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a, 2x-1 thuộc ước của 2,rồi giải ra
b,c tương tự
d\(\frac{x^2-64-123}{x+8}=\frac{\left(x+8\right)\left(x-8\right)-123}{x+8}=x-8-\frac{123}{X+8}\) .........rồi làm tương tự như câu a,,,,,,,,,,,,còn câu e cũng gần giống câu d
a) \(\dfrac{2x+3}{x-5}=\dfrac{2\left(x-5\right)+13}{x-5}=2+\dfrac{13}{x-5}\)
Để \(2+\dfrac{13}{x-5}\in Z\)
thì \(\dfrac{13}{x-5}\in Z\Rightarrow13⋮x-5\)
\(\Rightarrow x-5\inƯ\left(13\right)\)
\(\Rightarrow x-5\in\left\{\pm1;\pm13\right\}\)
Xét các trường hợp...
b) \(\dfrac{x^3-x^2+2}{x-1}=\dfrac{x^2\left(x-1\right)+2}{x-1}=x^2+\dfrac{2}{x-1}\)
Tương tự câu a)
c) \(\dfrac{x^3-2x^2+4}{x-2}=\dfrac{x^2\left(x-2\right)+4}{x-2}=x^2+\dfrac{4}{x-2}\)
...
d) \(\dfrac{2x^3+x^2+2x+2}{2x+1}=\dfrac{x^2\left(2x+1\right)+2x+2}{2x+1}=x^2+\dfrac{2x+2}{2x+1}\)
Khi đó lí luận cho \(2x+2⋮2x+1\)
\(\Rightarrow\left(2x+1\right)+1⋮2x+1\)
\(\Rightarrow1⋮2x+1\)
\(\Rightarrow2x+1\inƯ\left(1\right)\)
...
e) \(\dfrac{3x^3-7x^2+11x-1}{3x-1}=\dfrac{x^2\left(3x-1\right)-2x\left(3x-1\right)+3\left(3x-1\right)+2}{3x-1}\)
\(=\dfrac{\left(x^2-2x+3\right)\left(3x-1\right)+2}{3x-1}=\left(x^2-2x+3\right)+\dfrac{2}{3x-1}\)
...
f) \(\dfrac{x^4-16}{x^4-4x^3+8x^2-16x+16}=\dfrac{\left(x^2\right)^2-4^2}{\left(x-2\right)^2\left(x^2+4\right)}\)
\(=\dfrac{\left(x^2-4\right)\left(x^2+4\right)}{\left(x-2\right)^2\left(x^2+4\right)}=\dfrac{x^2-4}{\left(x-2\right)^2}=\dfrac{x+2}{x-2}=\dfrac{\left(x-2\right)+4}{x-2}=1+\dfrac{4}{x-2}\)
....
a: =>-4x>16
=>x<-4
c: =>20x-25<=21-3x
=>23x<=46
=>x<=2
d: =>20(2x-5)-30(3x-1)<12(3-x)-15(2x-1)
=>40x-100-90x+30<36-12x-30x+15
=>-50x-70<-42x+51
=>-8x<121
=>x>-121/8
a) \(\frac{6-x}{3}-\frac{x}{4}=\frac{3+2x}{2}-1\)
\(\frac{4\left(6-x\right)}{12}-\frac{3x}{12}=\frac{3+2x}{2}-\frac{2}{2}\)
\(\frac{24-4x-3x}{12}=\frac{3+2x-2}{2}\)
\(\frac{24-7x}{12}=\frac{2x+1}{2}\)
\(\Rightarrow2\left(24-7x\right)=12\left(2x+1\right)\)
\(\Rightarrow48-14x=24x+12\)
\(\Rightarrow24x+14x=48-12\)
\(\Rightarrow38x=36\)
\(\Rightarrow x=\frac{18}{19}\)
b) \(-7x-\frac{x-3}{5}-\frac{x}{2}=x+\frac{2x+1}{3}\)
\(\frac{-70x}{10}-\frac{2\left(x-3\right)}{10}-\frac{5x}{10}=\frac{3x}{3}+\frac{2x+1}{3}\)
\(\frac{-70x-2x+6-5x}{10}=\frac{3x+2x+1}{3}\)
\(\frac{-77x+6}{10}=\frac{5x+1}{3}\)
\(\Rightarrow3\left(-77x+6\right)=10\left(5x+1\right)\)
\(\Leftrightarrow-231x+18=50x+10\)
\(\Leftrightarrow50x+231x=18-10\)
\(\Leftrightarrow281x=8\)
\(\Leftrightarrow x=\frac{8}{281}\)
Mấy câu kia tương tự
a: \(\Leftrightarrow4\left(6-x\right)-3x=6\left(2x+3\right)-12\)
=>24-4x-3x=12x+18-12
=>12x+6=-7x+24
=>19x=18
=>x=18/19
b: \(\Leftrightarrow-210x-6\left(x-3\right)-15x=30x+10\left(2x+1\right)\)
=>-225x-6x+18=30x+20x+10
=>-231x+18-50x-10=0
=>-281x=-8
=>x=8/281
c: \(\Leftrightarrow36-2\left(x+3\right)=-4x+1-x\)
=>36-2x-6=-5x+1
=>3x=1+6-36=5-36=-31
=>x=-31/3
d: \(\Leftrightarrow-30\left(x-3\right)+10\left(2x-7\right)=6\left(6-x\right)\)
=>-30x+90+20x-70=36-6x
=>-10x+20=36-6x
=>-4x=16
=>x=-4
\(D=\dfrac{2x+4}{3x-1}\\ =>3D=\dfrac{6x+12}{3x-1}=\dfrac{2\left(3x-1\right)+14}{3x-1}=2+\dfrac{14}{3x-1}\)
Để 3D nguyên thì : \(\dfrac{14}{3x-1}\in Z\)
\(=>14⋮\left(3x-1\right)\\ =>3x-1\inƯ\left(14\right)=\left\{\pm1;\pm2;\pm7;\pm14\right\}\)
\(=>3x\in\left\{2;0;3;-1;8;-6;15;-13\right\}\\ =>x\in\left\{\dfrac{2}{3};0;1;-\dfrac{1}{3};\dfrac{8}{3};-2;5;-\dfrac{13}{3}\right\}\)
Mà x nguyên \(=>x\in\left\{0;1;-2;5\right\}\)
Do những giá trị trên chỉ là 3D nguyên nên chưa chắc D đã nguyên
Vậy thử lại thay từng giá trị x vào bt D
Kết luận : \(x\in\left\{0;1;-2;5\right\}\)