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\(\Leftrightarrow3\left(x-3\right)+5\left(2x+1\right)=90\)
=>3x-9+10x+5=90
=>13x-4=90
=>13x=94
hay x=94/13
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\(ĐKXĐ:x\ne2\)
\(\frac{x+5}{3x-6}-\frac{1}{2}=\frac{2x-3}{2x-4}\)
\(\Leftrightarrow\frac{x+5}{3\left(x-2\right)}-\frac{1}{2}-\frac{2x-3}{2\left(x-2\right)}=0\)
\(\Leftrightarrow\frac{2\left(x+5\right)-3\left(x-2\right)-3\left(2x-3\right)}{6\left(x-2\right)}=0\)
\(\Leftrightarrow2x+10-3x+6-6x+9=0\)
\(\Leftrightarrow-7x+25=0\)
\(\Leftrightarrow x=\frac{25}{7}\)(tm)
Vậy tập nghiệm của phương trình là \(S=\left\{\frac{25}{7}\right\}\)
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đk : x khác -3 ; 3
\(\Rightarrow-12+2x+6+3x-9=x^2-9\Leftrightarrow5x-15=x^2-9\)
\(\Leftrightarrow x^2-5x+6=0\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\Leftrightarrow x=2;x=3\left(ktm\right)\)
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\(x^4-3x^2=5\left(3-x^2\right)\)
=>\(x^2\left(x^2-3\right)-5\left(3-x^2\right)=0\)
=>\(x^2\left(x^2-3\right)+5\left(x^2-3\right)=0\)
=>\(\left(x^2-3\right)\left(x^2+5\right)=0\)
=>\(x^2-3=0\)
=>\(x^2=3\)
=>\(x=\pm\sqrt{3}\)
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\(\Leftrightarrow\left(x+1\right)\left(x+2\right)-5\left(x-2\right)=10+x^2-4\)
\(\Leftrightarrow x^2+3x+2-5x+10-10-x^2+4=0\)
=>-2x+6=0
hay x=3(nhận)
đk : x khác 2 ; -2
<=> x^2 + 3x + 2 - 5x + 10 = 10 + x^2 - 4
<=> x^2 - 2x + 12 = x^2 + 6
<=> -2x + 6 =0 <=> x = 3 (tm)
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\(40x-20+36x-24=60x-45\Leftrightarrow16x=-1\Leftrightarrow x=-\dfrac{1}{16}\)
\(\Leftrightarrow\left(2x-1\right)\left(x+3\right)-x\left(x+3\right)=0\)
=>(x+3)(x-1)=0
=>x=-3 hoặc x=1
\(\text{x(x+3)= (2x-1)(x+3)}\)
\(\Leftrightarrow x^2+3x=2x^2+6x-x-3\)
\(\Leftrightarrow x^2+3x-2x^2-6x+x+3=0\)
\(\Leftrightarrow-x^2-2x+3=0\)
\(a=-1\)
\(b=-2\)
\(c=3\)
\(\Delta=b^2-4ac=\left(-2\right)^2-4.\left(-1\right).3=-3< 0\)
Vậy \(S=\varnothing\)