}\cdot\dfrac{1}{2}=-\dfrac{1}{4}\\ =>\dfrac{3}{5}\left(3x^3-\dfrac{8}{9}\right)-\dfrac{1}{4}=-\dfrac{1}{4}\\ =>\dfrac{3}{5}\left(3x^3-\dfrac{8}{9}\right)=0\\ =>3x^3-\dfrac{8}{9}=0\\ =>3x^3=\dfrac{8}{9}\\ =>x^3=\dfrac{8}{9}:3=\dfrac{8}{27}\\ =>x^3=\left(\dfrac{2}{3}\right)^3\\ =>x=\dfrac{2}{3}\)

a: \(3^{2x-1}+2\cdot9^{x-1}=405\)

=>\(\dfrac{3^{2x}}{3}+2\cdot3^{2x-2}=405\)

=>\(\dfrac{1}{3}\cdot3^{2x}+2\cdot3^{2x}\cdot\dfrac{1}{9}=405\)

=>\(3^{2x}\cdot\left(\dfrac{1}{3}+\dfrac{2}{9}\right)=405\)

=>\(3^{2x}\cdot\dfrac{5}{9}=405\)

=>\(3^{2x}=405:\dfrac{5}{9}=405\cdot\dfrac{9}{5}=81\cdot9=3^6\)

=>2x=6

=>x=3

b: \(\left(\dfrac{1}{3}\right)^{x-1}+5\cdot\left(\dfrac{1}{3}\right)^{x+1}=\dfrac{14}{9^3}\)

=>\(\left(\dfrac{1}{3}\right)^x\cdot3+5\cdot\left(\dfrac{1}{3}\right)^x\cdot\dfrac{1}{3}=\dfrac{14}{9^3}\)

=>\(\left(\dfrac{1}{3}\right)^x\cdot\left(3+\dfrac{5}{3}\right)=\dfrac{14}{9^3}\)

=>\(\left(\dfrac{1}{3}\right)^x=\dfrac{14}{3^6}:\dfrac{14}{3}=\dfrac{3}{3^6}=\dfrac{1}{3^5}\)

=>x=5

c: \(\dfrac{3}{5}\left(3x^3-\dfrac{8}{9}\right)-\dfrac{1}{2}\left(\dfrac{3}{2}-1\right)=-\dfrac{1}{4}\)

=>\(\dfrac{9}{5}x^3-\dfrac{24}{45}-\dfrac{1}{2}\cdot\dfrac{1}{2}+\dfrac{1}{4}=0\)

=>\(\dfrac{9}{5}x^3=\dfrac{24}{45}=\dfrac{8}{15}\)

=>\(x^3=\dfrac{8}{15}:\dfrac{9}{5}=\dfrac{8}{15}\cdot\dfrac{5}{9}=\dfrac{40}{135}=\dfrac{8}{27}=\left(\dfrac{2}{3}\right)^3\)

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

d: \(\dfrac{7}{x}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{41\cdot45}=\dfrac{29}{45}\)

=>\(\dfrac{7}{x}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}=\dfrac{29}{45}\)

=>\(\dfrac{7}{x}+\dfrac{9}{45}-\dfrac{1}{45}=\dfrac{29}{45}\)

=>\(\dfrac{7}{x}=\dfrac{29}{45}-\dfrac{8}{45}=\dfrac{21}{45}=\dfrac{7}{15}\)

=>x=15

e: \(\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{\left(2x+1\right)\left(2x+3\right)}=\dfrac{5}{31}\)

=>\(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{\left(2x+1\right)\left(2x+3\right)}=\dfrac{10}{31}\)

=>\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2x+1}-\dfrac{1}{2x+3}=\dfrac{10}{31}\)

=>\(\dfrac{1}{3}-\dfrac{1}{2x+3}=\dfrac{10}{31}\)

=>\(\dfrac{1}{2x+3}=\dfrac{1}{3}-\dfrac{10}{31}=\dfrac{1}{93}\)

=>2x+3=93

=>2x=90

=>x=45

b: \(\dfrac{2}{5}-\left(\dfrac{4}{3}+\dfrac{4}{5}\right)-\left(-\dfrac{1}{9}-0,4\right)+\dfrac{11}{9}\)

\(=\dfrac{2}{5}-\dfrac{4}{3}-\dfrac{4}{5}+\dfrac{1}{9}+\dfrac{2}{5}+\dfrac{11}{9}\)

\(=\left(\dfrac{2}{5}-\dfrac{4}{5}+\dfrac{2}{5}\right)+\left(-\dfrac{4}{3}+\dfrac{1}{9}+\dfrac{11}{9}\right)\)

\(=-\dfrac{4}{3}+\dfrac{12}{9}=0\)

c: \(\dfrac{11}{8}\cdot\left[\left(-\dfrac{5}{11}:\dfrac{13}{8}-\dfrac{5}{11}:\dfrac{13}{5}\right)+\dfrac{-6}{33}\right]+\dfrac{3}{4}\)

\(=\dfrac{11}{8}\cdot\left[-\dfrac{5}{11}\cdot\dfrac{8}{13}-\dfrac{5}{11}\cdot\dfrac{5}{13}+\dfrac{-2}{11}\right]+\dfrac{3}{4}\)

\(=\dfrac{11}{8}\cdot\left[-\dfrac{5}{11}\left(\dfrac{8}{13}+\dfrac{5}{13}\right)-\dfrac{2}{11}\right]+\dfrac{3}{4}\)

\(=\dfrac{11}{8}\cdot\dfrac{-7}{11}+\dfrac{3}{4}=-\dfrac{7}{8}+\dfrac{3}{4}=-\dfrac{1}{8}\)

a:\(\widehat{BAC}+\widehat{xAC}=180^0\)(hai góc kề bù)

=> \(\widehat{BAC}+70^0=180^0\)

=>\(\widehat{BAC}=110^0\)

Ta có: \(\widehat{BAC}+\widehat{ABD}=180^0\)

mà hai góc này là hai góc ở vị trí trong cùng phía

nên AC//BD

b: Vì AC//BD

nên \(\widehat{yCx}=\widehat{CDB}\)(hai góc đồng vị)

=>\(\widehat{yCx}=60^0\)

Ta có: \(\widehat{yCx}+\widehat{ACD}=180^0\)(hai góc kề bù)

=>\(\widehat{ACD}+60^0=180^0\)

=>\(\widehat{ACD}=120^0\)

Ta có: \(\widehat{BAC}+\widehat{ABD}=180^0\)(AC//BD)

=>\(\widehat{BAC}+70^0=180^0\)

=>\(\widehat{BAC}=110^0\)

e: \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)

=>\(\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)=\left(x+1\right)\left(\dfrac{1}{13}+\dfrac{1}{14}\right)\)

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

=>x+1=0

=>x=-1

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...

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)}

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)

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

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