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\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)
\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)
\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+3\right)-x\left(x-3\right)}{x\left(x+3\right)}=0\)
\(\Leftrightarrow x^2-9-x^2+3x=0\)
\(\Leftrightarrow3x-9=0\)
\(\Leftrightarrow3x=9\)
\(\Leftrightarrow x=3\left(n\right)\)
Vậy \(S=\left\{3\right\}\)
\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)
\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)
\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)
\(\Leftrightarrow12x-9-12x+20+2x-7>0\)
\(\Leftrightarrow2x+4>0\)
\(\Leftrightarrow2x>-4\)
\(\Leftrightarrow x>-2\)
a) \(PT\Leftrightarrow3x-2x=2-3\Leftrightarrow x=-1\)
Vậy: \(S=\left\{-1\right\}\)
b) \(PT\Leftrightarrow-2x+3x=-7+22\Leftrightarrow x=15\)
Vậy: \(S=\left\{15\right\}\)
c) \(PT\Leftrightarrow8x-5x=3+12\Leftrightarrow3x=15\Leftrightarrow x=5\)
Vậy: \(S=\left\{5\right\}\)
d) \(PT\Leftrightarrow x+4x-2x=12+25-1\Leftrightarrow3x=36\Leftrightarrow x=12\)
Vậy: \(S=\left\{12\right\}\)
e) \(PT\Leftrightarrow x+2x+3x-3x=19+5\Leftrightarrow3x=24\Leftrightarrow x=8\)
Vậy: \(S=\left\{8\right\}\)
a)3x-2=2x-3
=>x=-1
b)7-2x=22-3x
=>x=15
c)8x-3=5x+12
=>3x=15
=>x=5
d)x-12+4x=25+2x-1
=>3x=12
=>x=4
e)x+2x+3x-19=3x+5
=>3x=24
=>x=8
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) 7x - 35 = 0
<=> 7x = 0 + 35
<=> 7x = 35
<=> x = 5
b) 4x - x - 18 = 0
<=> 3x - 18 = 0
<=> 3x = 0 + 18
<=> 3x = 18
<=> x = 5
c) x - 6 = 8 - x
<=> x - 6 + x = 8
<=> 2x - 6 = 8
<=> 2x = 8 + 6
<=> 2x = 14
<=> x = 7
d) 48 - 5x = 39 - 2x
<=> 48 - 5x + 2x = 39
<=> 48 - 3x = 39
<=> -3x = 39 - 48
<=> -3x = -9
<=> x = 3
a) \(3x-2=2x-3\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
b) \(3-4y+24+6y=y+27+3y\)
\(\Leftrightarrow-2y=0\Leftrightarrow y=0\)
c) \(7-2x=22-3x\)
\(\Leftrightarrow x-15=0\)
\(\Leftrightarrow x=15\)
d) \(8x-3=5x+12\)
\(\Leftrightarrow3x-15=0\Leftrightarrow x=5\)
a: =>-x+2x=3-7
=>x=-4
b: =>6x+2+2x-5=0
=>8x-3=0
hay x=3/8
c: =>5x+2x-2-4x-7=0
=>3x-9=0
hay x=3
d: =>10x2-10x2-15x=15
=>-15x=15
hay x=-1
a) 4 ( x + 5 )( x + 6 )( x + 10 )( x + 12 ) = 3x2
Do x = 0 không là nghiệm pt nên chia 2 vế pt cho \(x^2\ne0\), ta được :
\(\frac{4}{x^2}\left(x^2+60+17x\right)\left(x^2+60+16x\right)=3\)
\(\Leftrightarrow4\left(x+\frac{60}{x}+17\right)\left(x+\frac{60}{x}+16\right)=3\)
Đến đây ta đặt \(x+\frac{60}{x}+16=t\left(1\right)\)
Ta được :
\(4t\left(t+1\right)=3\Leftrightarrow4t^2+4t-3=0\Leftrightarrow\left(2t+3\right)\left(2t-1\right)=0\)
Từ đó ta lắp vào ( 1 ) tính được x