(a) \(f\left(x\right)⋮g\left(x\right)\Rightarrow\dfrac{x^2-5x+9}{x-3}\in Z\)

Ta có: \(\dfrac{x^2-5x+9}{x-3}\left(x\ne3\right)=\dfrac{x\left(x-3\right)-2\left(x-3\right)+3}{x-3}=x-2+\dfrac{3}{x-3}\)nguyên khi và chỉ khi: \(\left(x-3\right)\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(\Rightarrow\left[{}\begin{matrix}x-3=1\\x-3=-1\\x-3=3\\x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\\x=6\\x=0\end{matrix}\right.\) (thỏa mãn).

Vậy: \(x\in\left\{0;2;4;6\right\}\).

(b) \(f\left(x\right)⋮g\left(x\right)\Rightarrow\dfrac{2x^3-x^2+6x+2}{2x-1}\in Z\left(x\ne\dfrac{1}{2}\right)\)

Ta có: \(\dfrac{2x^3-x^2+6x+2}{2x-1}=\dfrac{x^2\left(2x-1\right)+3\left(2x-1\right)+5}{2x-1}=x^2+3+\dfrac{5}{2x-1}\)

nguyên khi và chỉ khi: \(\left(2x-1\right)\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

\(\Rightarrow\left[{}\begin{matrix}2x-1=1\\2x-1=-1\\2x-1=5\\2x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=0\\x=3\\x=-2\end{matrix}\right.\) (thỏa mãn).

Vậy: \(x\in\left\{-2;0;1;3\right\}\).

a: f(x) chia hết cho g(x)

=>x^2-3x-2x+6+3 chia hết cho x-3

=>3 chia hết cho x-3

=>x-3 thuộc {1;-1;3;-3}

=>x thuộc {4;2;6;0}

b: f(x) chia hết cho g(x)

=>2x^3-x^2+6x-3+5 chia hết cho 2x-1

=>5 chia hết cho 2x-1

=>2x-1 thuộc {1;-1;5;-5}

=>x thuộc {2;0;3;-2}

MK ko biế đúng ko nữa , sai thì ý kiến

a)

b)

Chúc các bn hok tốt

Tham khảo nhé

| x -1 | + | 2x - 2 | + | 6 - 6x | = 63

TH1: x-1+2(x-1)+6(x-1)=63

<=>9(x-1)=63

<=>x-1=7

<=> x=8

TH2: 1-x+2(1-x)+6(1-x)=63

<=>9(1-x)=63

<=>1-x=7

<=>x=-6

Vậy \(x\in\left\{8;-6\right\}\)

tr1

=>9(x-1)=63

=> x= 8

th2 

=>1-x = 7

=>x=-6

??????????????

?????????

Xin lỗi phải là các cặp số nguyên x,y

\(\dfrac{5}{x}+1+\dfrac{4}{x}+1=\dfrac{3}{-13}\\ \Rightarrow\dfrac{9}{x}+2=-\dfrac{3}{13}\\ \Rightarrow\dfrac{9}{x}=-\dfrac{59}{13}\\ \Rightarrow x=-\dfrac{207}{59}\)

a. \(\dfrac{5}{x+1}+\dfrac{4}{x+1}=\dfrac{-3}{13}\)

ĐKXĐ: x ≠ -1

⇔ \(\dfrac{65}{13\left(x+1\right)}+\dfrac{52}{13\left(x+1\right)}=\dfrac{-3\left(x+1\right)}{13\left(x+1\right)}\)

⇔ 65 + 52 = -3(x + 1)

⇔ 117 = -3x - 3

⇔ 117 + 3 = -3x

⇔ 120 = -3x 

⇔ x = \(\dfrac{120}{-3}=-40\) (TM)

b. -x + 2 + 2x + 3 + x + \(\dfrac{1}{4}\) + 2x + \(\dfrac{1}{6}\) = \(\dfrac{8}{3}\)

⇔ -x + 2x + x + 2x = \(\dfrac{8}{3}-\dfrac{1}{6}-\dfrac{1}{4}-3-2\)

⇔ 4x = -2,75

⇔ x = \(\dfrac{-2,75}{4}=\dfrac{-11}{16}\)

c. \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+2}\) = \(\dfrac{12}{26}\)

⇔  \(\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{2\left(3x+1\right)}=\dfrac{12}{26}\)

⇔ \(\dfrac{312\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\) + \(\dfrac{520\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\) - \(\dfrac{312\left(2x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\)

= \(\dfrac{48\left(2x+1\right)\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\)

⇔ 312(3x +1) + 520(3x + 1) - 312(2x + 1) = 48(2x + 1)(3x + 1)

⇔ 936x + 312 + 1560x + 520 - 624x - 312 = (96x + 48)(3x + 1)

⇔ 936x + 312 + 1560x + 520 - 624x - 312 = 288x² + 96x + 144x + 48

⇔ 936x + 1560x - 624x - 96x - 144x - 288x² = 48 - 312 - 520 + 312

⇔ 1632x - 288x² = -472

⇔ -288x² + 1632x + 472 = 0 (Tự giải tiếp, dùng phương pháp tách hạng tử)

⇔ x = 5,942459684 \(\approx\) 6

c: Ta có: \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\)

\(\Leftrightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)

\(\Leftrightarrow2x+1=13\)

hay x=6

6x+5/2x+1 nguyên 

=> 6x+5 chia hết cho 2x + 1

=> 6x+3 + 2 chia hết cho 2x + 1

6x+3 chia hết cho 2x+1

=> 2 chia hết cho 2x + 1

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

2x + 1 = -2 => 2x = -3;x=-3/2

2x + 1 = -1 => 2x = -2 ; x =-1

2x + 1 = 1 ; 2x = 0  ; x = 0

2x + 1 = 2 => 2x = 1 ; x = 1/2

Mà x nguyên => x \(\in\left\{-1;0\right\}\)

\(\frac{6x+5}{2x+1}=\frac{3.\left(2x+1\right)+2}{2x+1}=3+\frac{2}{2x+1}\) nguyên

<=> 2x + 1 thuộc Ư(2)

<=> 2x + 1 thuộc {-2;-1;1;2}

Vì x nguyên nên x thuộc {-1; 0}

a) \(\frac{-x}{2}+\frac{2x}{3}+x+\frac{1}{4}+2x+\frac{1}{6}=\frac{3}{8}.\)

\(\frac{-x}{2}+\frac{2x}{3}+3x+\frac{5}{12}=\frac{3}{8}\)

\(x.\left(-\frac{1}{2}+\frac{2}{3}+3\right)+\frac{5}{12}=\frac{3}{8}\)

\(x\cdot\frac{19}{6}=-\frac{1}{24}\)

x = -1/76

b) \(\frac{3}{2x+1}+\frac{10}{4x+2}-\frac{6}{6x+3}=\frac{12}{26}\)

\(\frac{3}{2x+1}+\frac{2.5}{2.\left(2x+1\right)}-\frac{2.3}{3.\left(2x+1\right)}=\frac{6}{13}\)

\(\frac{3}{2x+1}+\frac{5}{2x+1}-\frac{2}{2x+1}=\frac{6}{13}\)

\(\frac{3+5-2}{2x+1}=\frac{6}{13}\)

\(\frac{6}{2x+1}=\frac{6}{13}\)

=> 2x + 1 = 13

2x = 12

x = 6

\(a,-\dfrac{x}{2}+\dfrac{2x}{3}+\dfrac{x+1}{4}+\dfrac{2x+1}{6}=\dfrac{8}{3}\)

\(\Rightarrow-\dfrac{6x}{12}+\dfrac{8x}{12}+\dfrac{3\left(x+1\right)}{12}+\dfrac{2\left(2x+1\right)}{12}=\dfrac{8}{3}\)

\(\Rightarrow\dfrac{-6x+8x+3x+3+4x+2}{12}=\dfrac{8}{3}\)

\(\Rightarrow\dfrac{9x+5}{12}=\dfrac{8}{3}\)

\(\Rightarrow27x+15=96\)

\(\Rightarrow27x=81\)

\(\Rightarrow x=3\left(tm\right)\)

\(b,\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\)

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{3\left(2x+1\right)}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{3+5-2}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{6}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow2x+1=13\)

\(\Rightarrow2x=12\)

\(\Rightarrow x=6\left(tm\right)\)

#Toru

a) \(-\dfrac{x}{2}+\dfrac{2x}{3}+\dfrac{x+1}{4}+\dfrac{2x+2}{6}=\dfrac{8}{3}\) 

\(\Rightarrow\dfrac{-6x}{12}+\dfrac{8x}{12}+\dfrac{3\left(x+1\right)}{12}+\dfrac{2\left(2x+1\right)}{12}=\dfrac{4\cdot8}{12}\)

\(\Rightarrow-6x+8x+3x+3+4x+2=32\)

\(\Rightarrow9x+5=32\)

\(\Rightarrow9x=32-5\)

\(\Rightarrow9x=27\)

\(\Rightarrow x=\dfrac{27}{9}\)

\(\Rightarrow x=3\)

b) \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\) (ĐK: \(x\ne-\dfrac{1}{2}\)) 

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{3\left(2x+1\right)}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{6}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow2x+1=13\)

\(\Rightarrow2x=12\)

\(\Rightarrow x=\dfrac{12}{2}\)

\(\Rightarrow x=6\left(tm\right)\)