giải bpt 2x-x(3x+1)≤15-3x(x+2)
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DT
1
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 4 2021
ĐKXĐ: \(x^2+x-1\ge0\)
\(\Rightarrow3x^2-x+1>3\sqrt{\left(x^2-x+1\right)\left(x^2+x-1\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x^2+x-1}=b\ge0\end{matrix}\right.\)
\(\Rightarrow2a^2+b^2>3ab\)
\(\Leftrightarrow\left(2a-b\right)\left(a-b\right)>0\)
\(\Rightarrow\left[{}\begin{matrix}2a< b\\a>b\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}2\sqrt{x^2-x+1}< \sqrt{x^2+x-1}\\\sqrt{x^2-x+1}>\sqrt{x^2+x-1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4\left(x^2-x+1\right)< x^2+x-1\\x^2-x+1>x^2+x-1\end{matrix}\right.\)
\(\Leftrightarrow...\) (nhớ kết hợp ĐKXĐ ban đầu)
TT
1
HP
20 tháng 2 2021
ĐK: \(x\ne\dfrac{1}{2};x\ne-\dfrac{1}{3}\)
\(\dfrac{x+2}{3x+1}\ge\dfrac{x-2}{2x-1}\)
\(\Leftrightarrow\dfrac{\left(x+2\right)\left(2x-1\right)-\left(x-2\right)\left(3x+1\right)}{\left(3x+1\right)\left(2x-1\right)}\ge0\)
\(\Leftrightarrow\dfrac{2x^2+3x-2-3x^2+5x+2}{6x^2-x-1}\ge0\)
\(\Leftrightarrow\dfrac{-x^2+8x}{6x^2-x-1}\ge0\)
\(\Leftrightarrow\left\{{}\begin{matrix}-x^2+8x\ge0\\6x^2-x-1>0\end{matrix}\right.\left(1\right)\) hoặc \(\left\{{}\begin{matrix}-x^2+8x\le0\\6x^2-x-1< 0\end{matrix}\right.\left(2\right)\)
\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}0\le x\le8\\\left[{}\begin{matrix}x>\dfrac{1}{2}\\x< -\dfrac{1}{3}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\dfrac{1}{2}< x\le8\)
\(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\le0\\x\ge8\end{matrix}\right.\\-\dfrac{1}{3}< x< \dfrac{1}{2}\end{matrix}\right.\Leftrightarrow-\dfrac{1}{3}< x\le0\)
Vậy ...
17 tháng 3 2023
a: =>\(\dfrac{x^2+2x-13-x+1}{x-1}< 0\)
=>\(\dfrac{x^2+x-12}{x-1}< 0\)
=>\(\dfrac{\left(x+4\right)\left(x-3\right)}{x-1}< 0\)
=>1<x<3 hoặc x<-4
b: =>\(\dfrac{3x^2+4x-3x-4}{x-1}< 3\)
=>3x+4<3
=>3x<-1
=>x<-1/3
c: TH1: 2x^2-3x+1>0 và x+2>0
=>(2x-1)(x-1)>0 và x+2>0
=>x>1
TH2: (2x-1)(x-1)<0 và x+2<0
=>x<-2 và 1/2<x<1
=>Loại
12 tháng 2 2020
(2x-1)(x-3)-3x+1≤(x-1)(x+3)+x2-5
<=> 2x2-6x-x+3-3x+1≤x2+3x-x-3+x2-5
<=> -12x≤-6
<=>x≥\(\frac{1}{2}\)
Vậy nghiệm của bpt là S=[\(\frac{1}{2}\);+∞)
7 tháng 7 2020
a, \(\frac{x+9}{x^2-3x-10}-\frac{x+15}{x^2-25}=\frac{1}{x+2}\left(ĐKXĐ:x\ne\pm2;\pm5\right)\)
\(\frac{x+9}{\left(x-5\right)\left(x+2\right)}-\frac{x+15}{\left(x+5\right)\left(x-5\right)}=\frac{1}{x+2}\)
\(\frac{\left(x+9\right)\left(x+5\right)}{\left(x-5\right)\left(x+2\right)\left(x+5\right)}-\frac{\left(x+15\right)\left(x+2\right)}{\left(x+5\right)\left(x-5\right)\left(x+2\right)}=\frac{\left(x+5\right)\left(x-5\right)}{\left(x+2\right)\left(x+5\right)\left(x-5\right)}\)
Khử mẫu : \(\left(x+9\right)\left(x+5\right)-\left(x+15\right)\left(x+2\right)=\left(x+5\right)\left(x-5\right)\)
\(x^2+14x+45-x^2-17x-30=x^2-25\)
\(-3x+15-x^2+25=0\)
\(-3x-x^2+40=0\)( giải delta ta đc )
\(x_1=-5;x_2=8\)
b, \(\frac{1}{3x-1}+\frac{2x+2}{x-1}-\frac{3x^2+1}{3x^2-4x+1}=1ĐKXĐ\left(x\ne1;\frac{1}{3}\right)\)
\(\frac{1}{3x-1}+\frac{2x+2}{x-1}-\frac{3x^2+1}{\left(3x-1\right)\left(x-1\right)}=1\)
\(\frac{x-1}{\left(3x-1\right)\left(x-1\right)}+\frac{\left(2x+2\right)\left(3x-1\right)}{\left(x-1\right)\left(3x-1\right)}-\frac{3x^2+1}{\left(3x-1\right)\left(x-1\right)}=\frac{\left(3x-1\right)\left(x-1\right)}{\left(3x-1\right)\left(x-1\right)}\)
Khửi mẫu \(x-1+\left(2x+2\right)\left(3x-1\right)-3x^2-1=\left(3x-1\right)\left(x-1\right)\)( bn tự nốt nhé)
c, \(\left(x+3\right)^2-10\ge\left(x+3\right)\left(x+2\right)-4\)
\(x^2+6x+9-10\ge x^2+5x+6-4\)
\(x-3\ge0\Leftrightarrow x\ge3\)
24 tháng 7 2020
a) \(\frac{x+9}{x^2-3x-10}-\frac{x+15}{x^2-25}=\frac{1}{x+2}\); ĐKXĐ: x # -2; x # +-5
<=> \(\frac{x+9}{\left(x+2\right)\left(x-5\right)}-\frac{x+15}{\left(x-5\right)\left(x+5\right)}=\frac{1}{x+2}\)
<=> \(\frac{\left(x+9\right)\left(x+5\right)-\left(x+15\right)\left(x+2\right)}{\left(x+2\right)\left(x-5\right)\left(x+5\right)}=\frac{\left(x-5\right)\left(x+5\right)}{\left(x+2\right)\left(x-5\right)\left(x+5\right)}\)
<=> (x + 9)(x + 5) - (x + 15)(x + 2) = (x - 5)(x + 5)
<=> -3x + 15 = x^2 - 25
<=> -3x + 15 - x^2 + 25 = 0
<=> -3x + 40 - x^2 = 0
<=> x^2 + 3x - 40 = 0
<=> (x - 5)(x + 8) = 0
<=> x - 5 = 0 hoặc x + 8 = 0
<=> x = 5 (ktm0 hoặc x = -8 (tm)
b) \(\frac{1}{3x-1}+\frac{2x+2}{x-1}-\frac{3x^2+1}{3x^2-4x+1}=1\); ĐKXĐ: x # 1/3; x # 1
<=> \(\frac{1}{3x-1}+\frac{2\left(x+1\right)}{x-1}-\frac{3x^2+1}{x\left(3x-1\right)-\left(3x-1\right)}=1\)
<=> \(\frac{1}{3x-1}+\frac{2\left(x+1\right)}{x-1}-\frac{3x^2+1}{\left(x-1\right)\left(3x-1\right)}=1\)
<=> \(\frac{x-1}{\left(x-1\right)\left(3x-1\right)}+\frac{2\left(x+1\right)\left(3x-1\right)}{\left(x-1\right)\left(3x-1\right)}-\frac{3x^2+1}{\left(x-1\right)\left(3x-1\right)}=\frac{\left(x-1\right)\left(3x-1\right)}{\left(x-1\right)\left(3x-1\right)}\)
<=> x - 1 + 2(x + 1)(3x - 1) - 3x^2 + 1 = (x - 1)(3x - 1)
<=> 5x - 4 + 3x^2 = 3x^2 - 4x + 1
<=> 5x - 4 = -4x + 1
<=> 5x + 4x = 1 + 4
<=> 9x = 5
<=> x = 5/9 (tm)
c) (x + 3)^2 - 10 >= (x + 3)(x + 2) - 4
<=> x^2 + 3x + 3x + 9 - 10 >= x^2 + 2x + 3x + 6 - 4
<=> x^2 + 6x + 9 - 10 >= x^2 + 5x + 6 - 4
<=> x^2 + 6x - 1 >= x^2 + 5x + 2
<=> x^2 + 6x - 1 - x^2 - 5x - 2 >= 0
<=> x - 3 >= 0
<=> x >= 3
22 tháng 4 2022
Bài 1:
c) |2x - 1| = x + 2
<=> 2x - 1 = +(x + 2) hoặc -(x + 2)
* 2x - 1 = x + 2
<=> 2x - x = 2 + 1
<=> x = 3
* 2x - 1 = -(x + 2)
<=> 2x - 1 = x - 2
<=> 2x - x = -2 + 1
<=> x = -1
Vậy.....
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
2x-x(3x+1)≤15-3x(x+2)
2x-3x2-x≤15-3x2 -6x
2x-3x2-x+3x2 +6x≤15
7x≤15
x≤15/7