Tìm số nguyên x biết:
a,(x-4)(x+3)>hoặc= 0
b,(3x-6).3=3^4
c,5^x+2-5^x-1=3100
d,3^x+1-3^x-2=702
e,(2-x)(x+1)<hoặc= 0
f,(x-1)(x^2+4)< 0
g,3|2x-5|-7=20
K
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Những câu hỏi liên quan
5 tháng 10 2021
a) \(\Rightarrow3x\left(x-5\right)-2\left(x-5\right)=0\)
\(\Rightarrow\left(x-5\right)\left(3x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{2}{3}\end{matrix}\right.\)
b) \(\Rightarrow x^3+6x^2+12x+8-x^3+6x^2=4\)
\(\Rightarrow12x^2+12x+4=0\)
\(\Rightarrow x\in\varnothing\)(do \(12x^2+12x+4=12\left(x^2+x+\dfrac{1}{4}\right)+1=12\left(x+\dfrac{1}{2}\right)^2+1\ge1>0\))
HP
21 tháng 9 2021
a. 9x2 - 6x - 3 = 0
<=> 3(3x2 - 2x - 1) = 0
<=> 3(3x2 - 3x + x - 1) = 0
<=> \(3\left[3x\left(x-1\right)+\left(x-1\right)\right]=0\)
<=> 3(3x + 1)(x - 1) = 0
<=> \(\left[{}\begin{matrix}3x+1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=1\end{matrix}\right.\)
b. (2x + 1)2 - 4(x + 2)2 = 9
<=> (2x + 1)2 - \(\left[2\left(x+2\right)\right]^2=9\)
<=> (2x + 1 - 2x - 4)(2x + 1 + 2x + 4) = 9
<=> -3(4x + 5) = 9
<=> 4x + 5 = -3
<=> 5 + 3 = -4x
<=> -4x = 8
<=> -x = 2
<=> x = -2
21 tháng 9 2021
a) \(\Leftrightarrow\left(9x^2-6x+1\right)-4=0\)
\(\Leftrightarrow\left(3x-1\right)^2-4=0\)
\(\Leftrightarrow3\left(x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
b) \(\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\)
\(\Leftrightarrow12x=-24\Leftrightarrow x=-2\)
c) \(\Leftrightarrow3x^2-6x+3-3x^2+15x=21\)
\(\Leftrightarrow9x=18\Leftrightarrow x=2\)
d) \(\Leftrightarrow x^2+6x+9-x^2-4x+32=1\)
\(\Leftrightarrow2x=-40\Leftrightarrow x=-20\)
Bài 1.
24 tháng 9 2021
b: Ta có: \(\left(x-2\right)^3-x^2\left(x-6\right)=4\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+6x^2=4\)
\(\Leftrightarrow12x=12\)
hay x=2
d: Ta có: \(3\left(x-1\right)^2-3x\left(x-5\right)=1\)
\(\Leftrightarrow3x^2-6x+3-3x^2+15x=1\)
\(\Leftrightarrow9x=-2\)
hay \(x=-\dfrac{2}{9}\)
LH
1
20 tháng 10 2021
a: \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
b: \(\left(x+1\right)^2-4\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
21 tháng 1 2022
\(a,\left(2x-5\right)+17=6\\ \Rightarrow2x-5=-11\\ \Rightarrow2x=-6\\ \Rightarrow x=-3\\ b,10-2\left(4-3x\right)=-4\\ \Rightarrow2\left(4-3x\right)=14\\ \Rightarrow4-3x=7\\ \Rightarrow3x=-3\\ \Rightarrow x=-1\\ c,24:\left(3x-2\right)=-3\\ \Rightarrow3x-2=-8\\ \Rightarrow3x=-6\\ \Rightarrow x=-2\\ d,5-2x=-17+12\\ \Rightarrow5-2x=-5\\ \Rightarrow2x=10\\ \Rightarrow x=5\)
21 tháng 1 2022
a: =>2x-5=-11
=>2x=-6
hay x=-3
b: =>2(4-3x)=14
=>4-3x=7
=>3x=-3
hay x=-1
c: =>3x-2=-8
=>3x=-6
hay x=-2
13 tháng 9 2021
a) \(\left(x+3\right)\left(2x-1\right)-\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow2x^2+5x-3-x^2+2x+3=0\)
\(\Leftrightarrow x^2+7x=0\Leftrightarrow x\left(x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-7\end{matrix}\right.\)
b) \(\left(x+4\right)\left(2x-3\right)-3\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow2x^2+5x-12-3x^2+12=0\)
\(\Leftrightarrow x^2-5x=0\Leftrightarrow x\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
KG
1
26 tháng 7 2023
a: =>(x-5)(x+5)+(x-5)(3x-15)=0
=>(x-5)(x+5+3x-15)=0
=>(x-5)(4x-10)=0
=>x=5 hoặc x=5/2
c: =>x^3-3x^2+2x^2-6x-8x+24=0
=>(x-3)(x^2+2x-8)=0
=>(x-3)(x+4)(x-2)=0
=>\(x\in\left\{3;-4;2\right\}\)
23 tháng 10 2021
e: ta có: \(4x^2+4x-6=2\)
\(\Leftrightarrow4x^2+4x-8=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
f: Ta có: \(2x^2+7x+3=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
