Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{x-2}{18}-\frac{2x+5}{12}>\frac{x+6}{9}-\frac{x-3}{6}\)
\(\Leftrightarrow\frac{2\left(x-2\right)}{36}-\frac{3\left(2x+5\right)}{36}>\frac{4\left(x+6\right)}{36}-\frac{6\left(x-3\right)}{36}\)
\(\Leftrightarrow2x-4-6x-15>4x+24-6x+18\)
\(\Leftrightarrow2x-6x-4x+6x>24+18+4+15\)
\(\Leftrightarrow-2x>61\)
\(\Leftrightarrow x< -\frac{61}{2}\)
Vậy nghiệm của bất phương trình là \(x< -\frac{61}{2}\)
Bài b và c làm cách mình thì dễ hiểu hơn nhiều :3
\(\left(2x-2\right)\left(2x+3\right)\le0\)
TH1 : \(\hept{\begin{cases}2x-3\le0\\2x+3\ge0\end{cases}< =>\hept{\begin{cases}2x\le3\\2x\ge-3\end{cases}}}\)
\(< =>\hept{\begin{cases}x\le\frac{3}{2}\\x\ge-\frac{3}{2}\end{cases}}\)
TH2 : \(\hept{\begin{cases}2x-3\ge0\\2x+3\le0\end{cases}< =>\hept{\begin{cases}2x\ge3\\2x\le-3\end{cases}}}\)
\(< =>\hept{\begin{cases}x\ge\frac{3}{2}\\x\le-\frac{3}{2}\end{cases}}\)
Vậy ...
a, Ta có\(\left(x+3\right)^2+3\left(x-1\right)\ge x^2-4\)
\(\Leftrightarrow x^2+6x+9+3x-3\ge x^2-4\)
\(\Leftrightarrow x^2+9x+6\ge x^2-4\)
\(\Leftrightarrow9x+10\ge0\Leftrightarrow x\ge-\frac{10}{9}\)
\(\left(x+3\right)^2+3\left(x-1\right)\ge x^2-4\)
\(\Leftrightarrow x^2+6x+9+3x-3\ge x^2-4\)
\(\Leftrightarrow x^2+6x+3x-x^2\ge-4-9+3\)
\(\Leftrightarrow9x\ge-10\)
\(\Leftrightarrow x\ge-\frac{10}{9}\)
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, \(\left(x+3\right)^2-10\ge\left(x+3\right)\left(x+2\right)-4\)
\(x^2+6x+9-10\ge x^2+5x+6-4\)
\(x-3\ge0\Leftrightarrow x\ge3\)
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
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
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
đề = x-1>=0 \(\rightarrow\)x>=1
2x-3>=0\(\rightarrow\)x>=1,5
so sánh điều kiện S=(1;1,5)
ta thay đấu() = đấu ngoặc nhọn
a ) \(\left|x+5\right|=3x+1\) ( 1 )
+ ) \(x+5=x+5.\) Khi \(x\ge-5\)
\(\left(1\right)\Leftrightarrow x+5=3x+1\)
\(\Leftrightarrow-2x=-4\Leftrightarrow x=2\) ( TM )
+ ) \(x+5=-x-5.\) Khi \(x< -5\)
\(\left(1\right)\Leftrightarrow-x-5=3x+1\)
\(\Leftrightarrow-4x=6\)
\(\Leftrightarrow x=-\dfrac{3}{2}\)( KTM )
Vậy ..........
b ) \(\dfrac{3\left(x-1\right)}{4}+1\ge\dfrac{x+2}{3}\)
\(\Leftrightarrow9x-9+12\ge4x+8\)
\(\Leftrightarrow5x\ge5\)
\(\Leftrightarrow x\ge1\)
Vậy ...........
c ) \(\dfrac{x-2}{x+1}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\left(1\right)\)
ĐKXĐ : \(x\ne2;x\ne-2.\)
\(\left(1\right)\Leftrightarrow\dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow\left(x-2\right)^2-3\left(x+2\right)=2x-22\)
\(\Leftrightarrow x^2-4x+4-3x-6-2x+22=0\)
\(\Leftrightarrow x^2-9x+20=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\end{matrix}\right.\left(TMĐKXĐ\right)\)
Vậy .........
\(\Leftrightarrow\)
\(\frac{3x-1}{x-1}-\frac{2x+5}{x+3}+\frac{1}{x^2+2x-3}=1.\)
\(ĐK:\hept{\begin{cases}x-1\ne0\\x+3\ne\\x^2+2x-3\ne0\end{cases}0}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne\Leftrightarrow-3\end{cases}}\)
\(\Leftrightarrow\left(3x-1\right)\left(x+3\right)-\left(2x+5\right)\left(x-1\right)+4-x^2-2x+3=0\)
\(\Leftrightarrow3x^2+9x-x-3-2x^2+2x-5x+5+4-x^2-2x+3=0\)
\(\Leftrightarrow3x+9=0\)
\(\Leftrightarrow3x=-9\Leftrightarrow x=-3\) (loại)
Vậy pt vô No
câu a là nhân 2 hay mũ 2
* la the nao