\(B=\dfrac{x-5}{x+1}\inℤ\)
\(C=\dfrac{2x+3}{x-2}\inℤ\)
\(D=\dfrac{6x-12}{2x+1}\inℤ\)
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Dora
10 tháng 1 2023
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DD
Đoàn Đức Hà
Giáo viên
1 tháng 8 2021
a) \(\frac{1-x}{x+4}=\frac{5-4-x}{x+4}=\frac{5}{x+4}-1\inℤ\Leftrightarrow\frac{5}{x+4}\inℤ\)
mà \(x\inℤ\Rightarrow x+4\inƯ\left(5\right)=\left\{-5,-1,1,5\right\}\)
\(\Leftrightarrow x\in\left\{-9,-5,-3,1\right\}\)
b) \(\frac{11-2x}{x-5}=\frac{1+10-2x}{x-5}=\frac{1}{x-5}-2\inℤ\Leftrightarrow\frac{1}{x-5}\inℤ\)
mà \(x\inℤ\Rightarrow x-5\inƯ\left(1\right)=\left\{-1,1\right\}\Leftrightarrow x\in\left\{4,6\right\}\)
c) \(\frac{x+1}{2x+1}\inℤ\Rightarrow\frac{2\left(x+1\right)}{2x+1}=\frac{2x+1+1}{2x+1}=1+\frac{1}{2x+1}\inℤ\Leftrightarrow\frac{1}{2x+1}\inℤ\)
mà \(x\inℤ\Rightarrow2x+1\inƯ\left(1\right)=\left\{-1,1\right\}\Leftrightarrow x\in\left\{-1,0\right\}\).
Thử lại đều thỏa mãn.
19 tháng 12 2021
\(a,=\dfrac{x^2+4x+3-2x^2+2x+x^2-4x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ b,=\dfrac{1-2x+3+2y+2x-4}{6x^3y}=\dfrac{2y}{6x^3y}=\dfrac{1}{x^2}\\ c,=\dfrac{75y^2+18xy+10x^2}{30x^2y^3}\\ d,=\dfrac{5x+8-x}{4x\left(x+2\right)}=\dfrac{4\left(x+2\right)}{4x\left(x+2\right)}=\dfrac{1}{x}\\ c,=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)
17 tháng 7 2021
a) Để y nguyên thì \(6x-4⋮2x+3\)
\(\Leftrightarrow-13⋮2x+3\)
\(\Leftrightarrow2x+3\in\left\{1;-1;13;-13\right\}\)
\(\Leftrightarrow2x\in\left\{-2;-4;10;-16\right\}\)
hay \(x\in\left\{-1;-2;5;-8\right\}\)
5 tháng 1 2021
a) ĐKXĐ: x≠-5
Ta có: \(\dfrac{2x-5}{x+5}=4\)
\(\Leftrightarrow2x-5=4\left(x+5\right)\)
\(\Leftrightarrow2x-5=4x+20\)
\(\Leftrightarrow2x-5-4x-20=0\)
\(\Leftrightarrow-2x-25=0\)
\(\Leftrightarrow-2x=25\)
hay \(x=\dfrac{-25}{2}\)(nhận)
Vậy: \(S=\left\{-\dfrac{25}{2}\right\}\)
b) ĐKXĐ: x≠0
Ta có: \(\dfrac{x^2-4}{x}=\dfrac{2x+3}{2}\)
\(\Leftrightarrow2\left(x^2-4\right)=x\left(2x+3\right)\)
\(\Leftrightarrow2x^2-8=2x^2+3x\)
\(\Leftrightarrow2x^2-8-2x^2-3x=0\)
\(\Leftrightarrow-3x-8=0\)
\(\Leftrightarrow-3x=8\)
hay \(x=\dfrac{-8}{3}\)(nhận)
Vậy: \(S=\left\{-\dfrac{8}{3}\right\}\)
c) ĐKXĐ: \(x\notin\left\{\dfrac{1}{2};-5\right\}\)
Ta có: \(\dfrac{2x+3}{2x-1}=\dfrac{x-3}{x+5}\)
\(\Leftrightarrow\left(2x+3\right)\left(x+5\right)=\left(2x-1\right)\left(x-3\right)\)
\(\Leftrightarrow2x^2+10x+3x+15=2x^2-6x-x+3\)
\(\Leftrightarrow2x^2+13x+15=2x^2-7x+3\)
\(\Leftrightarrow2x^2+13x+15-2x^2+7x-3=0\)
\(\Leftrightarrow20x+12=0\)
\(\Leftrightarrow20x=-12\)
hay \(x=-\dfrac{3}{5}\)(nhận)
Vậy: \(S=\left\{-\dfrac{3}{5}\right\}\)
d) ĐKXĐ: \(x\notin\left\{-7;\dfrac{3}{2}\right\}\)
Ta có: \(\dfrac{3x-2}{x+7}=\dfrac{6x+1}{2x-3}\)
\(\Leftrightarrow\left(3x-2\right)\left(2x-3\right)=\left(x+7\right)\left(6x+1\right)\)
\(\Leftrightarrow6x^2-9x-4x+6=6x^2+x+42x+7\)
\(\Leftrightarrow6x^2-13x+6=6x^2+43x+7\)
\(\Leftrightarrow6x^2-13x+6-6x^2-43x-7=0\)
\(\Leftrightarrow-56x-1=0\)
\(\Leftrightarrow-56x=1\)
hay \(x=-\dfrac{1}{56}\)(nhận)
Vậy: \(S=\left\{-\dfrac{1}{56}\right\}\)
13 tháng 6 2023
b: =>1/4x+4/5-x-5=1/3x+1-1/2x+1
=>-3/4x+1/6x=2+5-4/5=24/5
=>x=-288/35
c: =>6x^2+3x-30x-15=6x^2+10x-21x-35
=>-27x-15=-11x-35
=>-16x=-20
=>x=5/4
23 tháng 7 2022
b: Đặt \(x^2-6x-2=a\)
Theo đề, ta có: \(a+\dfrac{14}{a+9}=0\)
=>(a+2)(a+7)=0
\(\Leftrightarrow\left(x^2-6x\right)\left(x^2-6x+5\right)=0\)
=>x(x-6)(x-1)(x-5)=0
hay \(x\in\left\{0;1;6;5\right\}\)
c: \(\Leftrightarrow\dfrac{-8x^2}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(2x+1\right)}\)
\(\Leftrightarrow-32x^2=8x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)\)
\(\Leftrightarrow-32x^2=16x^2+8x-3\left(16x^2-8x+2x-1\right)\)
\(\Leftrightarrow-48x^2=8x-48x^2+18x+3\)
=>26x=-3
hay x=-3/26
14 tháng 5 2022
a: \(=\dfrac{2x-2x+y}{2\left(2x-y\right)}=\dfrac{y}{2\left(2x-y\right)}\)
b: \(=\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{6x+2-x^2-x}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x^2+5x+2}{2\left(x-1\right)\left(x+1\right)}\)
c: \(=\dfrac{1}{x+2}+\dfrac{x+8}{3x\left(x+2\right)}\)
\(=\dfrac{3x+x+8}{3x\left(x+2\right)}=\dfrac{4x+8}{3x\left(x+2\right)}=\dfrac{4}{3x}\)
d: \(=\dfrac{4x+6-2x^2+3x+2x+1}{\left(2x-3\right)\left(2x+3\right)}\)
\(=\dfrac{-2x^2+9x+7}{\left(2x-3\right)\left(2x+3\right)}\)
19 tháng 1 2022
Câu 1:
\(\left(x-2\right)\left(x^2+2x+4\right)+25x=x\left(x+5\right)\left(x-5\right)+8\)
\(\Leftrightarrow x^3-8+25x=x\left(x^2-25\right)+8\)
\(\Leftrightarrow x^3-8+25x=x^3-25x+8\)
\(\Leftrightarrow x^3-8+25x-x^3+25x-8=0\)
\(\Leftrightarrow50x-16=0\)
\(\Leftrightarrow50x=16\)
\(\Leftrightarrow x=\dfrac{8}{25}\)
B
19 tháng 1 2022
Câu 2 :
\(\dfrac{x+5}{4}+\dfrac{3+2x}{3}=\dfrac{6x-1}{3}-\dfrac{1-2x}{12}\)
<=> \(\dfrac{3\left(x+5\right)}{12}+\dfrac{4\left(3+2x\right)}{12}=\dfrac{4\left(6x-1\right)}{12}-\dfrac{1-2x}{12}\)
<=>\(\dfrac{3x+15+12+8x}{12}=\dfrac{24x-4-1+2x}{12}\)
<=> 3x + 15 + 12 + 8x = 24x - 4 - 1 +2x
<=> 11x+27 = 26x -5
<=> ( 26x - 5 ) - ( 11x + 27 ) = 0
<=> 15x - 32 = 0
<=> 15x = 32
<=> x = \(\dfrac{32}{15}\)
