M=-(\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{1-100^2}{100^2}\))

=-(\(\frac{1.3}{2.2}.\frac{2.4}{3.3}\frac{3.5}{4.4}...\frac{99.100}{100.100}\))

=-(\(\frac{1.2.3...99}{2.3.4...100}.\frac{3.4.5...100}{2.3.4..100}\))

=-(\(\frac{1}{100}.\frac{1}{2}\))

=\(\frac{-1}{200}\)

thanks you very very much

32ⁿ - 16ⁿ = 2048

=> (2⁵)ⁿ - (2⁴)ⁿ = 2¹¹

=> 2ⁿ(2⁵ - 2⁴) = 2¹¹

=> 2ⁿ.2³ = 2¹¹

=> 2ⁿ = 2⁸

\(80-\left(4.5^2-3.2^3\right)=2^{10}-\left(x-4\right)\) )

=> \(80-\left[2^2\left(5^2-3.2\right)\right]=2^{10}-x+4\)

=> \(80-\left[2^2\left(25-6\right)\right]=2^{10}-x+4\)

=> \(80-\left(4.19\right)=2^{10}-x+4\)

=> \(80-76-4=2^{10}-x\)

=> \(2^{10}-x=0\)

=> \(x=1024\)

Vì \(\left(\frac{1}{2}x-5\right)^{10}\ge0\)và \(\left(y^2-\frac{1}{4}\right)^{20}\ge0\)

nên \(\left(\frac{1}{2}x-5\right)^{10}+\left(y^2-\frac{1}{4}\right)^{20}=0\)

<=>\(\hept{\begin{cases}\frac{1}{2}x-5=0\\y^2-\frac{1}{4}=0\end{cases}}\)<=>\(\hept{\begin{cases}x=10\\y=\pm\frac{1}{2}\end{cases}}\)

Ta có:\(\hept{\begin{cases}\left\{\frac{1}{2}x-5\right\}^{10}\ge0\forall x\\\left\{y^2-\frac{1}{4}\right\}^{20}\ge0\forall y\end{cases}}\)

Mà \(\left\{\frac{1}{2}x-5\right\}^{10}+\left\{y^2-\frac{1}{4}\right\}^{20}\le0\)

\(\Rightarrow\left\{\frac{1}{2}x-5\right\}^{10}+\left\{y^2-\frac{1}{4}\right\}^{20}=0\)

\(\Leftrightarrow\hept{\begin{cases}\left\{\frac{1}{2}x-5\right\}^{10}=0\\\left\{y^2-\frac{1}{4}\right\}^{20}=0\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{1}{2}x-5=0\\y^2-\frac{1}{4}=0\end{cases}\Leftrightarrow}\hept{\begin{cases}\frac{1}{2}x=5\\y^2=\frac{1}{4}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=10\\y=\pm\frac{1}{2}\end{cases}}}\)

Vậy \(x=10;y=\pm\frac{1}{2}\)

Ta có :

\(3a+2b⋮17\)

\(\Rightarrow9\left(3a+2b\right)⋮17\)

\(\Rightarrow27a+18b⋮17\)

\(\Rightarrow\left(17a+17b\right)+\left(10a+b\right)⋮17\)

\(\Rightarrow10a+b⋮17\)(1)

Ta có :

\(10a+b⋮17\)

\(\Rightarrow2\left(10a+b\right)⋮17\)

\(\Rightarrow20a+2b⋮17\)

\(\Rightarrow17a+3a+2b⋮17\)

\(\Rightarrow3a+2b⋮17\)(2)

Từ (1) và (2) \(\Rightarrow3a+2b⋮17\Leftrightarrow10a+b⋮17\)(đpcm)

_Chúc bạn học tốt_

\(9^{11}+1=\left(9^2\right)^5.9+1=81^5.9+1=\left(...1\right).9+1=\left(...9\right)+1=\left(...0\right)⋮2,5\\ \)

\(\left(x+5\right)\left(x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+5=0\\x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-5\\x=4\end{cases}}\)

Vậy x = -5 hoặc x = 4

_Chúc bạn học tốt_

(x+5).(x-4)=0 

<=> x+5 =0  ; x -4 =0

<=> x= -5 ; x= 4 

~~~ hok tốt nha ~~~