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a: \(\Leftrightarrow7\left(7-3x\right)+12\left(5x+2\right)=84\left(x+13\right)\)
\(\Leftrightarrow49-21x+60x+24=84x+1092\)
\(\Leftrightarrow39x-84x=1092-73\)
=>-45x=1019
hay x=-1019/45
b: \(\Leftrightarrow21\left(x+3\right)-14=4\left(5x+9\right)-7\left(7x-9\right)\)
=>21x+63-14=20x+36-49x+63
=>21x+49=-29x+99
=>50x=50
hay x=1
c: \(\Leftrightarrow7\left(2x+1\right)-3\left(5x+2\right)=21x+63\)
=>14x+7-15x-6-21x-63=0
=>-22x-64=0
hay x=-32/11
d: \(\Leftrightarrow35\left(2x-3\right)-15\left(2x+3\right)=21\left(4x+3\right)-17\cdot105\)
=>70x-105-30x-45=84x+63-1785
=>40x-150-84x+1722=0
=>-44x+1572=0
hay x=393/11
a: =>3,6-1,7x=2,3-1,4-4=0,9-4=-3,1
=>1,7x=6,7
hay x=67/17
b: \(\Leftrightarrow30\left(5x+4\right)-15\left(3x+5\right)=24\left(4x+9\right)-40\left(x-9\right)\)
=>150x+120-45x-75=96x+216-40x+360
=>105x+45=56x+576
=>49x=531
hay x=531/49
\(1,\left(dk:x\ne0,-1,4\right)\)
\(\Leftrightarrow\dfrac{9}{x+1}+\dfrac{2}{x-4}-\dfrac{11}{x}=0\)
\(\Leftrightarrow\dfrac{9x\left(x-4\right)+2x\left(x+1\right)-11\left(x+1\right)\left(x-4\right)}{x\left(x+1\right)\left(x-4\right)}=0\)
\(\Leftrightarrow9x^2-36x+2x^2+2x-11x^2+44x-11x+44=0\)
\(\Leftrightarrow-x=-44\)
\(\Leftrightarrow x=44\left(tm\right)\)
\(2,\left(đk:x\ne4\right)\)
\(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{2+x}{x-4}-\dfrac{3}{2\left(x-4\right)}+\dfrac{5}{6}=0\)
\(\Leftrightarrow\dfrac{14.2-6\left(2+x\right)-3.3+5\left(x-4\right)}{6\left(x-4\right)}=0\)
\(\Leftrightarrow28-12-6x-9+5x-20=0\)
\(\Leftrightarrow-x=13\)
\(\Leftrightarrow x=-13\left(tm\right)\)
a: =>2x^2+8x-3x-12<2x^2+2
=>5x<14
=>x<14/5
b: =>\(\dfrac{9x-3-\left(5x+1\right)\left(x-2\right)}{3\left(x-2\right)}-4>0\)
=>\(\dfrac{9x-3-5x^2+10x-x+2-12\left(x-2\right)}{3\left(x-2\right)}>0\)
=>\(\dfrac{-5x^2+18x-1-12x+24}{3\left(x-2\right)}>0\)
=>\(\dfrac{-5x^2+6x+23}{x-2}>0\)
TH1: x-2>0 và -5x^2+6x+23>0
=>x>2 và \(\dfrac{3-2\sqrt{31}}{5}< x< \dfrac{3+2\sqrt{31}}{5}\)
=>\(2< x< \dfrac{3+2\sqrt{31}}{5}\)
TH2: x-2<0 và -5x^2+6x+23<0
=>x<2 và \(\left[{}\begin{matrix}x< \dfrac{3-2\sqrt{31}}{5}\\x>\dfrac{3+2\sqrt{31}}{5}\end{matrix}\right.\)
=>\(x< \dfrac{3-2\sqrt{31}}{5}\)
1: Ta có: \(\dfrac{3}{x-3}+\dfrac{4}{x+3}=\dfrac{3x-7}{x^2-9}\)
\(\Leftrightarrow\dfrac{3x+9}{\left(x-3\right)\left(x+3\right)}+\dfrac{4x-12}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x-7}{\left(x-3\right)\left(x+3\right)}\)
Suy ra: \(3x+9+4x-12=3x-7\)
\(\Leftrightarrow4x=-7+12-9=-4\)
hay \(x=-1\left(nhận\right)\)
2: Ta có: \(\dfrac{3}{x-4}-\dfrac{4}{x+4}=\dfrac{3x-4}{x^2-16}\)
\(\Leftrightarrow\dfrac{3x+12}{\left(x-4\right)\left(x+4\right)}-\dfrac{4x-16}{\left(x+4\right)\left(x-4\right)}=\dfrac{3x-4}{\left(x-4\right)\left(x+4\right)}\)
Suy ra: \(3x+12-4x+16=3x-4\)
\(\Leftrightarrow28-4x=-4\)
\(\Leftrightarrow4x=32\)
hay \(x=8\left(tm\right)\)
3: Ta có: \(\dfrac{5x^2-12}{x^2-1}+\dfrac{3}{x-1}=\dfrac{5x}{x+1}\)
Suy ra: \(5x^2-12+3x+3=5x^2-5x\)
\(\Leftrightarrow3x-9+5x=0\)
\(\Leftrightarrow8x=9\)
hay \(x=\dfrac{9}{8}\left(nhận\right)\)
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
1: Ta có: \(\dfrac{-3}{x-4}-\dfrac{3-5x}{x^2-16}=\dfrac{1}{x+4}\)
Suy ra: \(-3\left(x+4\right)-3+5x=x-4\)
\(\Leftrightarrow-3x-12-3+5x-x+4=0\)
\(\Leftrightarrow x=11\left(nhận\right)\)
2. ĐKXĐ: $x\neq \pm 2$
PT \(\Leftrightarrow \frac{3(x-2)}{(2+x)(x-2)}-\frac{x-1}{(x-2)(x+2)}=\frac{2(x+2)}{(x-2)(x+2)}\)
\(\Leftrightarrow \frac{3(x-2)-(x-1)}{(x-2)(x+2)}=\frac{2(x+2)}{(x-2)(x+2)}\)
\(\Rightarrow 3(x-2)-(x-1)=2(x+2)\)
\(\Leftrightarrow 2x-5=2x+4\Leftrightarrow 9=0\) (vô lý)
Vậy pt vô nghiệm
a:=>3x=15
=>x=5
b: =>8-11x<52
=>-11x<44
=>x>-4
c: \(VT=\left(\dfrac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\right)\cdot\dfrac{x\left(x+6\right)}{2x-6}+\dfrac{x}{6-x}\)
\(=\dfrac{12x-36}{2x-6}\cdot\dfrac{1}{x-6}-\dfrac{x}{x-6}=\dfrac{6}{x-6}-\dfrac{x}{x-6}=-1\)
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 \(S=\varnothing\)
a: =>x^2-8x+16<x^2-8x
=>16<0(loại)
b: =>\(x+\dfrac{1}{2}>=\dfrac{5x-3}{3}\)
=>x+1/2>=5/3x-1
=>-2/3x>=-3/2
=>x<=3/2:2/3=9/4
c: =>\(\dfrac{7-x}{4}< =\dfrac{2x-4}{3}\)
=>21-3x<=8x-16
=>-11x<=-37
=>x>=37/11