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c) TH1 : \(\left|2x+1\right|+\left|3x-4\right|=\left(2x+1\right)+\left(3x-4\right)=5x-3=5\)
\(\Rightarrow x=\frac{8}{5}\)
TH2 : \(\left|2x+1\right|+\left|3x-4\right|=\left(2x+1\right)+\left(-3x+4\right)=-x+5=5\)
\(\Rightarrow x=0\)
d) \(\Rightarrow\left|2x+\frac{4}{5}\right|-\left|x-\frac{3}{2}\right|=0\)
Tương tự xét 2 trường hợp như câu c. Ta sẽ tìm được x
c ) \(\left|2x+1\right|+\left|3x-4\right|=5\)
\(\Rightarrow\left[\begin{array}{nghiempt}\left|2x+1\right|=5\\\left|3x-4\right|=5\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x+1=5\\2x+1=-5\\3x-4=5\\3x-4=-5\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}2\\-3\\3\\-\frac{1}{3}\end{array}\right.\)
2: =>2x-1/4=5/6-1/2x
=>5/2x=5/6+1/4=13/12
=>x=13/30
3: =>3x-5/6=2/3-1/2x
=>3,5x=2/3+5/6=4/6+5/6=9/6=3,2
hay x=32/35
\(a,\frac{-24}{x}+\frac{18}{x}=\frac{-24+18}{x}=\frac{-6}{x}\)
\(\Leftrightarrow x\inƯ(-6)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
\(b,\frac{2x-5}{x+1}=\frac{2x+2-7}{x+1}=\frac{2(x+1)-7}{x+1}=2-\frac{7}{x+1}\)
\(\Leftrightarrow7⋮x+1\Leftrightarrow x+1\inƯ(7)=\left\{\pm1;\pm7\right\}\)
Xét các trường hợp rồi tìm được x thôi :>
\(c,\frac{3x+2}{x-1}-\frac{x-5}{x-1}=\frac{3x+2-x-5}{x-1}=\frac{2x+7}{x-1}=\frac{2x-2+9}{x-1}=\frac{2(x-1)+9}{x-1}=2+\frac{9}{x-1}\)
\(\Leftrightarrow9⋮x-1\Leftrightarrow x-1\inƯ(9)=\left\{\pm1;\pm3;\pm9\right\}\)
\(\Leftrightarrow x\in\left\{2;0;4;-2;10;-8\right\}\)
d, TT
a)\(\frac{x+3}{x+5}=7\Leftrightarrow x+3=7\left(x+5\right)\)
\(\Leftrightarrow x+3=7x+35\)
\(\Leftrightarrow-6x=32\)
\(\Leftrightarrow x=-\frac{16}{3}\)
b)\(\frac{2x-1}{3x+5}=-\frac{2}{3}\)
\(\Leftrightarrow3\left(2x-1\right)=-2\left(3x+5\right)\)
\(\Leftrightarrow6x-3=-6x-10\)
\(\Leftrightarrow12x=-7\)
\(\Leftrightarrow x=-\frac{7}{12}\)
c)\(\frac{x+1}{4}=\frac{9}{x+1}\Leftrightarrow\left(x+1\right)^2=36\)
\(\Leftrightarrow\left(x+1\right)^2=6^2\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=6\\x+1=-6\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-7\end{cases}}}\)
d)\(\frac{6x-1}{2x+3}=\frac{3x}{x+2}\)
\(\Leftrightarrow\left(6x-1\right)\left(x+2\right)=3x\left(2x+3\right)\)
\(\Leftrightarrow6x^2+12x-x-2=6x^2+9x\)
\(\Leftrightarrow2x=2\Leftrightarrow x=1\)
a)\(\left(\frac{4}{5}\right)^{2x+7}=\left(\frac{4}{5}\right)^4\)
=> 2x + 7 = 4
2x = 4 - 7
2x = -3
x = -3 : 2
x = -1,5
Vậy x = -1,5
c) pt<=> 2x+1+3x-4 =5 <=> x = 8/5 hoặc 2x+1+3x-4 = -5 <=> x = -2/5 hoặc -2x-1 +3x-4 = 5 <=> x = 10 hoặc 2x+1 -3x +4 = 5 <=> x = 0
vậy x = { -2/5 ; 0 ;8/5 ;10 }
d) pt <=> 2x + 4/5 = x-3/2 <=> x = -23/10 hoặc 2x +4/5 = -x+3/2 <=> x = 7/30
vậy x = { -23/10 ; 7/30}