a) \(\frac{1985.1987-1}{1980+1985.1986}=\frac{1985.1986+1985-1}{1980+1985.1986}=\frac{1985.1986+1984}{1985.1986+1980}>\frac{1985.1986+1980}{1985.1986+1980}=1\)

b) \(A=\frac{13^{15}+1}{13^{16}+1}=\frac{13^{15}+\frac{1}{13}+\frac{12}{13}}{13^{16}+1}=\frac{\frac{1}{13}\left(13^{16}+1\right)+\frac{12}{13}}{13^{16}+1}=\frac{1}{13}+\frac{12}{13\left(13^{16}+1\right)}\)

\(B=\frac{13^{16}+1}{13^{17}+1}=\frac{13^{16}+\frac{1}{13}+\frac{12}{13}}{13^{17}+1}=\frac{\frac{1}{13}\left(13^{17}+1\right)+\frac{12}{13}}{13^{17}+1}=\frac{1}{13}+\frac{12}{13\left(13^{17}+1\right)}\)

Có \(13^{16}+1< 13^{17}+1\)nên \(\frac{12}{13\left(13^{16}+1\right)}>\frac{12}{13\left(13^{17}+1\right)}\)

Vậy \(A>B\).

Ta có : \(\hept{\begin{cases}\left(3x-2y\right)^2\ge0\\\left|x^2+y^2-52\right|\ge0\end{cases}}\Rightarrow\left(3x-2y\right)^2+\left|x^2+y^2-52\right|\ge0\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}3x-2y=0\\x^2+y^2-52=0\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{x}{2}=\frac{y}{3}\\x^2+y^2=52\end{cases}}\)

Đặt \(\frac{x}{2}=\frac{y}{3}=k\Leftrightarrow\hept{\begin{cases}x=2k\\y=3k\end{cases}}\)

Khi đó x² + y² = 52

 <=> (2k)² + (3k)² = 52

<=> 13k² = 52 

<=> k² = 4

<=> \(k=\pm2\)

Khi k = 2 => x = 4 ; y = 6

Khi k = -2 => x = -4 ; y = -6

Vậy các cặp (x;y) thỏa mãn là (4;6) ; (-4;-6) 

i)  ta có tỉ lệ thức : 14/10=21/15

10/14=15/21

10/15=14/21

14/10=21/15 

i ) 14/10 = 21/15 ; 10/14 = 15/21 ; 15/10 = 21/14 ; 10/15 = 14/21

ii ) AB/2 = 3/CD ; 2/AB = CD/3 ; CD/2 = 3/AB ; 2/CD = AB/3 

iii ) AB/EF = GH/CD ; EF/AB = CD/GH ; CD/EF = GH/AB ; EF/CD = AB/GH

iv ) 4/5 = MN/AB ; 5/4 = AB/MN ; AB/5 = MN/4  ; 5/AB = 4/MN 

a, 2(x + 1) + 3(x - 4 ) = 0

=> 2x + 2 + 3x - 12 = 0

=> 5x - 10 = 0

=> x = 2

b, 9x^2 - 16 = 0

=> 9x^2 = 16

=> x^2 = 16/9

=> x = +- 4/3

c, 2x^2 + 7x - 9 = 0

=> 2x^2 - 2x + 9x - 9 = 0

=> 2x(x - 1) + 9(x - 1) = 0

=> (2x + 9)(x - 1) = 0

=> x = -9/2 hoac x = 1

bài 1

ta có : \(\left(x+\frac{1}{2}\right)^2\ge0\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{5}{4}\ge\frac{5}{4}\)

bài 2 . ta có :

\(M=x^2\left(x+y-3\right)-y\left(x+y-3\right)+x+y+2019=0-0+3+2019=2022\)

\(\left(20^{2016}+11^{2016}\right)^{2017}>\left(20^{2016}+11^{2016}\right)^{2016}.20^{2016}\)

\(=\left(20.20^{2016}+20.11^{2016}\right)^{2016}>\left(20^{2017}+11^{2017}\right)^{2016}\)

Ta có: 9⁹⁹ = 9^{11 . 9} = (9¹¹)⁹

Do 9¹¹ > 99 <=> (9¹¹)⁹ > 99⁹

<=> 9⁹⁹ > 99⁹.

Vậy 9⁹⁹ > 99⁹.

ta có :

\(\hept{\begin{cases}9^{99}=\left(9^{11}\right)^9\\99^9< \left(9^3\right)^3\end{cases}}\) vì vậy \(99^9< 9^{99}\)