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để \(\dfrac{x+1}{x-1}\)nguyên thì
(x+1)⋮(x-1)
=> (x+1)-(x-1)⋮(x-1)
=> (x+1-x+1)⋮(x-1)
=> 2⋮(x+1)
=> x+1 ∈Ư (2)={-2;-1;1;2}
ta có bảng sau
| x+1 | -2 | -1 | 1 | 2 |
| x | -3 | -2 | 0 |
1 |
vậy để \(\dfrac{x+1}{x-1}\)thì x ∈{-3;-2;0;1}
1: \(\Leftrightarrow3x+4=2\)
=>3x=-2
=>x=-2/3
2: \(\Leftrightarrow7x-7=6x-30\)
=>x=-23
3: =>\(5x-5=3x+9\)
=>2x=14
=>x=7
4: =>9x+15=14x+7
=>-5x=-8
=>x=8/5
a) \(\left|2x-3\right|-\dfrac{5}{2}=\dfrac{1}{3}\)
\(\left|2x-3\right|=\dfrac{1}{3}+\dfrac{5}{2}=\dfrac{2}{6}+\dfrac{15}{6}\)
\(\left|2x-3\right|=\dfrac{17}{6}\)
\(+)2x-3=\dfrac{17}{6}\Rightarrow2x=\dfrac{35}{6}\Rightarrow x=\dfrac{35}{12}\)
\(+)2x-3=\dfrac{-17}{6}\Rightarrow2x=\dfrac{1}{6}\Rightarrow x=\dfrac{1}{12}\)
vậy...
\(\left|x-1\right|+3x=1\\ \Rightarrow\left|x-1\right|=1-3x\\ \Rightarrow\left\{{}\begin{matrix}x-1=1-3x\\x-1=-1+3x\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}4x=2\\-2x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\x=0\end{matrix}\right.\)
Dấu ngoặc vuông nhé
thánh bấm nhầm
câu E
\(\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left(2x-5\right)\left(5-2x\right)=-\left(\dfrac{3}{2}\right)^4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left|2x-5\right|=\left(\dfrac{3}{2}\right)^2\end{matrix}\right.\)
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{5}{2}\\2x-5=-\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{11}{8}< \dfrac{5}{2}\left(n\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{5}{2}\\2x-5=\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{29}{8}>\dfrac{5}{2}\left(n\right)\end{matrix}\right.\end{matrix}\right.\)
câu F (bạn cho vào lớp 7.2=lớp 14 nhé. )
a: \(\dfrac{-0.2}{x}=\dfrac{x}{-0.8}\)
\(\Leftrightarrow x^2=\dfrac{1}{5}\cdot\dfrac{4}{5}=\dfrac{4}{25}\)
=>x=2/5 hoặc x=-2/5
c: \(\dfrac{x-1}{x-2}=\dfrac{-3}{4}\)
=>4(x-1)=-3(x-2)
=>4x-4=-3x+6
=>7x=10
hay x=10/7
d: \(\dfrac{2-x}{5-x}=\dfrac{x+3}{x+2}\)
\(\Leftrightarrow\dfrac{x+3}{x+2}=\dfrac{x-2}{x-5}\)
\(\Leftrightarrow\left(x+3\right)\left(x-5\right)=\left(x-2\right)\left(x+2\right)\)
\(\Leftrightarrow x^2-2x-15=x^2-4\)
=>-2x=11
hay x=-11/2
b: =>(3x-1)(3x+1)(2x+3)=0
hay \(x\in\left\{\dfrac{1}{3};-\dfrac{1}{3};-\dfrac{3}{2}\right\}\)
c: \(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{5}{6}+\dfrac{3}{4}=\dfrac{19}{12}\)
=>2x-1/3=19/12 hoặc 2x-1/3=-19/12
=>2x=23/12 hoặc 2x=-15/12=-5/4
=>x=23/24 hoặc x=-5/8
d: \(\Leftrightarrow-\dfrac{5}{6}\cdot x+\dfrac{3}{4}=-\dfrac{3}{4}\)
=>-5/6x=-3/2
=>x=3/2:5/6=3/2*6/5=18/10=9/5
e: =>2/5x-1/2=3/4 hoặc 2/5x-1/2=-3/4
=>2/5x=5/4 hoặc 2/5x=-1/4
=>x=5/4:2/5=25/8 hoặc x=-1/4:2/5=-1/4*5/2=-5/8
f: =>14x-21=9x+6
=>5x=27
=>x=27/5
h: =>(2/3)^2x+1=(2/3)^27
=>2x+1=27
=>x=13
i: =>5^3x*(2+5^2)=3375
=>5^3x=125
=>3x=3
=>x=1
b: 2x-3<0
=>2x<3
hay x<3/2
c: \(\left(2x-4\right)\left(9-3x\right)>0\)
=>(x-2)(x-3)<0
=>2<x<3
d: \(\dfrac{2}{3}x-\dfrac{3}{4}>0\)
=>2/3x>3/4
hay x>9/8
a) \(P=\dfrac{2x+5}{x+3}\inℤ\left(x\inℤ;x\ne-3\right)\)
\(\Rightarrow2x+5⋮x+3\)
\(\Rightarrow2x+5-2\left(x+3\right)⋮x+3\)
\(\Rightarrow2x+5-2x-6⋮x+3\)
\(\Rightarrow-1⋮x+3\)
\(\Rightarrow x+3\in\left\{-1;1\right\}\)
\(\Rightarrow x\in\left\{-4;-2\right\}\)
b) \(P=\dfrac{3x+4}{x+1}\inℤ\left(x\inℤ;x\ne-1\right)\)
\(\Rightarrow3x+4⋮x+1\)
\(\Rightarrow3x+4-3\left(x+1\right)⋮x+1\)
\(\Rightarrow3x+4-3x-3⋮x+1\)
\(\Rightarrow1⋮x+1\)
\(\Rightarrow x+1\in\left\{-1;1\right\}\)
\(\Rightarrow x\in\left\{-2;0\right\}\)
c) \(P=\dfrac{4x-1}{2x+3}\inℤ\left(x\inℤ;x\ne-\dfrac{3}{2}\right)\)
\(\Rightarrow4x-1⋮2x+3\)
\(\Rightarrow4x-1-2\left(2x+3\right)⋮2x+3\)
\(\Rightarrow4x-1-4x-6⋮2x+3\)
\(\Rightarrow-7⋮2x+3\)
\(\Rightarrow2x+3\in\left\{-1;1;-7;7\right\}\)
\(\Rightarrow x\in\left\{-2;-1;-5;2\right\}\)
a) P=\(\dfrac{2x+5}{x+3}=\dfrac{2\left(x+3\right)-2}{x+3}=\dfrac{2\left(x+3\right)}{x+3}-\dfrac{2}{x+3}=2-\dfrac{2}{x+3}\)
để \(P\inℤ\) thì \(\dfrac{2}{x+3}\inℤ\) hay 2 ⋮ (x-3) ⇒x+3 ϵ Ư2= (2,-2,1,-1)
ta có bảng sau:
Vậy x \(\in-1,-2,-5,-4\)