\(\left(2x+5\right)\left(y-3\right)=22\\ \Rightarrow\left(2x+5\right);\left(y-3\right)\inƯ\left(22\right)=\left\{1;2;11;22\right\}\\ TH1:2x+5=1\Rightarrow x=-2\left(loại\right);\left(y-3\right)=22\Rightarrow y=25\\ TH2:2x+5=2\Rightarrow x=-\dfrac{3}{2}\left(loại\right);\left(y-3\right)=11\Rightarrow y=14\\ TH3:2x+5=11\Rightarrow x=3;\left(y-3\right)=2\Rightarrow y=5\\ TH4:2x+5=22\Rightarrow x=\dfrac{17}{2}\left(loại\right);\left(y-3\right)=1\Rightarrow y=4\\Vậy:\left(x;y\right)=\left(3;5\right)\)

(2x + 5)(y - 3) = 22

Ư(22) = {-1,-2,-11,-22,1,2,11,22}

=> Ta có bảng:

Vậy ta có cặp x,y nguyên dương thỏa mãn là: (3;5)

0

0

(0,5 )² .4= ( 0,5 . 2 )² = 1² = 1 

( 0,5)³ . 8 = ( 0,5 . 2 )³ = 1³ = 1 

(0,5)³ . 32 = ( 0,5 . 2 )³ .2² = 1³ .2² = 1.4 = 4 

( 0,5)⁶ . 64 = ( 0,5 . 2 )⁶  = 1⁶ = 1 

5, 0,25² .16 = (0,25.4)² = 1² = 1

6,(0,25)³ .64 = (0,25 .4 )³ = 1³ =1

7,(0,2)² .25 = ( 0,2 .5 )² = 1² = 1 

8,( 0,2 )³ .125 = ( 0,2 . 5 )³ = 1³ = 1

\(a,36-4x^2+20xy-25y^2\\ =36-\left(4x^2-20xy+25y^2\right)\\ =6^2-\left[\left(2x\right)^2-2.2x.5y+\left(5y\right)^2\right]\\ =6^2-\left(2x-5y\right)^2\\ =\left[6-\left(2x-5y\right)\right]\left[6+\left(2x-5y\right)\right]\\ =\left(6-2x+5y\right).\left(6+2x-5y\right)\)

a/

\(=6^2-\left[\left(2x\right)^2-2.2x.5y+\left(5y\right)^2\right]=\)

\(6^2-\left(2x-5y\right)^2=\left[6-\left(2x-5y\right)\right].\left[6+\left(2x-5y\right)\right]\)

\(\left(x-2\right)\left(4x-20\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\4x-20=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\4x=20\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ \left(x-5\right)\left(25-5x?\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\25-5x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\5x=25\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=5\end{matrix}\right.\\ \left(x-4\right)\left(2x-8\right)\\ \Rightarrow\left[{}\begin{matrix}x-4=0\\2x-8=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4\\2x=8\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=4\end{matrix}\right.\)

a,(x-2)(4x-20)=0

=>x-2=0 hoặc 4x-20=0

=>x=2 hoặc x=5

b,(x-5)(25-5)=0

=>x-5=0  ( vì 25-5 ≠0)

=>x=5

c,(x-4)(2x-8)=0

=>x-4=0 hoặc 2x-8=0

=>x=4 

\(A=\dfrac{3}{1^2+2^2}+\dfrac{5}{2^2+3^2}+...+\dfrac{19}{9^2+10^2}\) (sửa \(1^22^2\) thành \(1^2+2^2\))

Ta có : \(\left(1+2\right)^2=1^2+2^2+2.1.2\Rightarrow1^2+2^2< \left(1+2\right)^2\)

\(\Rightarrow1^2+2^2< 3^2=3.3\)

\(\Rightarrow\dfrac{3}{1^2+2^2}< \dfrac{1}{3}< 1\)

Tương tự \(\dfrac{5}{2^2+3^2}< \dfrac{1}{5}< 1\)

\(.....\)

\(\dfrac{9}{9^2+10^2}< \dfrac{1}{19}< 1\)

\(\Rightarrow A=\dfrac{3}{1^2+2^2}+\dfrac{5}{2^2+3^2}+...+\dfrac{19}{9^2+10^2}< 1.9=9< 1\)

\(\Rightarrow dpcm\)

\(\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)...\left(1+\dfrac{1}{99.101}\right)\)

\(=\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}....\dfrac{100^2}{99.101}\)

\(=\dfrac{2.3.4...100}{1.2.3.4...99}.\dfrac{2.3.4...100}{3.4.5....101}\)

\(=\dfrac{100}{1}.\dfrac{2}{101}\)

\(=\dfrac{200}{101}\)

\(\left(2x-\dfrac{3}{2}\right)^2=\dfrac{49}{81}\\ \Rightarrow\left(2x-\dfrac{3}{2}\right)^2=\left(\dfrac{7}{9}\right)^2\\ \Rightarrow2x-\dfrac{3}{2}=\pm\dfrac{7}{9}\\ \Rightarrow\left[{}\begin{matrix}2x-\dfrac{3}{2}=\dfrac{7}{9}\\2x-\dfrac{3}{2}=-\dfrac{7}{9}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=\dfrac{41}{18}\\2x=\dfrac{13}{18}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{41}{36}\\x=\dfrac{13}{36}\end{matrix}\right.\)

(2x-3/2)²=49/81=(7/9)²

=>2x-3/2=7/9 hoặc 2x-3/2=-7/9

=>x=41/36 hoặc x=13/36.

\(A=3^{100}-3^{99}+3^{98}-...-3+1\\ \Rightarrow\dfrac{1}{3}A=3^{99}-3^{98}+3^{97}-...-1+\dfrac{1}{3}\\ \Rightarrow\dfrac{4}{3}A=3^{100}+\dfrac{1}{3}\\ \Rightarrow A=\dfrac{3^{101}}{4}+\dfrac{1}{4}\)

Vế 2 thiếu dấu " ( " bạn ơi.