Ta có

   \(28602+x-2682=25920+x=2880.9+x\)

  Vì \(9⋮9\Rightarrow2880.9⋮9\Rightarrow25920⋮9\)

Để \(25920+x⋮9\Rightarrow x⋮9\)

Mà \(42\le x\le50\)

Nên x = \(45\)

   Vậy x = 45

Ta xét phương trình \(4x-5y-6xy+7=0\Leftrightarrow2x\left(2-3y\right)=5y-7\)

\(\Leftrightarrow2x=\frac{5y-7}{2-3y}\Leftrightarrow x=\frac{5y-7}{4-6y}\)

Để x nguyên thì \(\frac{5y-7}{4-6y}\)nguyên hay \(5y-7⋮4-6y\)

\(\Leftrightarrow6\left(5y-7\right)⋮4-6y\Leftrightarrow30y-42⋮4-6y\)

\(\Leftrightarrow-22-5\left(4-6y\right)⋮4-6y\)

Mà \(-5\left(4-6y\right)⋮4-6y\)nên \(-22⋮4-6y\)hay \(4-6y\inƯ\left(22\right)=\left\{\pm1;\pm2;\pm11;\pm22\right\}\)

Mà 4 - 6y chẵn nên \(4-6y\in\left\{\pm2;\pm22\right\}\)

Lập bảng:

Vậy phương trình có 2 nghiệm \(\left(x,y\right)\in\left\{\left(1;1\right);\left(-1;-3\right)\right\}\)

Ta có : \(A=\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{ab}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{4}{2ab}\)

Sử dụng BĐT Bunhiacopxki ta có :

\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{4}{2ab}=\frac{1^2}{a^2}+\frac{1^2}{b^2}+\frac{2^2}{2ab}\ge\frac{\left(1+1+2\right)^2}{a^2+b^2+2ab}\)

\(=\frac{4^2}{\left(a+b\right)^2}=\frac{16}{2^2}=\frac{16}{4}=4\)

Dấu = xảy ra khi và chỉ khi \(a=b=1\)

Vậy \(A_{min}=4\)khi \(a=b=1\)

\(A=\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{ab}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{4}{2ab}\)

\(\ge\frac{\left(1+1+2\right)^2}{a^2+2ab+b^2}=\frac{16}{\left(a+b\right)^2}=\frac{16}{4}=4\)

Dấu "=" xảy ra <=> a = b = 1

Trả lời hộ mình mn ưi 

1, Businessman is going to that Hanoi

2, Is he driving his truck ?

Study well ^^__^^

1 That businessman is going to Ha Noi

2 Is he driving his truck ?

Hok tốt !!!!!!!!!!!!!

mọi người giúp mk vs

a. \(3^9-8=\left(3^3\right)^3-2^3=27^3-2^3\)

\(=\left(27-2\right)\left(27^2+54+4\right)=25\left(27^2+58\right)⋮25\)( đpcm )

b. Đặt số lẻ đó là 2k + 1

Theo đề ta có : ( 2k + 1 )² - 1 chia hết cho 8

=> ( 2k + 1 - 1 ) ( 2k + 1 + 1 )

=> 2k ( 2k + 2 )

=>4k² + 4k

Vì 4k² chia hết cho 4 ; 4k chia hết cho 2

=> 4k² + 4k chia hết cho 8 

=> Đpcm

a) \(\frac{6x-5}{-7}=\frac{5x-3}{-5}\)

=> -5(6x - 5) = -7(5x - 3)

=> -30x + 25 = -35x + 21

=> -30x + 25  + 35x - 21 = 0

=> (-30x + 35x) + (25 - 21) = 0

=> 5x + 4 = 0

=> 5x = -4

=> x = -4/5

b) \(\frac{12-7x}{-13}=\frac{4-3x}{-5}\)

=> -5(12 - 7x) = -13(4 - 3x)

=> -60 + 35x = -52 + 39x

=> -60 + 35x + 52 - 39x = 0

=> (-60 + 52) + (35x - 39x) = 0

=> -8 - 4x = 0

=> -8 = 4x

=> x = -2

c) \(\frac{2x+4}{7}=\frac{4x-2}{15}\)

=> 15(2x + 4) = 7(4x - 2)

=> 30x + 60 = 28x - 14

=> 30x + 60 - 28x + 14 = 0

=> 2x + 74 = 0

=> 2x = -74

=> x = -37