(3x-5)(2y+7)=100

=>(3x-5;2y+7)\(\in\){(1;100);(100;1);(-1;-100);(-100;-1);(2;50);(50;2);(-2;-50);(-50;-2);(4;25);(25;4);(-4;-25);(-25;-4);(5;20);(20;5);(-5;-20);(-20;-5);(10;10);(-10;-10)}

=>(3x;2y)\(\in\){(6;93);(105;-6);(4;-107);(-95;-8);(7;43);(55;-5);(3;-57);(-45;-9);(9;18);(30;-3);(1;-32);(-20;-11);(10;13);(25;-2);(0;-27);(-15;-12);(15;3);(-5;-17)}

=>(x;y)\(\in\){(2;93/2);(35;-3);(4/3;-107/2);(-95/3;-4);(7/3;43/2);(55/3;-5/2);(1;-57/2);(-15;-9/2);(3;9);(10;-3/2);(1/3;-16);(-20/3;-11/2);(10/3;13/2);(25/3;-1);(0;-27/2);(-5;-6);(5;3/2);(-5/3;-17/2)}

(3x - 5)(2y + 7) = 100

Ta có: 100 = 1 x 100 = 2 x 50 = 4 x 25

Do 2y + 7 là số lẻ nên 2y + 7 chỉ có thể = 1 hoặc 25

Trường hợp 1: 2y + 7 = 1

⇒ 2y = 1 - 7

⇒ 2y = -6

⇒ y = (-6) : 2

⇒ y = -3

Vậy 3x - 5 = 100

⇒ 3x = 100 + 5

⇒ 3x = 105

⇒ x = 105 : 3

⇒ x = 35

Trường hợp 2: 2y + 7 = 25

⇒ 2y = 25 - 7

⇒ 2y = 18

⇒ y = 18 : 2

⇒ y = 9

Vậy 3x - 5 = 4

⇒ 3x = 4 + 5

⇒ 3x = 9

⇒ x = 9 : 3

⇒ x = 3

Vậy (x; y) ϵ {(35; -3); (3; 9)}

`H = 1 - 5 + 5^2 - 5^3 + ... + 5^100`

`5H = 5 - 5^2 + 5^3 - 5^4 + ... + 5^101`

`5H + H = (5 - 5^2 + 5^3 - 5^4 + ... + 5^101) + (1 - 5 + 5^2 - 5^3 + ... + 5^100)`

`6H = 1 + (5 - 5) + (5^2 - 5^2) + ... + (5^100 - 5^100) + 5^101`

`6H = 1 +  5^101`

`H = (1 +  5^101)/6`

ΔEHF vuông tại H

=>\(HE^2+HF^2=EF^2\)

=>\(HE=\sqrt{5^2-3^2}=4\left(cm\right)\)

Xét ΔHEG vuông tại H và ΔHFE vuông tại H có

\(\widehat{HEG}=\widehat{HFE}\left(=90^0-\widehat{G}\right)\)

Do đó: ΔHEG~ΔHFE

=>\(\dfrac{HE}{HF}=\dfrac{HG}{HE}\)

=>\(HE^2=HF\cdot HG\)

=>\(HG=\dfrac{4^2}{3}=\dfrac{16}{3}\left(cm\right)\)

ΔEHG vuông tại H

=>\(HE^2+HG^2=EG^2\)

=>\(EG=\sqrt{\left(\dfrac{16}{3}\right)^2+4^2}=\dfrac{8\sqrt{13}}{3}\left(cm\right)\)

`12x + 6/7 : 2 = 9x + 8`

`=> 12x + 3/7 - 9x  =  8`

`=> 3x   =  8 -  3/7`

`=> 3x = 53/7`

`=> x = 53/7 : 3`

`=> x = 53/7 . 1/3`

`=> x=  53/21`

Vậy ..

\(12x+\dfrac{6}{7}\times\dfrac{1}{2}=9x+8\)

\(12x+\dfrac{3}{7}=9x+8\)

\(12x=9x+8-\dfrac{3}{7}\)

\(12x=9x+\dfrac{53}{7}\)

\(\Rightarrow\) \(\dfrac{53}{7}=3x\)

\(\Rightarrow\dfrac{53}{7}\div3=x\)

\(\Rightarrow\dfrac{53}{21}=x\)

Vậy \(x=\dfrac{53}{21}\)

I am down with a cold and fever, so I can’t go to school

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