\(x^2+2x+4⋮x+1\)

\(\Leftrightarrow\left(x^2+x\right)+\left(x+1\right)+3⋮x+1\)

\(\Leftrightarrow x\left(x+1\right)+\left(x+1\right)+3⋮x+1\)

\(\Leftrightarrow3⋮x+1\)

\(\Leftrightarrow x+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

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

Ta có: \(x^2+2x+4\)

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

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

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

Để \(x^2+2x+4\) chia hết cho x + 1 thì 3 phải chia hết cho x + 1

\(\Rightarrow\left(x+1\right)\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

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

Bài làm:

Ta có: \(xy+2x+y=9\)

\(\Leftrightarrow\left(xy+2x\right)+\left(y+2\right)=11\)

\(\Leftrightarrow x\left(y+2\right)+\left(y+2\right)=11\)

\(\Leftrightarrow\left(x+1\right)\left(y+2\right)=11\)

Mà \(11=1.11=\left(-1\right).\left(-11\right)\) nên ta xét:

+ Nếu: \(\hept{\begin{cases}x+1=1\\y+2=11\end{cases}}\Rightarrow\hept{\begin{cases}x=0\\y=9\end{cases}}\)

+ Nếu: \(\hept{\begin{cases}x+1=11\\y+2=1\end{cases}}\Rightarrow\hept{\begin{cases}x=10\\y=-1\end{cases}}\)

+ Nếu: \(\hept{\begin{cases}x+1=-1\\y+2=-11\end{cases}}\Rightarrow\hept{\begin{cases}x=-2\\y=-13\end{cases}}\)

+ Nếu: \(\hept{\begin{cases}x+1=-11\\y+2=-1\end{cases}}\Rightarrow\hept{\begin{cases}x=-12\\y=-3\end{cases}}\)

Vậy \(\left(x;y\right)\in\left\{\left(0;9\right);\left(10;-1\right);\left(-2;-13\right);\left(-12;-3\right)\right\}\)

Mệt-.-

:v Thôi thì làm cho bạn nè :)

Ta có: xy + 2x + y = 9

=> xy + 2x + y + 2 = 11

=> x(y + 2) + (y + 2) = 11

=> (y + 2) (x + 1) = 11

=> y - 2, x + 1 thuộc Ư(11) = {-11 ; -1 ; 1 ; 11}

Ta có bảng sau: ...

Bạn tự lập bảng nhé, tương tự như bài trước thôi ạ.

a, ( 44 - x ) / 3 = ( x - 12 ) / 5

=> 5 ( 44 - x  ) = 3 ( x - 12 )

     220 - 5x     = 3x  - 36

     - 5x - 3x     = - 36 - 220

      - 8 x          = - 256

           x          = 32

b , ( 3 - x ) / 4 = ( 2x + 7 ) / 5

=> 5 ( 3 - x )   = 4 ( 2x + 7 )

     15 - 5x      = 8 x  + 28

     - 5 x - 8 x  = 28 - 15

        - 13 x     = 13

               x     = -1

a, \(\frac{\left(44-x\right)}{3}=\frac{\left(x-12\right)}{5}\)

 => (44 - x) . 5 = (x - 12) . 3

 => 44 - x . 5   = x - 12 .3

 => 44 - x . 5   = x - 36

 => x5 + x        = - 36 - 44

 => x5 + x        = - 80

=> x . (5 + 1)    = - 80

=> x . 6           = - 80

=> x                = - 80 : 6

=> x                = - 13,3

b, \(\frac{\left(3-x\right)}{4}=\frac{\left(2x+7\right)}{5}\)

=> (3 - x) . 5 = (2x + 7) . 4

=> 3 - x . 5   = 2x + 7 . 4

=> 3 - x . 5   = 2x + 28

=> -x . 5 + 2x = 28 - 3

=> -x . 5 + 2x = 25

=>  x . 5 + 2x = 25

=>  x . (5 + 2) = 25

=>  x . 7         = 25

=>  x              = 25 : 7

=>  x              = 3,57

a)(44-x).5=(x-12).3

44.5-x.5=x.3-12.3

220-x.5=x.3-36

220-x.5-x.3=-36

x.5-x.3=220+36

x.(5-3)=256

x.2=256

x=256:2

x=128

a) \(A=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2x+10}{\left(x+5\right)\left(x-5\right)}\)

\(A=\dfrac{x-5+2x+10-2x-10}{\left(x+5\right)\left(x-5\right)}=\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}=\dfrac{1}{x+5}\)

b) \(A=-3\Rightarrow\dfrac{1}{x+5}=-3\)

\(\Leftrightarrow x+5=-\dfrac{1}{3}\Leftrightarrow x=-\dfrac{1}{3}-5=\dfrac{-16}{3}\)

\(9x^2-42x+49=\left(3x-7\right)^2=\left(3.\dfrac{-16}{3}-7\right)^2=\left(-23\right)^2=529\) \(\left(x=\dfrac{-16}{3}\right)\)

\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right).\left(2x+3\right)}=\frac{15}{93}\)

\(2.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right).\left(2x+3\right)}\right)=2.\frac{15}{93}\)

\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{\left(2x+1\right).\left(2x+3\right)}=\frac{30}{93}\)

\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)

\(\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)

\(\frac{1}{2x+3}=\frac{1}{3}-\frac{10}{31}\)

\(\frac{1}{2x+3}=\frac{1}{93}\)

=> 2x + 3 = 93

=> 2x = 93 - 3

=> 2x = 90

=> x = 90 : 2

=> x = 45

Vậy x = 45

sai rồi

Đặt \(A=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x+1\right).\left(2x+3\right)}\)

\(2A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2x+1\right).\left(2x+3\right)}\)

\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}\)

\(2A=\frac{1}{3}-\frac{1}{2x+3}\)

\(A=\frac{2x}{3\left(2x+3\right)}:2\)

\(\Rightarrow\frac{2x}{3\left(2x+3\right)}:2=\frac{15}{93}\)

\(\frac{2x}{3\left(2x+3\right)}=\frac{15}{93}.2=\frac{30}{93}=\frac{10}{31}\)

\(\frac{2x}{6x+9}=\frac{10}{31}\)

\(\Rightarrow2x.31=10.\left(6x+9\right)\)

\(\Rightarrow62x=60x+90\)

\(62x-60x=90\)

\(2x=90\)

\(x=45\)

Vậy x = 45

Ủng hộ mk nha !!! ^_^

99 - 97 + 95 - 93 + 91 - 89 + ... + 7 - 5

= ( 99 - 97 ) + ( 95 - 93 ) + ( 91 - 89 ) + ... + ( 7 - 5 )

= 2 + 2 + 2 + ... + 2

Dãy số trên có số cặp là :

       [ ( 99 - 5 ) : 2 + 1 ] : 2 = 24 ( cặp )

\(\Rightarrow\) 2 . 24 = 48

Vậy phép tính trên = 48 .

- Giúp mình với 

có số số hạng là:

(99-1):2+1=50

có số cặp là: 50:2=25cặp

(99-97)+(95-93)...+(7-5)-(3-1)=2 nhân 25 =50 vậy kết quả là 50

k mình nha công của mình cả đấy

Ta có 6 chia hết cho x-1 

=> x-1 thuộc Ư(6)

=> Ư(6)={1;2;3;6)

=> X=2;3;4;7

Câu b đâu bạn ?

A=9-4x-x²

  =-(9+4x+x²)

  =-((x+2)²+5) 

  =-(x+2)²-5          Mặt khác: -(x+2)²\(\le\)0

                                            =>-(x+2)²-5\(\le\)-5      Vậy MAX (A)=-5

B=2x-x²

B-1=2x-x²-1

B-1=-(-2x+x²+1)

B-1=-(x-1)²

B=-(x-1)²+1                      Mặt khác: -(x-1)²\(\le\)0

                                                       =>-(x-1)²+1\(\le\)1