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21 tháng 4 2018
Bài 1:
\(a,3\left(x-11\right)-2\left(x+11\right)=2011\)
\(\Leftrightarrow3x-33-2x-22=2011\)
\(\Leftrightarrow x-55=2011\)
\(\Leftrightarrow x=2066\)
Vậy pt có nghiệm x = 2066
\(b,\left(x-1\right)\left(3x-7\right)=\left(x-1\right)\left(x+30\right)\)
\(\Leftrightarrow\left(x-1\right)\left(3x-7\right)-\left(x-1\right)\left(x+30\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-7-x-30\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-37\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x-37=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{37}{2}\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{1;\dfrac{37}{2}\right\}\)
\(c,\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x^2-2x}\) (1)
ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\x\ne2\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\dfrac{x\left(x+2\right)}{x\left(x-2\right)}-\dfrac{x-2}{x\left(x-2\right)}=\dfrac{2}{x\left(x-2\right)}\)
\(\Rightarrow x^2+2x-x+2-2=0\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{0;-1\right\}\)
\(d,\left|2x-3\right|=x+1\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=x+1\\2x-3=-x-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-x=1+3\\2x+x=-1+3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\3x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{4;\dfrac{2}{3}\right\}\)
Bài 2:
\(a,2\left(x-1\right)< x+1\)
\(\Leftrightarrow2x-2< x+1\)
\(\Leftrightarrow2x-x< 1+2\)
\(\Leftrightarrow x< 3\)
Vậy bpt có nghiệm x < 3
b, Đề bài ko rõ
21 tháng 4 2018
x-\(\dfrac{x+2}{3}\)nhỏ hơn hoặc bằng 3x+\(\dfrac{x}{2}+5\)
BG
2
PT
6 tháng 2 2019
ta có : x^5+2x^4+3x^3+3x^2+2x+1=0
\(\Leftrightarrow\)x^5+x^4+x^4+x^3+2x^3+2x^2+x^2+x+x+1=0
\(\Leftrightarrow\)(x^5+x^4)+(x^4+x^3)+(2x^3+2x^2)+(x^2+x)+(x+1)=0
\(\Leftrightarrow\)x^4(x+1)+x^3(x+1)+2x^2(x+1)+x(x+1)+(x+1)=0
\(\Leftrightarrow\)(x+1)(x^4+x^3+2x^2+x+1)=0
\(\Leftrightarrow\)(x+1)(x^4+x^3+x^2+x^2+x+1)=0
\(\Leftrightarrow\)(x+1)[x^2(x^2+x+1)+(x^2+x+1)]=0
\(\Leftrightarrow\)(x+1)(x^2+x+1)(x^2+1)=0
VÌ x^2+x+1=(x+\(\dfrac{1}{2}\))^2+\(\dfrac{3}{4}\)\(\ne0\) và x^2+1\(\ne0\)
\(\Rightarrow\)x+1=0
\(\Rightarrow\)x=-1
CÒN CÂU B TỰ LÀM (02042006)
14 tháng 2 2023
b: x^4+3x^3-2x^2+x-3=0
=>x^4-x^3+4x^3-4x^2+2x^2-2x+3x-3=0
=>(x-1)(x^3+4x^2+2x+3)=0
=>x-1=0
=>x=1
HT
0
28 tháng 4 2023
a: =>17x-5x-15-2x-5=0
=>10x-20=0
=>x=2
b: =>\(\dfrac{3x-6-5x-10}{\left(x+2\right)\left(x-2\right)}=\dfrac{11x+23}{\left(x+2\right)\left(x-2\right)}\)
=>11x+23=-2x-16
=>13x=-39
=>x=-3(nhận)
c: =>5x+7>=3x-3
=>2x>=-10
=>x>=-5
d: =>5(3x-1)=-2(x+1)
=>15x-5=-2x-2
=>17x=3
=>x=3/17
e: =>4x^2-1-4x^2-3x-2=0
=>-3x-3=0
=>x=-1
g: =>7x-5-8x+2-7<0
=>-x-10<0
=>x+10>0
=>x>-10
28 tháng 4 2023
a: =>17x-5x-15-2x-5=0
=>10x-20=0
=>x=2
b: =>\(\dfrac{3x-6-5x-10}{\left(x+2\right)\left(x-2\right)}=\dfrac{11x+23}{\left(x+2\right)\left(x-2\right)}\)
=>11x+23=-2x-16
=>13x=-39
=>x=-3(nhận)
c: =>5x+7>=3x-3
=>2x>=-10
=>x>=-5
d: =>5(3x-1)=-2(x+1)
=>15x-5=-2x-2
=>17x=3
=>x=3/17
e: =>4x^2-1-4x^2-3x-2=0
=>-3x-3=0
=>x=-1
g: =>7x-5-8x+2-7<0
=>-x-10<0
=>x+10>0
=>x>-10
BG
2
3 tháng 2 2019
a) \(x^5+2x^4+3x^3+3x^2+2x+1=0\)
\(\Leftrightarrow x^5+x^4+x^4+x^3+2x^3+2x^2+x^2+x+x+1=0\)
\(\Leftrightarrow x^4\left(x+1\right)+x^3\left(x+1\right)+2x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^4+x^3+2x^2+x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^4+x^3+x^2+x^2+x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left[x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+x+1\right)\left(x^2+1\right)=0\)
Dễ thấy \(x^2+x+1>0\forall x;x^2+1>0\forall x\)
\(\Rightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Vậy....
3 tháng 2 2019
b) \(x^4+3x^3-2x^2+x-3=0\)
\(\Leftrightarrow x^4-x^3+4x^3-4x^2+2x^2-2x+3x-3=0\)
\(\Leftrightarrow x^3\left(x-1\right)+4x^2\left(x-1\right)+2x\left(x-1\right)+3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+4x^2+2x+3\right)=0\)
...
\(\Leftrightarrow x=1\)
p/s: có bác nào giải đc pt \(x^3+4x^2+2x+3=0\)thì giúp nhé :))
PH
1
30 tháng 8 2021
a, \(5\left|2x-1\right|-3=7\Leftrightarrow5\left|2x-1\right|=10\Leftrightarrow\left|2x-1\right|=2\)
TH1 : \(2x-1=2\Leftrightarrow x=\frac{3}{2}\)
TH2 : \(2x-1=-2\Leftrightarrow x=-\frac{1}{2}\)
b, \(\left(2x+3\right)\left(x-2\right)-x^2+4=0\Leftrightarrow\left(2x+3\right)\left(x-2\right)-\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x+3-x-2\right)=0\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow x=-1;x=2\)
c, \(\frac{2x-3}{2}< \frac{1-3x}{-5}\Leftrightarrow\frac{2x-3}{2}+\frac{1-3x}{5}< 0\)
\(\Leftrightarrow\frac{10x-15+2-6x}{10}< 0\Rightarrow4x-13< 0\Leftrightarrow x< \frac{13}{4}\)
11 tháng 4 2020
a/ \(3x(2x-3)=5(3-2x) \Leftrightarrow 3x(2x-3)+5(2x-3)=0 \\\ \Leftrightarrow (2x-3)(3x+5)=0 \)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\3x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=-\frac{5}{3}\end{matrix}\right.\)
KL: .............
b/ \(\left(x^2+1\right)\left(2x+5\right)=\left(x-1\right)\left(x^2+1\right)\Leftrightarrow\left(x^2+1\right)\left(2x+5\right)-\left(x-1\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(2x+5-x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x^2+1=0\\x+6=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\x=-6\end{matrix}\right.\)
KL: .............
c/ \(3x^3=x^2+3x-1\Leftrightarrow3x^3-x^2-3x+1=0\Leftrightarrow x^2\left(3x-1\right)-\left(3x-1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2-1\right)=0\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=1\\x=-1\end{matrix}\right.\)
KL: ..........
d/ \(x^2-9x+20=0\Leftrightarrow x^2-5x-4x+20=0\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\end{matrix}\right.\)
KL: .............
\(\frac{2x-1}{3}-\frac{5-3x}{2}=x-2\)
\(\Leftrightarrow\frac{2\left(2x-1\right)}{6}-\frac{3\left(5-3x\right)}{6}=\frac{6\left(x-2\right)}{6}\)
\(\Rightarrow2\left(2x-1\right)-3\left(5-3x\right)=6\left(x-2\right)\)
\(\Leftrightarrow4x-2-15+9x=6x-12\)
\(\Leftrightarrow4x-2-15+9x-6x+12=0\)
\(\Leftrightarrow7x-5=0\)
\(\Leftrightarrow7x=5\)
\(\Leftrightarrow x=\frac{5}{7}\)
CHÚC EM HỌC TỐT!!!