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a: 3x-5>15-x
=>4x>20
hay x>5
b: \(3\left(x-2\right)\left(x+2\right)< 3x^2+x\)
=>3x2+x>3x2-12
=>x>-12
a: 3x-5>15-x
=>3x+x>15+5
=>4x>20
=>x>5
b: \(3\left(x-2\right)\left(x+2\right)<3x^2+x\)
=>\(3\left(x^2-4\right)<3x^2+x\)
=>\(3x^2-12-3x^2-x<0\)
=>-x-12<0
=>x+12>0
=>x>-12
c: \(\left(2x+1\right)^2+3x\left(1-x\right)\le\left(x+2\right)^2\)
=>\(4x^2+4x+1+3x-3x^2\le x^2+4x+4\)
=>\(x^2+7x+1\le x^2+4x+4\)
=>7x+1<=4x+4
=>7x-4x<=4-1
=>3x<=3
=>x<=1
d: \(\frac{5x-20}{3}-\frac{2x^2+x}{2}>\frac{x\left(1-3x\right)}{3}-\frac{5x}{4}\)
=>\(\frac{4\left(5x-20\right)-6\left(2x^2+x\right)}{12}>\frac{4x\left(1-3x\right)-15x}{12}\)
=>\(4\left(5x-20\right)-6\left(2x^2+x\right)>4x\left(1-3x\right)-15x\)
=>\(20x-80-12x^2-6x>4x-12x^2-15x\)
=>14x-80>-11x
=>25x>80
=>\(x>\frac{80}{25}=\frac{16}{5}\)
e: 4-2x<=3x-6
=>-2x-3x<=-6-4
=>-5x<=-10
=>x>=2
f: \(\left(x+4\right)\left(5x-1\right)>5x^2+16x+2\)
=>\(5x^2-x+20x-4>5x^2+16x+2\)
=>19x-4>16x+2
=>3x>6
=>x>2
g: \(x\left(2x-1\right)-8<5-2x\left(1-x\right)\)
=>\(2x^2-x-8<5-2x+2x^2\)
=>-x-8<-2x+5
=>-x+2x<5+8
=>x<13
h: \(\frac{3x-1}{4}-\frac{3\left(x-2\right)}{8}-1>\frac{5-3x}{2}\)
=>\(\frac{2\left(3x-1\right)}{8}-\frac{3\left(x-2\right)}{8}-\frac88>\frac{4\left(5-3x\right)}{8}\)
=>2(3x-1)-3(x-2)-8>4(5-3x)
=>6x-2-3x+6-8>20-12x
=>3x-4>20-12x
=>15x>24
=>\(x>\frac{24}{15}\)
=>x>1,6
1: =>2(x+2)>3x+1
=>2x+4-3x-1>0
=>-x+3>0
=>-x>-3
=>x<3
2: =>12x^2-2x>12x^2+9x-8x-6
=>-2x>-x-6
=>-x>-6
=>x<6
3: =>4(x+1)-12>=3(x-2)
=>4x+4-12>=3x-6
=>4x-8>=3x-6
=>x>=2
4: =>-5x<=15
=>x>=-3
5: =>3(x+2)-5(x-2)<30
=>3x+6-5x+10<30
=>-2x+16<30
=>-2x<14
=>x>-7
6: =>5(x+2)<3(3-2x)
=>5x+10<9-6x
=>11x<-1
=>x<-1/11
a:=>3x=15
=>x=5
b: =>8-11x<52
=>-11x<44
=>x>-4
c: \(VT=\left(\dfrac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\right)\cdot\dfrac{x\left(x+6\right)}{2x-6}+\dfrac{x}{6-x}\)
\(=\dfrac{12x-36}{2x-6}\cdot\dfrac{1}{x-6}-\dfrac{x}{x-6}=\dfrac{6}{x-6}-\dfrac{x}{x-6}=-1\)
1: \(\Leftrightarrow x^2+6x+9-6x+3>x^2-4x\)
=>-4x<12
hay x>-3
2: \(\Leftrightarrow6+2x+2>2x-1-12\)
=>8>-13(đúng)
4: \(\dfrac{2x+1}{x-3}\le2\)
\(\Leftrightarrow\dfrac{2x+1-2x+6}{x-3}< =0\)
=>x-3<0
hay x<3
6: =>(x+4)(x-1)<=0
=>-4<=x<=1
Muốn ăn tát k ?
=V
Đưa mặt đêy
1.\(\dfrac{x+2}{x-3}+\dfrac{x}{x+2}=\dfrac{x^2+6}{x^2-x-6}\)
\(\Leftrightarrow\dfrac{x+2}{x-3}+\dfrac{x}{x+2}=\dfrac{x^2+6}{\left(x+2\right)\left(x-3\right)}\)
\(ĐK:x\ne3;-2\)
\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x+2\right)+x\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}=\dfrac{x^2+6}{\left(x+2\right)\left(x-3\right)}\)
\(\Leftrightarrow\left(x+2\right)\left(x+2\right)+x\left(x-3\right)=x^2+6\)
\(\Leftrightarrow x^2+4x+4+x^2-3x-x^2-6=0\)
\(\Leftrightarrow x^2+x-2=0\)
\(\Leftrightarrow\left(x^2-x\right)+\left(2x-2\right)=0\)
\(\Leftrightarrow x\left(x-1\right)+2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-2\left(ktm\right)\end{matrix}\right.\)
Vậy \(S=\left\{1\right\}\)
b.\(\left(x+1\right)^2+\left|x-1\right|=x^2+4\)
\(\Leftrightarrow\) \(\left(x+1\right)^2+x-1=x^2+4\) hoặc \(\left(x+1\right)^2+1-x=x^2+4\)
Xét \(\left(x+1\right)^2+x-1=x^2+4\)
\(\Leftrightarrow x^2+2x+1+x-1-x^2-4=0\)
\(\Leftrightarrow3x-4=0\)
\(\Leftrightarrow x=\dfrac{4}{3}\)
Xét \(\left(x+1\right)^2+1-x=x^2+4\)
\(\Leftrightarrow x^2+2x+1+1-x-x^2-4=0\)
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
Vậy \(S=\left\{\dfrac{4}{3};2\right\}\)
2.\(1-\dfrac{x-1}{3}< \dfrac{x+3}{3}-\dfrac{x-2}{2}\)
\(\Leftrightarrow\dfrac{6-2\left(x-1\right)}{6}< \dfrac{2\left(x+3\right)-3\left(x-2\right)}{6}\)
\(\Leftrightarrow6-2\left(x-1\right)< 2\left(x+3\right)-3\left(x-2\right)\)
\(\Leftrightarrow6-2x+2< 2x+6-3x+6\)
\(\Leftrightarrow-x< 4\)
\(\Leftrightarrow x>4\)
Vậy \(S=\left\{x|x>4\right\}\)
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