tìm số nguyên x biết:
a) |x+5| = 0
b) 2x+ 2x+1 = ( -2 ) + 50
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11 tháng 7 2018
ĐKXĐ: \(x\ne\pm1;-2\)
\(P=\left(\frac{x+1}{x-1}+\frac{2}{x^2-1}-\frac{x}{x+1}\right).\frac{x-1}{x+2}\)
\(=\left(\frac{\left(x+1\right)^2}{\left(x-1\right).\left(x+1\right)}+\frac{2}{\left(x-1\right).\left(x+1\right)}-\frac{x\left(x-1\right)}{\left(x-1\right).\left(x+1\right)}\right).\frac{x-1}{x+2}\)
\(=\left(\frac{x^2+2x+1}{\left(x-1\right).\left(x+1\right)}+\frac{2}{\left(x-1\right).\left(x+1\right)}-\frac{x^2-x}{\left(x-1\right).\left(x+1\right)}\right).\frac{x-1}{x+2}\)
\(=\left(\frac{x^2+2x+1+2-x^2+x}{\left(x-1\right).\left(x+1\right)}\right).\frac{x-1}{x+2}\)
\(=\frac{3x+3}{\left(x-1\right).\left(x+1\right)}.\frac{x-1}{x+2}=\frac{3.\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}.\frac{x-1}{x+2}=\frac{3}{x+2}\)
c. \(x^2-3x=0\Leftrightarrow x.\left(x-3\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
Nếu x=0 thì: \(P=\frac{3}{x+2}=\frac{3}{0+2}=\frac{3}{2}\)
Nếu x=3 thì: \(P=\frac{3}{x+2}=\frac{3}{3+2}=\frac{3}{5}\)
d. Ta có: \(P=\frac{3}{x+2}\inℤ\)
Vì \(x\inℤ\Rightarrow x+2\inℤ\Rightarrow x+2\inƯ\left\{3\right\}\Rightarrow x+2\in\left\{\pm1;\pm3\right\}\Leftrightarrow x\in\left\{-3;-1;1;-5\right\}\)
Kết hợp ĐKXĐ \(\Rightarrow x\in\left\{-3;-5\right\}\)
11 tháng 7 2018
ĐKXĐ: \(x\ne-5;0\)
\(A=\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50-5x}{2x.\left(x+5\right)}\)
\(=\frac{\left(x^2+2x\right).x}{2x.\left(x+5\right)}+\frac{2.\left(x+5\right).\left(x-5\right)}{2x.\left(x+5\right)}+\frac{50-5x}{2x\left(x+5\right)}\)
\(=\frac{x^3+2x^2}{2x\left(x+5\right)}+\frac{2.\left(x^2-25\right)}{2x\left(x+5\right)}+\frac{50-5x}{2x\left(x+5\right)}=\frac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}\)
\(=\frac{x^3+4x^2-5x}{2x\left(x+5\right)}=\frac{x\left(x^2+4x-5\right)}{2x\left(x+5\right)}=\frac{x\left(x+5\right)\left(x-1\right)}{2x\left(x+5\right)}=\frac{x-1}{2}\)
b. \(A=0\Leftrightarrow\frac{x-1}{2}=0\Rightarrow x-1=0\Leftrightarrow x=1\)
\(A=\frac{1}{4}\Leftrightarrow\frac{x-1}{2}=\frac{1}{4}\Leftrightarrow4x-4=2\Leftrightarrow4x-6=0\Leftrightarrow x=\frac{3}{2}\)
c. Với x=0 thì \(A=\frac{0-1}{2}=-\frac{1}{2}\)
Với x=2 thì: \(A=\frac{2-1}{2}=\frac{1}{2}\)
d. \(A>0\Leftrightarrow\frac{x-1}{2}>0\Rightarrow\left(x-1\right).2>0\Rightarrow x-1>0\Leftrightarrow x>1\)
\(A< 0\Leftrightarrow\frac{x-1}{2}< 0\Leftrightarrow\left(x-1\right).2< 0\Leftrightarrow x-1< 0\Leftrightarrow x< 1;x\ne-5,0\)
e. \(A=\frac{x-1}{2}\inℤ\Rightarrow x-1\in Z\Rightarrow x\inℤ\)
Và \(\left(x-1\right)⋮2\Rightarrow x:2dư1\)
Vậy \(A\in Z\Leftrightarrow x\inℤ\)và x chia 2 dư 1
MT
1
16 tháng 2 2023
\(a.\left(x-1\right)=0\Leftrightarrow x=1\)
\(b.2x+5-x-9=0\Leftrightarrow x=4\)
LT
0
NH
6 tháng 9 2019
a, \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x-1=0\\2x-\frac{1}{3}=0\end{cases}\Rightarrow}\orbr{\begin{cases}5x=1\\2x=\frac{1}{3}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{6}\end{cases}}\)
b. \(\left(x^2+1\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2+1=0\\x-4=0\end{cases}\Rightarrow}\orbr{\begin{cases}x^2=-1\left(Voly\right)\\x=4\end{cases}\Rightarrow x=4}\)
c, \(2x^2-\frac{1}{3}x=0\)
\(\Leftrightarrow x\left(2x-\frac{1}{3}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\2x-\frac{1}{3}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{6}\end{cases}}\)
d, \(\left(\frac{4}{5}\right)^{5x}=\left(\frac{4}{5}\right)^7\)
\(\Rightarrow5x=7\)
\(\Rightarrow x=\frac{7}{5}\)
e, Ta có: \(A=\frac{x+5}{x-2}=\frac{\left(x-2\right)+7}{x-2}=1+\frac{7}{x-2}\)
Để A ∈ Z <=> (x - 2) ∈ Ư(7) = { ±1; ±7 }
| x - 2 | 1 | -1 | 7 | -7 |
| x | 3 | 1 | 9 | -5 |
Vậy....
6 tháng 9 2019
a) \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x-1=0\\2x-\frac{1}{3}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}5x=1\\2x=\frac{1}{3}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{6}\end{cases}}\)
Vậy : ....
b) \(\left(x^2+1\right)\left(x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x-4=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=-1\left(loại\right)\\x=4\end{cases}}\)
c) \(2x^2-\frac{1}{3}x=0\)
\(\Leftrightarrow x\left(2x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x-\frac{1}{3}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{6}\end{cases}}\)
Vậy :...
LP
7
NM
15 tháng 8 2023
a, \(\left(x-1\right).\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
b, \(\left(2x-4\right).\left(3x+9\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-4=0\\3x+9=0\end{matrix}\right.\left[{}\begin{matrix}2x=4\\3x=-9\end{matrix}\right.\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
AD
15 tháng 8 2023
a) TH1: x-1=0 => x=1
TH2: x+2=0 => x=-2
b) TH1: 2x-4=0 <=> 2x= 4 <=> x=2
TH2: 3x+9=0 <=> 3x=-9 <=> x= -3
a) \(|x+5|=0\)
\(\Leftrightarrow x=-5\)
b) \(2x+2x+1=\left(-2\right)+50\)
\(\Rightarrow2x+2x+1=48\)
\(\Rightarrow4x=47\)
\(\Leftrightarrow x=\frac{48}{4}=11,75\)
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x. lỗi
sửa thành 47 chia 4 nhé