\(\sqrt{2-x^2}+\sqrt{2-\frac{1}{x^2}}=4-\left(x+\frac{1}{x}\right)\)

\(\Rightarrow2-x^2+2-\frac{1}{x^2}+2\sqrt{\left(2-x^2\right)\left(2-\frac{1}{x^2}\right)}=16-8\left(x+\frac{1}{x}\right)+\left(x+\frac{1}{x}\right)^2\)

\(\Rightarrow4-\left(x^2+\frac{1}{x^2}\right)+2\sqrt{5-2\left(x^2+\frac{1}{x^2}\right)}=16-8\left(x+\frac{1}{x}\right)+\left(x+\frac{1}{x}\right)^2\)

\(\Rightarrow x^2+\frac{1}{x^2}+2\sqrt{5-2\left(x^2+\frac{1}{x^2}\right)}=8\left(x+\frac{1}{x}\right)-\left(x+\frac{1}{x}\right)^2-12\)

Đặt \(a=x+\frac{1}{x}\Rightarrow\left|a\right|=\left|x+\frac{1}{x}\right|=\left|x\right|+\frac{1}{\left|x\right|}\ge2\Rightarrow\left|a\right|\ge2\)

Phươn trình trở thành:

\(a^2-2+2\sqrt{5-2\left(a^2-2\right)}=8a-a^2-12\)

Tớ nghĩ là theo cách này có vẻ khả quan

x=3; y=5

(x+1)+(x+2)+(x+3)+...+(x+30)=795

30x+(1+..+30)=795

30x+(1+31)*30/2=795

30x+465=795

30x=795-465

30x=330

x=11

31x mà

a: \(2x+5⋮x+1\)

=>\(2x+2+3⋮x+1\)

=>\(3⋮x+1\)

=>\(x+1\in\left\{1;-1;3;-3\right\}\)

=>\(x\in\left\{0;-2;2;-4\right\}\)

b: \(5x+9⋮x+2\)

=>\(5x+10-1⋮x+2\)

=>\(-1⋮x+2\)

=>\(x+2\in\left\{1;-1\right\}\)

=>\(x\in\left\{-1;-3\right\}\)

c: \(2x+11⋮x+3\)

=>\(2x+6+5⋮x+3\)

=>\(5⋮x+3\)

=>\(x+3\in\left\{1;-1;5;-5\right\}\)

=>\(x\in\left\{-2;-4;2;-8\right\}\)

d: \(4x+9⋮2x+1\)

=>\(4x+2+7⋮2x+1\)

=>\(7⋮2x+1\)

=>\(2x+1\in\left\{1;-1;7;-7\right\}\)

=>\(2x\in\left\{0;-2;6;-8\right\}\)

=>\(x\in\left\{0;-1;3;-4\right\}\)

e: \(6x+7⋮3x+1\)

=>\(6x+2+5⋮3x+1\)

=>\(5⋮3x+1\)

=>\(3x+1\in\left\{1;-1;5;-5\right\}\)

=>\(3x\in\left\{0;-2;4;-6\right\}\)

=>\(x\in\left\{0;-\dfrac{2}{3};\dfrac{4}{3};-2\right\}\)

g: \(10x+13⋮5x+1\)

=>\(10x+2+11⋮5x+1\)

=>\(11⋮5x+1\)

=>\(5x+1\in\left\{1;-1;11;-11\right\}\)

=>\(5x\in\left\{0;-2;10;-12\right\}\)

=>\(x\in\left\{0;-\dfrac{2}{5};2;-\dfrac{12}{5}\right\}\)

A=\(\frac{2}{2x4}+\frac{2}{4x6}+.........+\frac{2}{2014x2016}\)

=\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+.............+\frac{1}{2014}-\frac{1}{2016}\)

=\(\frac{1}{2}-\frac{1}{2016}\)

=\(\frac{1008}{2016}-\frac{1}{2016}\)

=\(\frac{1007}{2016}\)

mình nghĩ nguyễn hung phat làm đúng sồi

39916800

1 x 2 x 3 x 4 x ....... x 11 = 39916800

1 x 2 x 2 x 4 x ..... x 11 = 39 916 800

kkkkkkk

kkkkkkk

ta có x+5 là ước của 102 và lớn hơn hoặc bằng 5 nên 

\(x+5\in\left\{17,34,51,102\right\}\)

từ đó ta tìm được các cặp (x,y) là \(\left(12,4\right);\left(29,1\right);\left(46,0\right)\)

x²+4x+4=25

(x+2)²=25

(x+2)=5

x=3

1 + 1 = 3

        Đúng ko

trời đương nhiên là = 2 ròi

bằng 2.0