a) \(\dfrac{2x+5}{2x+1}=\dfrac{2x+1+4}{2x+1}=\dfrac{2x+1}{2x+1}+\dfrac{4}{2x+1}=1+\dfrac{4}{2x+1}\)  

Để \(\dfrac{2x+5}{2x+1}\in Z\) thì \(\dfrac{4}{2x+1}\in Z\) 

\(\Rightarrow4\) ⋮ \(2x+1\)

\(\Rightarrow2x+1\inƯ\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)

\(\Rightarrow2x\in\left\{0;-2;1;-3;3;-5\right\}\)

\(\Rightarrow x\in\left\{0;-1;\dfrac{1}{2};-\dfrac{3}{2};\dfrac{3}{2};-\dfrac{5}{2}\right\}\)

Mà x nguyên \(\Rightarrow\text{x}\in\left\{0;-1\right\}\) 

b) \(\dfrac{3x+5}{x+1}=\dfrac{3x+3+2}{x+1}=\dfrac{3\left(x+1\right)+2}{x+1}=\dfrac{3\left(x+1\right)}{x+1}+\dfrac{2}{x+1}=3+\dfrac{2}{x+1}\) 

Để \(\dfrac{3x+5}{x+1}\in Z\) thì \(\dfrac{2}{x+1}\in Z\) 

\(\Rightarrow2\) ⋮ \(x+1\)

\(\Rightarrow x+1\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)

\(\Rightarrow x\in\left\{0;-2;1;-3\right\}\) 

c) \(\dfrac{3x+8}{x-1}=\dfrac{3x-3+11}{x-1}=\dfrac{3\left(x-1\right)+11}{x-1}=\dfrac{3\left(x-1\right)}{x-1}+\dfrac{11}{x-1}=3+\dfrac{11}{x-1}\)  

Để: \(\dfrac{3x+8}{x-1}\in Z\) thì \(\dfrac{11}{x-1}\in Z\)

\(\Rightarrow11\) ⋮ \(x-1\)

\(\Rightarrow x-1\inƯ\left(11\right)=\left\{1;-1;11;-11\right\}\)

\(\Rightarrow x\in\left\{2;0;12;-10\right\}\)

d) \(\dfrac{5x+12}{x-2}=\dfrac{5x-10+22}{x-2}=\dfrac{5\left(x-2\right)+22}{x-2}=\dfrac{5\left(x-2\right)}{x-2}+\dfrac{22}{x-2}=5+\dfrac{22}{x-2}\)

Để: \(\dfrac{5x+12}{x-2}\in Z\) thì \(\dfrac{22}{x-2}\in Z\)

\(\Rightarrow22\) ⋮ \(x-2\)

\(\Rightarrow x-2\inƯ\left(22\right)=\left\{1;-1;2;-2;11;-11;22;-22\right\}\)

\(\Rightarrow x\in\left\{3;1;4;0;13;-9;24;-20\right\}\)

e) \(\dfrac{7x-12}{x+16}=\dfrac{7x+112-124}{x+16}=\dfrac{7\left(x+16\right)-124}{x+16}=\dfrac{7\left(x+16\right)}{x+16}-\dfrac{124}{x+16}=7-\dfrac{124}{x+16}\)

Để \(\dfrac{7x-12}{x+16}\in Z\) thì \(\dfrac{124}{x+16}\in Z\) 

\(\Rightarrow124\) ⋮ \(x+16\)

\(\Rightarrow x+16\inƯ\left(124\right)=\left\{1;-1;2;-2;4;-4;31;-31;62;-62;124;-124\right\}\)

\(\Rightarrow x\in\left\{-15;-17;-14;-18;-12;-20;15;-47;46;-78;108;-140\right\}\)

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

b: \(\left(x+2\right)^2-x^2=4x+4\)

c: \(\left(x+3\right)\left(x^2-3x+9\right)-x^3=27\)

a) -16 + 23 + x = -16

=> 7 + x = -16

=> x = -16 - 7

=> x = -23

b) 3x + 17 = 12

=> 3x = 12 - 17

=> 3x = -5

=> x = -5/3

vì x là số nguyên => x ko có gtri thõa mãn đề bài

c) (2x - 5) + 17 = 16

=> 2x - 5 = 16 - 17

=> 2x - 5 = -1

=> 2x = -1 + 5

=> 2x = 4

=> x = 4 : 2

=> x = 2

\(-16+23+x=-16\)

\(\Rightarrow x=-16-23+16\)

\(\Rightarrow x=-23\)

\(3x+17=12\)

\(3x=-5\)

\(x=-\frac{5}{3}\)

\(24:\left(3x-2\right)=-3\)

\(\Rightarrow3x-2=-8\)

\(3x=-6\)

\(x=-2\)

\(\left(2x-5\right)+17=16\)

\(2x=16-17+5\)

\(x=2\)

\(-6< x< 3\)

\(\Rightarrow x\in\left\{-5;-4;-3;-2;-1;0;1;2\right\}\)

"Tự tính tổng "

a) ( x + 2 )( x + 3 ) - ( x - 2 )( x + 5 ) = 16

<=> x² + 5x + 6 - ( x² + 3x - 10 ) = 16

<=> x² + 5x + 6 - x² - 3x + 10 = 16

<=> 2x + 16 = 16

<=> 2x = 0

<=> x = 0

b) 3x( 2x - 4 ) - 2x( 3x + 5 ) = 44

<=> 6x² - 12x - 6x² - 10x = 44

<=> -22x = 44

<=> x = -2

c) 2( 5x - 8 - 3 )( 4x - 5 ) = 4( 3x - 4 )

<=> 2( 5x - 11 )( 4x - 5 ) = 4( 3x - 4 )

<=> 2( 20x² - 69x + 55 ) = 12x - 16

<=> 40x² - 138x + 110 = 12x - 16

<=> 40x² - 138x + 110 - 12x + 16 = 0

<=> 40x² - 150 + 126 = 0 ( chưa học nghiệm vô tỉ nên để vô nghiệm nha :) )

=> Vô nghiệm

a. (x+2) (x+3)-(x-2) (x+5)= 16

x²+5x+6-x²-3x+10=16

2x+16=16

2x=0

x=0

b,3x (2x-4)-2x (3x+5)= 44

6x²-12x-6x²-10x=44

-22x=44

x=-2

Ý c bạn tự lm,tương tự nhưa,b

a, \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=16\)

\(\Leftrightarrow x^2+3x+2x+6-\left(x^2+5x-2x-10\right)=16\)

\(\Leftrightarrow8x-4=16\Leftrightarrow x=\frac{20}{8}=\frac{5}{2}\)

b, \(3x\left(2x-4\right)-2x\left(3x+5\right)=44\)

\(\Leftrightarrow6x^2-12x-\left(6x^2-10x\right)=44\Leftrightarrow-2x=44\Leftrightarrow x=-22\)

c, \(2\left(5x-8-3\right)\left(4x-5\right)=4\left(3x-4\right)\)

\(\Leftrightarrow\left(10x-22\right)\left(4x-5\right)=4\left(3x-4\right)\Leftrightarrow40x^2-50x-88x+110=0\)

\(\Leftrightarrow40x^2-138x+110=0\)( vô nghiệm ) 

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Bài 3: 

\(\Leftrightarrow x^3+64-x^3+25x=264\)

hay x=8

\(1,C=6x^2+23x-55-6x^2-23x-21=-76\\ 2,=\left(2x^4-x^2+2x^3-x-6x^2+6-3\right):\left(2x^2-1\right)\\ =\left[\left(2x^2-1\right)\left(x^2+x-6\right)-3\right]:\left(2x^2-1\right)\\ =x^2+x-6\left(dư.-3\right)\\ 3,\Leftrightarrow x^3+64-x^3+25x=264\\ \Leftrightarrow25x=200\Leftrightarrow x=8\)

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

\(\Leftrightarrow\left(3x+1\right)\left(x-3\right)-\left(3x+1\right)\left(2x-5\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(x-3-2x+5\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(2-x\right)=0\)

\(\Leftrightarrow\left[\begin{matrix}3x+1=0\\2-x=0\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left[\begin{matrix}3x=-1\\x=2\end{matrix}\right.\)

\(\Leftrightarrow\left[\begin{matrix}x=-\frac{1}{3}\\x=2\end{matrix}\right.\)

Vậy tập nghiệm của pt là \(S=\left\{-\frac{1}{3};2\right\}\)

Có : \(\left(3x+1\right)\left(x-3\right)=\left(3x+1\right)\left(2x-5\right)\)

\(\Leftrightarrow\) \(\left(3x+1\right)\left(x-3\right)-\left(3x+1\right)\left(2x-5\right)=0\)

\(\Leftrightarrow\) \(\left(3x+1\right)\left(x-3-2x+5\right)=0\)

\(\Leftrightarrow\) \(\left(3x+1\right)\left(-x+2\right)=0\)

\(\Leftrightarrow\) \(\left[\begin{matrix}3x+1=0\\-x+2=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left[\begin{matrix}3x=-1\\-x=-2\end{matrix}\right.\) \(\Leftrightarrow\) \(\left[\begin{matrix}x=\frac{-1}{3}\\x=2\end{matrix}\right.\)

Vậy phương trình có tập nghiệm \(S=\left\{\frac{-1}{3};2\right\}\)

\(\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}=\frac{\left(x-3\right)\left(3-x\right)}{4}\)

\(\Leftrightarrow\frac{24\left(x-3\right)}{24}-\frac{4\left(x-3\right)\left(2x-5\right)}{24}=-\frac{6\left(x-3\right)\left(x-3\right)}{24}\)

\(\Leftrightarrow24\left(x-3\right)-4\left(x-3\right)\left(2x-5\right)+6\left(x-3\right)^2=0\)

\(\Leftrightarrow\left(x-3\right)\left[24-4\left(2x-5\right)+6\left(x-3\right)\right]=0\)

\(\Leftrightarrow\left(x-3\right)\left(24-8x+20+6x-18\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(26-2x\right)=0\)

\(\Leftrightarrow2\left(x-3\right)\left(13-x\right)=0\)

\(\Leftrightarrow\left[\begin{matrix}x-3=0\\13-x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[\begin{matrix}x=3\\x=13\end{matrix}\right.\)

Vậy tập nghiệm của pt là \(S=\left\{3;13\right\}\)