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Dạng 3 :
a) 3x - 10 = 2x + 13
=> 3x - 2x = 13 - 10
=> x = 3
b) x + 12 = -5 - x
=> x + x = -5 - 12
=> 2x = -17
=> x = -8,5
c) x + 5 = 10 - x
=> x + x = 10 - 5
=> 2x = 5
=> x = 2,5
d) 6x + 23 = 2x - 12
=> 2x - 6x = 23 + 12
=> -4x = 35
=> x = -8,75
e) 12 - x = x + 1
=> x + x = 12 - 1
=> 2x = 11
=> x = 5,5
f) 14 + 4x = 3x + 20
=> 4x - 3x = 20 - 14
=> x = 6
a) -2 /3 x + 1/5 = 3/10
-2/3x =1/10
x = -3/20
vậy x = -3/20
b) 25/9 - 12/13x = 7/
12/13x = 2
x = 13/6
c) (x) - 3/4 =5/3
(x) = 29/12
x = 29/12 ; -29/-12
d) x = 11/2
:
x−12−4+3x3=7−2x+2x−58the fraction with numerator x minus 1 and denominator 2 end-fraction minus the fraction with numerator 4 plus 3 x and denominator 3 end-fraction equals 7 minus 2 x plus the fraction with numerator 2 x minus 5 and denominator 8 end-fraction𝑥−12−4+3𝑥3=7−2𝑥+2𝑥−58
\(2x+4⋮x-1\Rightarrow2\left(x-1\right)+6⋮x-1\)
\(\Rightarrow6⋮x-1\Rightarrow x-1\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
\(\Rightarrow x\in\left\{2;0;3;-1;4;-2;7;-5\right\}\)
Vậy...........................................
\(2x^2+\left(-3\right)^2=41\)
\(\Rightarrow2x^2=41-9=32\)
\(\Rightarrow x^2=16\)
\(\Rightarrow x=\pm4\)
\(2\left(x-5\right)-3\left(x+7\right)=14\)
\(\Rightarrow2x-10-3x-21=14\)
\(\Rightarrow2x-3x=14+21+10\)
\(\Rightarrow-x=45\Rightarrow x=-45\)
\(-7\left(5-x\right)-2\left(x-10\right)=15\)
\(\Rightarrow-35+x-2x+20=15\)
\(\Rightarrow x-2x=15-20+35\)
\(\Rightarrow-x=30\Rightarrow x=-30\)
\(\left(x+\frac{1}{2}\right).\left(\frac{2}{3}-2x\right)=0\)
\(=>\orbr{\begin{cases}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{cases}=>\orbr{\begin{cases}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{cases}=>\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{1}{3}\end{cases}}}}\)
Vậy x thuộc {-1/2 ; 1/3}
\(x.3\frac{1}{4}+\frac{-7}{6}.x-1\frac{2}{3}=\frac{5}{12}.2\)
\(x.\frac{13}{4}+\frac{-7}{6}.x-\frac{5}{3}=\frac{5}{6}\)
\(x.\left(\frac{13}{4}+\frac{-7}{6}\right)=\frac{5}{6}+\frac{5}{3}\)
\(x.\left(\frac{39}{12}+\frac{-14}{12}\right)=\frac{5}{6}+\frac{10}{6}\)
\(x.\frac{25}{12}=\frac{5}{2}\)
\(x=\frac{5}{2}:\frac{25}{12}\)
\(x=\frac{5}{2}.\frac{12}{25}\)
\(x=\frac{6}{5}\)
a) Ta có:
\(\frac{3}{x+2}=\frac{5}{2x+1}\)
\(\Rightarrow3\left(2x+1\right)=5\left(x+2\right)\)
\(\Rightarrow6x+3=5x+10\)
\(\Rightarrow6x-5x=10-3\)
\(\Rightarrow x=7\)
b)Ta có:
\(\frac{5}{8x-2}=\frac{-4}{7-x}\)
\(\Rightarrow5\left(7-x\right)=-4\left(8x-2\right)\)
\(\Rightarrow35-5x=-32x+8\)
\(\Rightarrow-5x+32x=8-35\)
\(\Rightarrow27x=-27\)
\(\Rightarrow x=-1\)
c) Ta có:
\(\frac{4}{3}=\frac{2x-1}{x}\)
\(\Rightarrow4x=3\left(2x-1\right)\)
\(\Rightarrow4x=6x-3\)
\(\Rightarrow3=6x-4x=2x\)
\(\Rightarrow x=\frac{3}{2}\)
d)Ta có:
\(\frac{2x-1}{3}=\frac{3x+1}{4}\)
\(\Rightarrow4\left(2x-1\right)=3\left(3x+1\right)\)
\(\Rightarrow8x-4=9x+3\)
\(\Rightarrow8x-9x=3+4\)
\(\Rightarrow-x=7\Rightarrow x=-7\)
e)Ta có:
\(\frac{4}{x+2}=\frac{7}{3x+1}\)
\(\Rightarrow4\left(3x+1\right)=7\left(x+2\right)\)
\(\Rightarrow12x+4=7x+14\)
\(\Rightarrow12x-7x=14-4\)
\(\Rightarrow5x=10\)
\(\Rightarrow x=2\)
f)Ta có:
\(\frac{-3}{x+1}=\frac{4}{2-2x}\)
\(\Rightarrow-3\left(2-2x\right)=4\left(x+1\right)\)
\(\Rightarrow-6+6x=4x+4\)
\(\Rightarrow6x-4x=4+6\)
\(\Rightarrow2x=10\)
\(\Rightarrow x=5\)
\(1/\left(x+5\right)\left(x^2-4\right)=0\\ TH1:x+5=0\\ =>x=-5\\ TH2:x^2-4=0\\ =>x^2=4\\ =>x^2=\left(\pm2\right)^2\\ =>x=\pm2\\ 2/3x-10=2x+13\\ =>3x-2x=13+10\\ =>x=23\\ 3/3\left(4-x\right)-2\left(x-1\right)=x+2\\ =>12-3x-2x+2=x+2\\ =>14-5x=x+2\\ =>x+5x=14-2\\ =>6x=12\\ =>x=\dfrac{12}{6}=2\\ 4/2\left(x-1\right)+3\left(x-2\right)=x-4\\ =>2x-2+3x-6=x-4\\ =>5x-8=x-4\\ =>5x-x=-4+8\\ =>4x=4\\ =>x=\dfrac{4}{4}=1\\ 5/4\left(2x+7\right)-3\left(3x-2\right)=24\\ =>8x+28-9x+6=24\\ =>34-x=24\\ =>x=34-24=10\\ 6/6x+23=2x-12\\ =>6x-2x=-12-23\\ =>4x=-35\\ =>x=\dfrac{-35}{4}\)