\(\frac{x-2}{27}+\frac{x-3}{26}+\frac{x-4}{25}+\frac{x-5}{24}+\frac{x-44}{5}=1\)

\(\Leftrightarrow\left(\frac{x-2}{27}-1\right)+\left(\frac{x-3}{26}-1\right)+\left(\frac{x-4}{25}-1\right)+\left(\frac{x-5}{24}-1\right)\)\(+\left(\frac{x-44}{5}+3\right)=1-1\)

\(\Leftrightarrow\frac{x-29}{27}+\frac{x-29}{26}+\frac{x-29}{25}+\frac{x-29}{24}\)\(+\frac{x-29}{5}=0\)

\(\Leftrightarrow\left(x-29\right)\left(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{5}\right)=0\)

Mà \(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{5}\ne0\)

=> x - 29 = 0

=> x = 29.

khong biet

\(-4\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)

\(\Rightarrow-\frac{13}{3}.\left(\frac{3}{6}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{4}{12}-\frac{6}{12}-\frac{9}{12}\right)\)

\(\Rightarrow-\frac{13}{3}.\frac{2}{6}\le x\le-\frac{2}{3}.\frac{-11}{12}\)

\(\Rightarrow\frac{-13}{9}\le x\le\frac{11}{18}\)

\(\Rightarrow\frac{-26}{18}\le x\le\frac{11}{18}\)

=> -1,44444444444........... ≤ x ≤ 0,6111111111...........

Mà x ∈ Z

=> x ∈ { -1 ; 0 }

\(x\in\varnothing\) 

\(A=\left(\frac{x^2-16}{x-4}-1\right):\left(\frac{x-2}{x-3}+\frac{x+3}{x+1}+\frac{x+2-x^2}{x^2-2x-3}\right)\)ĐK : \(x\ne3;-1;4\)

\(=\left(\frac{\left(x-4\right)\left(x+4\right)}{x-4}-1\right):\left(\frac{\left(x-2\right)\left(x+1\right)}{\left(x-3\right)\left(x+1\right)}+\frac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x+1\right)}+\frac{x+2-x^2}{\left(x-3\right)\left(x+1\right)}\right)\)

\(=\left(x-3\right):\left(\frac{x^2-x-2+x^2-9+x+2-x^2}{\left(x-3\right)\left(x+1\right)}\right)=\left(x-3\right):\left(\frac{x^2-9}{\left(x-3\right)\left(x-1\right)}\right)\)thơm thế :))

\(=\left(x-3\right):\left(\frac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x-1\right)}\right)=\left(x-3\right).\frac{x-1}{x+3}=\frac{\left(x-3\right)\left(x-1\right)}{x+3}\)

1) đk: \(x\ne\left\{-1;3;4\right\}\)

Ta có:

\(A=\left(\frac{x^2-16}{x-4}-1\right)\div\left(\frac{x-2}{x-3}+\frac{x+3}{x+1}+\frac{x+2-x^2}{x^2-2x-3}\right)\)

\(A=\left[\frac{\left(x-4\right)\left(x+4\right)}{x-4}-1\right]\div\frac{\left(x-2\right)\left(x+1\right)+\left(x+3\right)\left(x-3\right)+x+2-x^2}{\left(x+1\right)\left(x-3\right)}\)

\(A=\left(x+4-1\right)\div\frac{x^2-x-2+x^2-9-x^2+x+2}{\left(x+1\right)\left(x-3\right)}\)

\(A=\left(x+3\right)\div\frac{x^2-9}{\left(x+1\right)\left(x-3\right)}\)

\(A=\left(x+3\right)\cdot\frac{\left(x+1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)

\(A=x+1\)

2) Ta có: \(\frac{A}{x^2+x+1}=\frac{x+1}{x^2+x+1}\)

Để \(\frac{A}{x^2+x+1}\) nguyên thì \(\left(x+1\right)⋮\left(x^2+x+1\right)\Leftrightarrow\left(x+1\right)^2⋮\left(x^2+x+1\right)\)

\(\Rightarrow\left(x+1\right)^2-\left(x^2+x+1\right)⋮\left(x^2+x+1\right)\)

\(\Rightarrow x⋮\left(x^2+x+1\right)\Rightarrow1⋮x^2+x+1\)

\(\Rightarrow x^2+x+1\in\left\{-1;1\right\}\Rightarrow x^2+x+1=1\Leftrightarrow x^2+x=0\Rightarrow\orbr{\begin{cases}x=-1\left(ktm\right)\\x=0\left(tm\right)\end{cases}}\)

Vậy x = 0

ĐKXĐ : \(\hept{\begin{cases}x\ne4\\x\ne3\\x\ne-1\end{cases}}\)

\(A=\left(\frac{x^2-16}{x-4}-\frac{x-4}{x-4}\right)\div\left(\frac{\left(x-2\right)\left(x+1\right)}{\left(x-3\right)\left(x+1\right)}+\frac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x+1\right)}+\frac{x+2-x^2}{\left(x-3\right)\left(x+1\right)}\right)\)

\(=\left(\frac{x^2-x-12}{x-4}\right)\div\left(\frac{x^2-x-2+x^2-9+x+2-x^2}{\left(x-3\right)\left(x+1\right)}\right)\)

\(=\frac{\left(x-4\right)\left(x+3\right)}{x-4}\div\frac{x^2-9}{\left(x-3\right)\left(x+1\right)}\)

\(=\frac{\left(x-4\right)\left(x+3\right)}{x-4}\times\frac{\left(x-3\right)\left(x+1\right)}{\left(x-3\right)\left(x+3\right)}=x+1\)

a.=>-3\(⋮\) x-1

x-1 thuộc ước của -3

x-1=1=>x=1+1=

x-1=-1=>....

x-1=3=>..

x-1=-3=>......

b. tương tự câu a

c.\(\frac{3x+7}{x-1}=\frac{3x-3+10}{x-1}=\frac{3\left(x-1\right)}{x-1}=\frac{10}{x-1}\)

Tự tính tiếp nha 

d.chịu 

a) Để  \(\frac{-3}{x-1}\) nguyên <=> x -1 \(\varepsilon\) Ư(-3)

                                 ta có   Ư(-3) = {-3 ; 3 ; 1; -1 }.

Với x -1 = 1 <=> x=2

Với x-1 =-1  <=> x= 0

Với x-1 =3   <=> x=4

Với x-1 =-3  <=> x=-2

Vậy.......

ý b bạn làm tương tự nhé có j hỏi mk thêm mk sẽ hướng dẫn ý c và d cho đỡ tồn thời gian

c) \(\frac{3x+7}{x-1}\) 

=\(\frac{3x-3+10}{X-1}=\frac{3\left(x-1\right)+10}{x-1}\)

 =\(\frac{3\left(x-1\right)}{x-1}+\frac{10}{x-1}\)  

= 3 +\(\frac{10}{x-1}\) 

Để \(\frac{3x+7}{x-1}\) nguyên <=> x -1\(\varepsilon\) Ư(10)

                                    ta có Ư(10) ={-1; 1 ; -2 ; 2 ; 5 ; -5 , 10 ; -10}.       

Với x -1 = -1 <=> x=0

Với x -1 = 1<=> x= 2

Với x-1=-2 <=> x= -1

Với x-1=2 <=> x= 3

Với x-1 =5 <=> x=5

Với x-1=-5<=>x=-4

Với x-1= 10<=>x=11

Với x-1=-10<=>x=-9

VẬY ...................................

D) \(\frac{4x-1}{3-x}\) 

  =\(\frac{4x-12+11}{3-x}\) 

  =\(\frac{4\left(x-3\right)+11}{3-x}\) 

  =\(\frac{4\left(x-3\right)}{-\left(x-3\right)}+\frac{11}{3-x}\) 

  =  -4+  \(\frac{11}{3-x}\)

Để \(\frac{4x-1}{3-x}\) nguyên <=> 3-x\(\varepsilon\) Ư(11)={-1 ; 1 ;-11 ;11 }.

Với 3 -x =-1 <=> x=4

Với 3 -x =1 <=> x=2

Với 3 -x = -11 <=> x=14

Với 3 -x = 11 <=> x = -8

 VẬY ........................

                       ĐÂY LÀ CACH GIẢI CHI TIẾT NHẤT ĐẤY . CHÚC BẠN NGÀY CÀNG HỌC GIỎI. NHỚ CHO MK NHÉ

a, |x-6|=( 2/11.13 + 2/13.15 +... + 2/19.21 ) . 462/5

    |x-6|=2.( 1/11.13 + 1/13.15+...+ 1/19.21 ) . 462/5

    |x-6|=2.1/2.( 1/11 - 1/13 + 1/13 - 1/15 +...+ 1/19 - 1/21 ) .462/5

    |x-6|= ( 1/11 - 1/21) . 462/5

    |x-6|= 10/231 . 462/5

    |x-6|= 4

    x-6= 4 suy ra x=10

    x-6= -4 suy ra x=2

 b,2/a + 1/30 = b/6

   =) 2/a= b/6 - 1/30 

   =) 2/a= 5b/30 - 1/30

   =) 2/a= 5b - 1/30

   =) 5b-1 thuộc Ư(30)

  Ư(30)={+-1;+-2;+-3;+-5;+-6;+-10;+-15;+-30)

=) ta có bảng sau: 

vì b thuộc Z =) b=0

với b=0 thay vào ta đc :

2/a = 5.0-1/30

=) 2/a = -1/30

=) -1.a = 2.30

=) -a = 60

=) a = -60

vậy a= -60 ; b= 0

Bằng =0 

nếu cần chi tiết xẽ có

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