a , 2x -7 \(⋮\)2x +1
b, 3x +1 \(⋮\)x +2
c, 5x +2 \(⋮\)3x +1
d, x .(x+2)\(⋮\)x+1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
4 tháng 2 2023
a: 3x^3+2x^2-7x+a chia hêt cho 3x-1
=>3x^3-x^2+3x^2-x-6x+2+a-2 chia hết cho 3x-1
=>a-2=0
=>a=2
c: =>2x^2-6x+(a+6)x-3a-18+3a+19 chia x-3 dư 4
=>3a+19=4
=>3a=-15
=>a=-5
d: 2x^3-x^2+ax+b chiahêt cho x^2-1
=>2x^3-2x-x^2+1+(a+2)x+b-1 chia hết cho x^2-1
=>a+2=0 và b-1=0
=>a=-2 và b=1
DX
12 tháng 4 2023
a.
\(\left|5x\right|=3x+8\Leftrightarrow\left[{}\begin{matrix}-5x=3x+8\\5x=3x+8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=4\end{matrix}\right.\)
b.
\(\left|-4x\right|=-2x+11\Leftrightarrow\left[{}\begin{matrix}-4x=-2x+11\\4x=-2x+11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=\dfrac{11}{6}\end{matrix}\right.\)
c.
\(\left|3x-1\right|=4x+1\Leftrightarrow\left[{}\begin{matrix}-3x+1=4x+1\\3x-1=4x+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
d.
\(\left|3-2x\right|=3x-7\Leftrightarrow\left[{}\begin{matrix}-3+2x=3x-7\\3-2x=3x-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
e.
\(9-\left|-5x\right|+2x=0\Leftrightarrow\left[{}\begin{matrix}9-5x+2x=0\\9+5x+2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{9}{7}\end{matrix}\right.\)
f.
\(\left(x+1\right)^2+\left|x+10\right|-x^2-12=0\Leftrightarrow\left[{}\begin{matrix}x^2+2x+1-x-10-x^2-12=0\\x^2+2x+1+x+10-x^2-12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=21\\x=\dfrac{1}{3}\end{matrix}\right.\)
10 tháng 1 2023
a: \(=\dfrac{x-2x-1}{x+1}=\dfrac{-\left(x+1\right)}{x+1}=-1\)
b: \(=\dfrac{2+2x}{x\left(x+1\right)}=\dfrac{2\left(x+1\right)}{x\left(x+1\right)}=\dfrac{2}{x}\)
c: \(=\dfrac{3x-1}{2\left(3x+1\right)}+\dfrac{3x+1}{2\left(3x-1\right)}-\dfrac{6x}{\left(3x-1\right)\left(3x+1\right)}\)
\(=\dfrac{9x^2-6x+1+9x^2+6x+1-12x}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{18x^2-12x+2}{2\left(3x-1\right)\left(3x+1\right)}\)
\(=\dfrac{2\left(3x-1\right)^2}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{3x-1}{3x+1}\)
10 tháng 2 2022
b: \(\Leftrightarrow\dfrac{-3x^2+36x+12}{3\left(x+4\right)\left(x-1\right)}=\dfrac{36\left(x-1\right)}{3\left(x+4\right)\left(x-1\right)}+\dfrac{12\left(x+4\right)}{3\left(x-1\right)\left(x+4\right)}\)
\(\Leftrightarrow-3x^2+36x+12=36x-36+12x+48\)
\(\Leftrightarrow-3x^2+36x+12-48x-12=0\)
\(\Leftrightarrow3x\left(x+4\right)=0\)
=>x=0(nhận) hoặc x=-4(loại)
22 tháng 10 2023
2:
a: \(x^2-12x+20\)
\(=x^2-2x-10x+20\)
=x(x-2)-10(x-2)
=(x-2)(x-10)
b: \(2x^2-x-15\)
=2x^2-6x+5x-15
=2x(x-3)+5(x-3)
=(x-3)(2x+5)
c: \(x^3-x^2+x-1\)
=x^2(x-1)+(x-1)
=(x-1)(x^2+1)
d: \(2x^3-5x-6\)
\(=2x^3-4x^2+4x^2-8x+3x-6\)
\(=2x^2\left(x-2\right)+4x\left(x-2\right)+3\left(x-2\right)\)
\(=\left(x-2\right)\left(2x^2+4x+3\right)\)
e: \(4y^4+1\)
\(=4y^4+4y^2+1-4y^2\)
\(=\left(2y^2+1\right)^2-\left(2y\right)^2\)
\(=\left(2y^2+1-2y\right)\left(2y^2+1+2y\right)\)
f; \(x^7+x^5+x^3\)
\(=x^3\left(x^4+x^2+1\right)\)
\(=x^3\left(x^4+2x^2+1-x^2\right)\)
\(=x^3\left[\left(x^2+1\right)^2-x^2\right]\)
\(=x^3\left(x^2-x+1\right)\left(x^2+x+1\right)\)
g: \(\left(x^2+x\right)^2-5\left(x^2+x\right)+6\)
\(=\left(x^2+x\right)^2-2\left(x^2+x\right)-3\left(x^2+x\right)+6\)
\(=\left(x^2+x\right)\left(x^2+x-2\right)-3\left(x^2+x-2\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x-3\right)\)
\(=\left(x^2+x-3\right)\left(x+2\right)\left(x-1\right)\)
h: \(\left(x^2+2x\right)^2-2\left(x+1\right)^2-1\)
\(=\left(x^2+2x+1-1\right)^2-2\left(x+1\right)^2-1\)
\(=\left[\left(x+1\right)^2-1\right]^2-2\left(x+1\right)^2-1\)
\(=\left(x+1\right)^4-2\left(x+1\right)^2+1-2\left(x+1\right)^2-1\)
\(=\left(x+1\right)^4-4\left(x+1\right)^2\)
\(=\left(x+1\right)^2\left[\left(x+1\right)^2-4\right]\)
\(=\left(x+1\right)^2\left(x+1+2\right)\left(x+1-2\right)\)
\(=\left(x+1\right)^2\cdot\left(x+3\right)\left(x-1\right)\)
i: \(x^2+4xy+4y^2-4\left(x+2y\right)+3\)
\(=\left(x+2y\right)^2-4\left(x+2y\right)+3\)
\(=\left(x+2y\right)^2-\left(x+2y\right)-3\left(x+2y\right)+3\)
\(=\left(x+2y\right)\left(x+2y-1\right)-3\left(x+2y-1\right)\)
\(=\left(x+2y-1\right)\left(x+2y-3\right)\)
j: \(x\cdot\left(x+1\right)\left(x+2\right)\left(x+3\right)-3\)
\(=\left(x^2-3x\right)\left(x^2-3x+2\right)-3\)
\(=\left(x^2-3x\right)^2+2\left(x^2-3x\right)-3\)
\(=\left(x^2-3x+3\right)\left(x^2-3x-1\right)\)
26 tháng 3 2023
a: =>5x>1
=>x>1/5
b: =>3x-3<2
=>3x<5
=>x<5/3
c: =>2x-3x^2-x<15-3x^2-6x
=>x<15-6x
=>7x<15
=>x<15/7
KV
18 tháng 4 2022
a) \(5x-3=7\)
\(\Leftrightarrow5x=7+3\)
\(\Leftrightarrow5x=10\)
\(\Leftrightarrow x=\dfrac{10}{5}\)
\(\Leftrightarrow x=2\)
Vậy \(S=\left\{2\right\}\)
b) \(\left(x+3\right)\left(x-4\right)=0\)
\(\Leftrightarrow x+3=0\) hoặc \(x-4=0\)
*) \(x+3=0\)
\(x=0-3\)
\(x=-3\)
*) \(x-4=0\)
\(x=0+4\)
\(x=4\)
Vậy \(S=\left\{-3;4\right\}\)
c) \(\left|x^2+2014\right|=1\)
\(\Leftrightarrow x^2+2014=1\) hoặc \(x^2+2014=-1\)
*) \(x^2+2014=1\)
\(\Leftrightarrow x^2=1-2014\)
\(\Leftrightarrow x^2=-2013\) (vô lý)
*) \(x^2+2014=-1\)
\(\Leftrightarrow x^2=-1-2014\)
\(\Leftrightarrow x^2=-2015\) (vô lý)
Vậy \(S=\varnothing\)
d) \(\dfrac{2}{x+1}-\dfrac{1}{x-3}=\dfrac{3x-11}{x^2-2x-3}\) (1)
ĐKXĐ: \(x\ne-1;x\ne3\)
\(\left(1\right)\Leftrightarrow2\left(x-3\right)-\left(x+1\right)=3x-11\)
\(\Leftrightarrow2x-6-x-1=3x-11\)
\(\Leftrightarrow-2x=-11+7\)
\(\Leftrightarrow-2x=-4\)
\(\Leftrightarrow x=2\) (nhận)
Vậy \(S=\left\{2\right\}\)
22 tháng 6 2023
b: =>2x^2-8x+3x-12+x^2-7x+10=3x^2-17x+20
=>-12x-2=-17x+20
=>5x=22
=>x=22/5
c: =>24x^2+16x-9x-6-4x^2-16x-7x-28=20x^2-4x+5x-1
=>-16x-34=x-1
=>-17x=33
=>x=-33/17
d: =>2x^2+3x^2-3=5x^2+5x
=>5x=-3
=>x=-3/5
e: =>8x+16-5x^2-10x+4x^2-4x-8=4-x^2
=>-6x+8=4
=>-6x=-4
=>x=2/3
f: =>4(x^2+4x-5)-x^2-7x-10=3x^2+3x-6
=>4x^2+16x-20-4x^2-10x+4=0
=>6x=16
=>x=8/3
12 tháng 8 2021
\(a,\dfrac{5}{-x^2+5x-6}+\dfrac{x+3}{2-x}=0\left(x\ne2;x\ne3\right)\\ \Leftrightarrow\dfrac{5}{\left(x-3\right)\left(x-2\right)}-\dfrac{x+3}{x-2}=0\\\Leftrightarrow\dfrac{5-\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x-2\right)}=0 \\ \Leftrightarrow5-x^2+9=0\\ \Leftrightarrow14-x^2=0\\ \Leftrightarrow x^2=14\\ \Leftrightarrow\left[{}\begin{matrix}x=\sqrt{14}\\x=-\sqrt{14}\end{matrix}\right.\)
\(b,\dfrac{x}{2x+2}-\dfrac{2x}{x^2-2x-3}=\dfrac{x}{6-2x}\left(x\ne-1;x\ne3\right)\\ \Leftrightarrow\dfrac{x}{2\left(x+1\right)}-\dfrac{2x}{\left(x-3\right)\left(x+1\right)}=\dfrac{x}{2\left(3-x\right)}\\ \Leftrightarrow\dfrac{x\left(x-3\right)-2x\cdot2}{2\left(x-3\right)\left(x+1\right)}=\dfrac{-x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}\\ \Leftrightarrow x^2-3x-4x=-x^2-x\\ \Leftrightarrow2x^2-6x=0\\ \Leftrightarrow2x\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
\(c,\dfrac{1}{x-1}-\dfrac{3x^2}{x^3-1}=\dfrac{2x}{x^2+x+1}\left(x\ne1\right)\\ \Leftrightarrow\dfrac{x^2+x+1-3x^2}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\\ \Leftrightarrow-2x^2+x+1=2x^2-2x\\ \Leftrightarrow4x^2-3x-1=0\\ \Leftrightarrow\left(x-1\right)\left(4x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{4}\end{matrix}\right.\)
\(d,\dfrac{x+25}{2x^2-50}-\dfrac{x+5}{x^2-5x}=\dfrac{5-x}{2x^2+10x}\left(x\ne5;x\ne-5\right)\\ \Leftrightarrow\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}-\dfrac{x+5}{x\left(x-5\right)}=\dfrac{5-x}{2x\left(x+5\right)}\\ \Leftrightarrow\dfrac{x^2+25x-2\left(x+5\right)^2}{2x\left(x-5\right)\left(x+5\right)}=\dfrac{\left(5-x\right)\left(x-5\right)}{2x\left(x+5\right)\left(x-5\right)}\\ \Leftrightarrow x^2+25x-2\left(x^2+10x+25\right)=-\left(x^2-10x+25\right)\\ \Leftrightarrow x^2+25x-2x^2-20x-50=-x^2+10x-25\\ \Leftrightarrow-5x=25\\ \Leftrightarrow x=-5\)
Tick nha
tìm x à ?