\(3^{x-1}+3^{x+1}-1=89\)

\(3^{x-1}.\left(1+3^2\right)=89+1\)

\(3^{x-1}.10=90\)

\(3^{x-1}=90:10\)

\(3^{x-1}=9\)

\(3^{x-1}=3^2\)

\(x-1=2\)

\(x=2+1\)

\(x=3\)

\(3^{x-1}+3^{x+1}-1=89\)

\(\Rightarrow3^x\cdot\left(3^{-1}+3^1\right)=89+1\)

\(\Rightarrow3^x\cdot\left(\dfrac{1}{3}+3\right)=90\)

\(\Rightarrow3^x\cdot\dfrac{10}{3}=90\)

\(\Rightarrow3^x=90:\dfrac{10}{3}\)

\(\Rightarrow3^x=27\)

\(\Rightarrow3^x=3^3\)

\(\Rightarrow x=3\)

.

a) \(\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{90}\right)\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{9\cdot10}\right)\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}-...+\dfrac{1}{9}-\dfrac{1}{10}\right)\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow\left(1-\dfrac{1}{10}\right)\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow\dfrac{9}{10}\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow90-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)=90-89\)

\(\Rightarrow\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)=1\)

\(\Rightarrow x+\dfrac{103}{50}=\dfrac{5}{2}\)

\(\Rightarrow x=\dfrac{11}{25}\)

b) \(x\cdot9,85+x\cdot0,15=0,1\)

\(\Rightarrow x\cdot\left(9,85+0,15\right)=0,1\)

\(\Rightarrow x\cdot10=0,1\)

\(\Rightarrow x=\dfrac{0,1}{10}\)

\(\Rightarrow x=0,01\)

c) \(\dfrac{2}{5}+2022x=\dfrac{4}{10}\)

\(\Rightarrow\dfrac{2}{5}+2022x=\dfrac{2}{5}\)

\(\Rightarrow2022x=\dfrac{2}{5}-\dfrac{2}{5}\)

\(\Rightarrow2022x=0\)

\(\Rightarrow x=\dfrac{0}{2022}\)

\(\Rightarrow x=0\)

a) \(\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{90}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\left(1\right)\)

Ta có :

\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{90}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)

\(\left(1\right)\Rightarrow\dfrac{9}{10}.100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right].2=89\)

\(\Rightarrow\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right].2=90-89\)

\(\Rightarrow\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)=\dfrac{1}{2}\)

\(\Rightarrow x+\dfrac{206}{100}=\dfrac{5}{2}:\dfrac{1}{2}\)

\(\Rightarrow x+\dfrac{103}{50}=\dfrac{5}{2}.\dfrac{2}{1}\)

\(\Rightarrow x+\dfrac{103}{50}=5\)

\(\Rightarrow x=5-\dfrac{103}{50}\)

\(\Rightarrow x=\dfrac{250}{50}-\dfrac{103}{50}\)

\(\Rightarrow x=\dfrac{147}{50}\)

1.

a) = 354

b) = -54

c) = -167

2.

A= -37

có tận cùng là 17 chữ số 0

Vì: 10 x 20 x .....x 80 có tận cùng là 8 số 0 

tích các số với chữ số chẵn với số tận cùng 5 đều có tận cùng là 0 mà từ 1 đến 89 có 9 số có tận cùng là chữ số 5 => có 9 chữ số 0

=>9 + 8 = 17

Thanks bạn!

bn chỉ cần tìm hiểu cách tính toán Gau-xơ

bn doc bai viet cua gau-xo trong [sgk}y nha bn

Dãy số trên được viết như sau :

1 + 2 + 3 + 5 + 8 + 13 + 21 + 34 + 55 + 89

Vậy tổng là : 229

Đáp số : 229

Giúp mình đi các bạn

♪ chia ra thôi ♪...đợi...a chia ko hết à?

= \(\dfrac{760284}{89}\)

= \(\dfrac{4578}{5}\)

= 1234

báo c

\(8765\div90=\dfrac{1753}{18}\)\(=97,4\)

\(1325\times65=86125\)

\(6543\div89=73,52\)