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a: =>-4x>16
=>x<-4
c: =>20x-25<=21-3x
=>23x<=46
=>x<=2
d: =>20(2x-5)-30(3x-1)<12(3-x)-15(2x-1)
=>40x-100-90x+30<36-12x-30x+15
=>-50x-70<-42x+51
=>-8x<121
=>x>-121/8
Câu 2:
ĐKXĐ: \(\left[{}\begin{matrix}1-9x^2\ne0\\1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Rightarrow \left[{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)
\(\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\left(1\right)\)
\(\left(1\right):\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}-\dfrac{\left(1-3x\right)\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\dfrac{\left(1+3x\right)\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}=0\)
\(\Leftrightarrow 12-\left(1-3x-3x+9x^2\right)+\left(1+3x+3x+9x^2\right)=0\)
\(\Leftrightarrow 12-1+3x+3x-9x^2+1+3x+3x+9x^2=0\)
\(\Leftrightarrow12x+12=0\\ \Leftrightarrow12x=-12\\ \Leftrightarrow x=-1\left(TM\right)\)
Vậy \(S=\left\{-1\right\}\)
a) 4x -8 ≥ 3(3x-1)-2x +1
⇒4x -8 ≥7x -2
⇒4x -7x ≥ -2 +8
⇒-3x ≥ 6
⇒x≤-2
Vậy bpt có nghiệm là:{x|x≤-2}
b) (x-3)(x+2)+(x+4)2≤ 2x (x+5)+4
⇔ x2+2x - 3x - 6 +x2 + 8x +16≤ 2x2 + 10x +4
⇔ x2 +2x - 3x + x2 + 8x - 2x2- 10x ≤ 4+6-16
⇔ -3x ≤ -6
⇔ x≥ 2
Vậy bpt có tập nghiệm là: {x|x≥2}
Mk thấy mấy cái này dễ mà, toàn trong sách giáo khoa hết á. Bạn cố gắng đọc và lm đi. Sắp lên lớp 9 rồi đó 
a)\(\dfrac{2x^2+10}{1-x}\le0\Rightarrow1-x< 0\Leftrightarrow x>1\)
b) \(\dfrac{3x-4}{x+2}\ge4\Leftrightarrow\dfrac{3x-4}{x+2}-\dfrac{4\left(x+2\right)}{x+2}\ge0\Leftrightarrow\dfrac{-x-12}{x+2}\ge0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}-x-12\le0\\x+2< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-12\\x< -2\end{matrix}\right.\Leftrightarrow-12\le x< -2}}\\\left\{{}\begin{matrix}-x-12\ge0\\x+2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le-12\\x>-2\end{matrix}\right.\end{matrix}\right.\)\(S=\left\{x|-12\le x< -2\right\}\)
c) \(\dfrac{1}{x+4}\le\dfrac{1}{x-2}\Leftrightarrow\dfrac{6}{\left(x+4\right)\left(x-2\right)}\le0\Rightarrow\left(x+4\right)\left(x-2\right)< 0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+4>0\\x-2< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>-4\\x< 2\end{matrix}\right.\Leftrightarrow-4< x< 2}}\\\left\{{}\begin{matrix}x+4< 0\\x-2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< -4\\x>2\end{matrix}\right.\end{matrix}\right.\)
\(S=\left\{x|-4< x< 2\right\}\)
a,\(x-\frac{5x+2}{6}=\frac{7-3x}{4}\)
=> \(\frac{12x}{12}-\frac{\left(5x+2\right)2}{12}=\frac{\left(7-3x\right)3}{12}\)
=>\(\frac{12x-10x-4}{12}=\frac{21-9x}{12}\)
=>(khử mẫu)
=>\(12x-10x-4=21-9x\)
=>11x=25
=>x=25/11
b: \(\Leftrightarrow3\left(10x+3\right)=36+4\left(8x+6\right)\)
=>30x+9=36+32x+24
=>32x+60=30x+9
=>2x=-51
=>x=-51/2
c: \(\Leftrightarrow2x-3\left(2x+1\right)=x+6x\)
=>7x=2x-6x-3
=>7x=-4x-3
=>11x=-3
=>x=-3/11
d: \(\Leftrightarrow4\left(x+2\right)-6x=3\left(1-2x+1\right)\)
=>4x+8-6x=3(-2x+2)
=>-2x+8+6x-6=0
=>4x+2=0
=>x=-1/2
a)\(x-\dfrac{5x+2}{6}=\dfrac{7-3x}{4}\)
\(\Leftrightarrow\dfrac{12x-10x-4}{12}=\dfrac{21-9x}{12}\)
\(\Leftrightarrow2x-4=21-9x\)
\(\Leftrightarrow2x-4-21+9x=0\)
\(\Leftrightarrow11x-25=0\)
\(\Leftrightarrow x=\dfrac{25}{11}\)
b)\(\dfrac{10x+3}{12}=1+\dfrac{6+8x}{9}\)
\(\Leftrightarrow\dfrac{30x+9}{36}=\dfrac{36+24+32x}{36}\)
\(\Leftrightarrow30x+9=60+32x\)
\(\Leftrightarrow30x+9-60-32x=0\)
\(\Leftrightarrow-2x-51=0\)
\(\Leftrightarrow x=-\dfrac{51}{2}\)
c)\(\dfrac{x}{3}-\dfrac{2x+1}{2}=\dfrac{x}{6}-6\)
\(\Leftrightarrow\dfrac{2x-6x-3}{6}=\dfrac{x-36}{6}\)
\(\Leftrightarrow-4x-3=x-36\)
\(\Leftrightarrow-4x-3-x+36=0\)
\(\Leftrightarrow-5x+33=0\)
\(\Leftrightarrow x=\dfrac{33}{5}\)
d)\(\dfrac{2+x}{3}-\dfrac{1}{2}x=\dfrac{1-2x}{4}+\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{8+4x-6x}{12}=\dfrac{3-6x+3}{12}\)
\(\Leftrightarrow8-2x=6-6x\)
\(\Leftrightarrow8-2x-6+6x=0\)
\(\Leftrightarrow4x+2=0\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
Tính lại xem đúng không nha 
a) \(x-\dfrac{5x+2}{6}=\dfrac{7-3x}{4}\)
\(\Leftrightarrow\dfrac{24x}{24}-\dfrac{4\left(5x+2\right)}{24}=\dfrac{6\left(7-3x\right)}{24}\)
\(\Leftrightarrow24x-4\left(5x+2\right)=6\left(7-3x\right)\)
\(\Leftrightarrow24x-20x-8=42-18x\)
\(\Leftrightarrow4x-8=42-18x\)
\(\Leftrightarrow4x+18x=42+8\)
\(\Leftrightarrow22x=50\)
\(\Leftrightarrow x=\dfrac{25}{11}\)
Vậy S\(=\left\{\dfrac{25}{11}\right\}\)
Mấy này bạn quy đồng lên cùng mẫu xong khử mẫu rồi giải. Dễ mà.
a) \(\dfrac{2-x}{3}-x-2\le\dfrac{x-17}{2}\) \(\Leftrightarrow\) \(6\left(\dfrac{2-x}{3}-x-2\right)\le6\left(\dfrac{x-17}{2}\right)\) \(\Leftrightarrow\) 4-2x-6x-12\(\le\)3x-51 \(\Leftrightarrow\) -2x-6x-3x\(\le\)-51-4+12 \(\Leftrightarrow\) -11x\(\le\)-43 \(\Rightarrow\) x\(\ge\)43/11.
b) \(\dfrac{2x+1}{3}-\dfrac{x-4}{4}\le\dfrac{3x+1}{6}-\dfrac{x-4}{12}\) \(\Leftrightarrow\) \(12\left(\dfrac{2x+1}{3}+\dfrac{4-x}{4}\right)\le12\left(\dfrac{3x+1}{6}+\dfrac{4-x}{12}\right)\) \(\Leftrightarrow\) 8x+4+12-3x\(\le\)6x+2+4-x \(\Leftrightarrow\) 8x-3x-6x+x\(\le\)2+4-4-12 \(\Leftrightarrow\) 0x\(\le\)-10 (vô lí).
a) \(\dfrac{2-x}{3}-x-2\le\dfrac{x-17}{2}\)
\(\Leftrightarrow2\left(2-x\right)-6\left(x+2\right)\le3\left(x-17\right)\)
\(\Leftrightarrow4-2x-6x-12\le3x-51\)
\(\Leftrightarrow-11x\le-43\)
\(\Leftrightarrow x\ge\dfrac{43}{11}\)
Vậy S = {\(x\) | \(x\ge\dfrac{43}{11}\) }
b) \(\dfrac{2x+1}{3}-\dfrac{x-4}{4}\le\dfrac{3x+1}{6}-\dfrac{x-4}{12}\)
\(\Leftrightarrow4\left(2x+1\right)-3\left(x-4\right)\le2\left(3x+1\right)-\left(x-4\right)\)
\(\Leftrightarrow8x+4-3x+12\le6x+2-x+4\)
\(\Leftrightarrow0x\le-10\) (vô lý)
Vậy \(S=\varnothing\)