1) \(M=\frac{x^2+y^2+7}{x^2+y^2+5}=1+\frac{2}{x^2+y^2+5}\)

Ta có: \(x^2+y^2\ge0,\forall x;y\)

=> \(x^2+y^2+5\ge5\) với mọi x; y 

=> \(\frac{2}{x^2+y^2+5}\le\frac{2}{5}\)

=> \(M\le1+\frac{2}{5}=\frac{7}{5}\)

Dấu "=" xảy ra <=> x = y = 0 

Vậy max M = 7/5 đạt tại x = y = 0 

2) \(f\left(x-1\right)=x^2-3x+5=x^2-x-2x+2+3\)

\(=x\left(x-1\right)-2\left(x-1\right)+3=x\left(x-1\right)-\left(x-1\right)-\left(x-1\right)+3\)

\(=\left(x-1\right)\left(x-1\right)-\left(x-1\right)+3\)

=> \(f\left(x\right)=x.x-x+3=x^2-x+3\)

x + y = x.y

=> xy - x - y = 0

=> (xy - x) - y + 1 = 1

=> x(y - 1) - (y - 1) = 1

=> (x - 1)(y - 1) = 1

=> x - 1 = y - 1 = 1 hoặc x - 1 = y - 1 = -1

=> x = y = 2 hoặc x = y = 0

`#3107.101107`

\(x(x+5)(x-5) - (x+2)(x^2-2x+4)=5\)

`<=> x(x^2 - 25) - (x^3 + 2^3) = 5`

`<=> x^3 - 25x - x^3 - 8 = 5`

`<=> -25x - 8 = 5`

`<=> -25x = 13`

`<=> x = -13/25`

Vậy, `x = -13/25`

_____

\((x+1)^3 - (x-1)^3 -6(x-1)^2 = -19\)

`<=> x^3 + 3x^2 + 3x + 1 - (x^3 - 3x^2 + 3x - 1) - 6(x^2 - 2x + 1) = -19`

`<=> x^3 + 3x^2 + 3x + 1 - x^3 + 3x^2 - 3x + 1 - 6x^2 + 12x - 6 = -19`

`<=> (x^3 - x^3) + (3x^2 + 3x^2 - 6x^2) + (3x - 3x + 12x) + (1 + 1 - 6) = -19`

`<=> 12x - 4 = -19`

`<=> 12x = -15`

`<=> x = -15/12 = -5/4`

Vậy, `x = -5/4.`

________

`@` Sử dụng các hđt:

`1)` `A^2 + B^2 = (A - B)(A + B)`

`2)` `A^3 + B^3 = (A + B)(A^2 - AB + B^2)`

`3)` `(A - B)^3 = A^3 - 3A^2B + 3AB^2 - B^3`

`4)` `(A + B)^3 = A^3 + 3A^2B + 3AB^2 + B^3`

`5)` `(A - B)^2 = A^2 - 2AB + B^2.`

a: \(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=5\)

=>\(x\left(x^2-25\right)-x^3-8=5\)

=>\(x^3-25x-x^3-8=5\)

=>-25x=13

=>\(x=-\dfrac{13}{25}\)

b: \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-19\)

=>\(x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-2x+1\right)=-19\)

=>\(6x^2+2-6x^2+12x-6=-19\)

=>12x-4=-19

=>12x=-15

=>x=-5/4

 20 . 2^x + 1 = 10.4^2 + 1 

 20 . 2^x + 1 = 10 . 16 + 1

20 . 2^x + 1 = 161

20 . 2^x = 161 - 1

20 . 2^x = 160

2^x = 8

2^x = 2^3

=> x = 3

Ban co giai dc phan 2 ko ?

( 4 - x : 2 )^3 - 1 = 2 . ( 2^3 - 5 : 2^0 )

( 4 - x : 2 )^3 - 1 = 2 . ( 8 - 5 : 1 )

( 4 - x : 2 )^3 - 1 = 2 . 3

( 4 - x : 2 )^3 - 1 = 6

( 4 - x : 2 )^3  = 7

=> ko tìm đc x

7 \(\times\) ( 2\(x\) - 5) - 5 \(\times\) (7\(x\) - 2) + 2 \(\times\) (5\(x\) - 7) = (\(x\) - 2) - (\(x\) +4)

14\(x\) - 35 - 35\(x\) + 10 + 10\(x\) - 14 = \(x\) - 2 - \(x\) - 4

(14\(x\) - 35\(x\) + 10\(x\)) - (35 - 10+ 14)  = -6

(- 21 \(x\) + 10\(x\)) - (25 + 14) = - 6

         -11\(x\) - 39 = - 6

          -11\(x\)       = - 6 + 39

          - 11\(x\)     = 33

                \(x\) = 33 : (-11)

                \(x\) =  - 3

ko đúng r bạn ơi

bạn xem đề bài đúng chưa? Vế phải là phép nhân hay phép trừ?

x(1+1+1+5)+7=891

x8=891-7=884

x=884:8=221/2=110.5

x +  x + x + 5 * x + 2 = 891

x * 1 + x * 1 + x * 1 + 5 * x + 2 = 891

x * ( 1 + 1 + 1 + 5 ) = 891 - 2

x * 8 = 889

x =889 :8 

x = 111,125