Tìm số nguyên x
a) |3x+1|>4
b)|4x-1|+2x
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MD
1
13 tháng 2 2020
\(a,3x-32>-5x+1\)
\(\Leftrightarrow3x+5x>32+1\)
\(\Leftrightarrow8x>33\)
\(\Leftrightarrow x>\frac{33}{8}=4,125\approx4\)
Vậy \(x>4,x\inℤ\)
\(b,15+4x< 2x-145\)
\(\Leftrightarrow4x-2x< -145-15\)
\(\Leftrightarrow2x< -160\)
\(\Leftrightarrow x< \frac{-160}{2}=-80\)
Vậy \(x< -80,x\inℤ\)
TH
1 tháng 9 2021
\(|-2x+1,5|=\dfrac{1}{4}\Rightarrow-2x+1,5=\pm\dfrac{1}{4}\)
\(-2x+1,5=\dfrac{1}{4}\Rightarrow-2x=1,5-0,25\Rightarrow-2x=1,25\Rightarrow x=1,25:\left(-2\right)\Rightarrow x=...\)
\(-2x+1,5=-\dfrac{1}{4}\Rightarrow-2x=-0,25-1,5\Rightarrow-2x=1,75\Rightarrow x=1,75:\left(-2\right)\Rightarrow x=...\)
TH
1 tháng 9 2021
\(\dfrac{3}{2}-|1.\dfrac{1}{4}+3x|=\dfrac{1}{4}\Rightarrow|1.\dfrac{1}{4}+3x|=\dfrac{3}{2}-\dfrac{1}{4}\Rightarrow|1.\dfrac{1}{4}+3x|=\dfrac{5}{4}\)
\(\Rightarrow1.\dfrac{1}{4}+3x=\pm\dfrac{5}{4}\)
\(1.\dfrac{1}{4}+3x=\dfrac{5}{4}\Rightarrow\dfrac{1}{4}+3x=\dfrac{5}{4}\Rightarrow3x=\dfrac{5}{4}-\dfrac{1}{4}\Rightarrow3x=1\Rightarrow x=3\)
\(1.\dfrac{1}{4}+3x=-\dfrac{5}{4}\Rightarrow\dfrac{1}{4}+3x=-\dfrac{5}{4}\Rightarrow3x=-\dfrac{5}{4}-\dfrac{1}{4}\Rightarrow3x=-\dfrac{3}{2}x=...\)
K
30 tháng 6 2021
a) 3x(4x-3)-2x(5-6x)=0
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow24x^2-19x=0\)
\(\Leftrightarrow x\left(24x-19\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)
Vậy x=0 hoặc x=\(\dfrac{19}{24}\)
30 tháng 7 2021
a) \(\left(2x-1\right)^2-25=0\)
⇔ \(\left(2x-1\right)^2-5^2=0\)
⇔ \(\left(2x-1-5\right)\left(2x-1+5\right)=0\)
⇒ \(2x-1-5=0\) hoặc \(2x-1+5=0\)
⇔ \(x=3\) hoặc \(x=-2\)
TC
30 tháng 7 2021
Bài 1: Tìm x
a) (2x-1) ² - 25 = 0
<=> (2x-1)2 = 25
<=> 2x-1 = 5 hay 2x-1 =-5
<=> 2x= 6 hay 2x=-4
<=> x=3 hay x= -2
Vậy S={3; -2}
b) 3x (x-1) + x - 1 = 0
<=> (x-1)(3x+1)=0
<=> x-1=0 hay 3x+1=0
<=> x=1 hay 3x=-1
<=> x=1 hay x=\(\dfrac{-1}{3}\)
Vậy S={1;\(\dfrac{-1}{3}\)}
c) 2(x+3) - x ² - 3x = 0
<=> 2(x+3)- x(x+3)=0
<=> (x+3)(2-x)=0
<=> x+3=0 hay 2-x=0
<=> x=-3 hay x=2
Vậy S={-3;2}
d) x(x - 2) + 3x - 6 = 0
<=> x(x-2)+3(x-2)=0
<=> (x-2)(x+3)=0
<=> x-2=0 hay x+3=0
<=> x=2 hay x=-3
Vậy S={2;-3}
e) 4x ² - 4x +1 = 0
<=> (2x-1)2=0
<=> 2x-1=0
<=> 2x=1
<=> x=\(\dfrac{1}{2}\)
Vậy S={\(\dfrac{1}{2}\)}
f) x +5x2 = 0
<=> x(1+5x)=0
<=>x=0 hay 1+5x=0
<=> x=0 hay 5x=-1
<=> x=0 hay x= \(\dfrac{-1}{5}\)
Vậy S={0;\(\dfrac{-1}{5}\)}
g) x ²+ 2x -3 = 0
<=> x2-x+3x-3=0
<=> x(x-1)+3(x-1)=0
<=> (x-1)(x+3)=0
<=> x-1=0 hay x+3=0
<=> x=1 hay x=-3
Vậy S={1;-3}
14 tháng 12 2021
\(a,\Leftrightarrow x^2+6x+9-x^2+3x+10=1\\ \Leftrightarrow9x=-18\Leftrightarrow x=-2\\ b,\Leftrightarrow4x^2-4x+1-4x^2+17x+15=3\\ \Leftrightarrow13x=-13\Leftrightarrow x=-1\\ c,\Leftrightarrow3x\left(x-2\right)+4\left(x-2\right)=0\\ \Leftrightarrow\left(3x+4\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=2\end{matrix}\right.\\ d,\Leftrightarrow2x\left(3x+5\right)-6\left(3x+5\right)=0\\ \Leftrightarrow\left(x-3\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{3}\end{matrix}\right.\)
H9
HT.Phong (9A5)
CTVHS
9 tháng 8 2023
a) \(3^{x+2}\cdot5^{y-3}=45^x\)
\(\Rightarrow3^{x+2}\cdot5^{y-3}=\left(3^2\right)^x\cdot5^x\)
\(\Rightarrow3^{x+2}\cdot5^{y-3}=3^{2x}\cdot5^x\)
\(\Rightarrow\left\{{}\begin{matrix}3^{x+2}=3^{2x}\\5^{y-3}=5^x\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+2=2x\\y-3=x\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=2\\y-3=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=2\\y=5\end{matrix}\right.\)
30 tháng 10 2023
a: ĐKXD: x<>0
\(\dfrac{14x^3+12x^2-14x}{2x}=\left(x+2\right)\left(3x-4\right)\)
=>\(\dfrac{2x\left(7x^2+6x-7\right)}{2x}=\left(x+2\right)\left(3x-4\right)\)
=>\(7x^2+6x-7=3x^2-4x+6x-8\)
=>\(7x^2+6x-7=3x^2+2x-8\)
=>\(4x^2+4x+1=0\)
=>\(\left(2x+1\right)^2=0\)
=>2x+1=0
=>x=-1/2(nhận)
b: \(\left(4x-5\right)\left(6x+1\right)-\left(8x+3\right)\left(3x-4\right)=15\)
=>\(24x^2+4x-30x-5-\left(24x^2-32x+9x-12\right)=15\)
=>\(24x^2-26x-5-24x^2+23x+12=15\)
=>-3x+7=15
=>-3x=8
=>\(x=-\dfrac{8}{3}\)
21 tháng 8 2023
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) \(P=\dfrac{4x-1}{2x+3}\inℤ\left(x\inℤ;x\ne-\dfrac{3}{2}\right)\)
\(\Rightarrow4x-1⋮2x+3\)
\(\Rightarrow4x-1-2\left(2x+3\right)⋮2x+3\)
\(\Rightarrow4x-1-4x-6⋮2x+3\)
\(\Rightarrow-7⋮2x+3\)
\(\Rightarrow2x+3\in\left\{-1;1;-7;7\right\}\)
\(\Rightarrow x\in\left\{-2;-1;-5;2\right\}\)
21 tháng 8 2023
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:
| x+3 | 2 | -2 | 1 | -1 |
| x | -1 | -5 | -2 | -4 |
Vậy x \(\in-1,-2,-5,-4\)